OP Malhotra Class 10 Maths Solutions Chapter 8 Matrices Exercise 8 (C)

S Chand Class 10 ICSE Maths Solutions Chapter 8 Matrices Exercise 8(C)

Question 1. If \( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, B = \begin{bmatrix} 6 & 1 \\ 1 & 1 \end{bmatrix}, C = \begin{bmatrix} -2 & -3 \\ 0 & 1 \end{bmatrix} \), find each of the following:
(a) AB, BA, AC, CA, BC, CB
(b) \( A^2, B^2, C^2 \)
Answer:
(a) We need to find the products AB, BA, AC, CA, BC, CB.
First, let's find AB:
\( AB = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 6 & 1 \\ 1 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 1 \times 6 + 2 \times 1 & 1 \times 1 + 2 \times 1 \\ 3 \times 6 + 4 \times 1 & 3 \times 1 + 4 \times 1 \end{bmatrix} \)
\( = \begin{bmatrix} 6 + 2 & 1 + 2 \\ 18 + 4 & 3 + 4 \end{bmatrix} \)
\( = \begin{bmatrix} 8 & 3 \\ 22 & 7 \end{bmatrix} \)

Next, let's find BA:
\( BA = \begin{bmatrix} 6 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \)
\( = \begin{bmatrix} 6 \times 1 + 1 \times 3 & 6 \times 2 + 1 \times 4 \\ 1 \times 1 + 1 \times 3 & 1 \times 2 + 1 \times 4 \end{bmatrix} \)
\( = \begin{bmatrix} 6 + 3 & 12 + 4 \\ 1 + 3 & 2 + 4 \end{bmatrix} \)
\( = \begin{bmatrix} 9 & 16 \\ 4 & 6 \end{bmatrix} \)

Now, let's find AC:
\( AC = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} -2 & -3 \\ 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 1 \times (-2) + 2 \times 0 & 1 \times (-3) + 2 \times 1 \\ 3 \times (-2) + 4 \times 0 & 3 \times (-3) + 4 \times 1 \end{bmatrix} \)
\( = \begin{bmatrix} -2 + 0 & -3 + 2 \\ -6 + 0 & -9 + 4 \end{bmatrix} \)
\( = \begin{bmatrix} -2 & -1 \\ -6 & -5 \end{bmatrix} \)

Then, let's find CA:
\( CA = \begin{bmatrix} -2 & -3 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \)
\( = \begin{bmatrix} -2 \times 1 + (-3) \times 3 & -2 \times 2 + (-3) \times 4 \\ 0 \times 1 + 1 \times 3 & 0 \times 2 + 1 \times 4 \end{bmatrix} \)
\( = \begin{bmatrix} -2 - 9 & -4 - 12 \\ 0 + 3 & 0 + 4 \end{bmatrix} \)
\( = \begin{bmatrix} -11 & -16 \\ 3 & 4 \end{bmatrix} \)

Next, let's find BC:
\( BC = \begin{bmatrix} 6 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} -2 & -3 \\ 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 6 \times (-2) + 1 \times 0 & 6 \times (-3) + 1 \times 1 \\ 1 \times (-2) + 1 \times 0 & 1 \times (-3) + 1 \times 1 \end{bmatrix} \)
\( = \begin{bmatrix} -12 + 0 & -18 + 1 \\ -2 + 0 & -3 + 1 \end{bmatrix} \)
\( = \begin{bmatrix} -12 & -17 \\ -2 & -2 \end{bmatrix} \)

Finally, let's find CB:
\( CB = \begin{bmatrix} -2 & -3 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 6 & 1 \\ 1 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} -2 \times 6 + (-3) \times 1 & -2 \times 1 + (-3) \times 1 \\ 0 \times 6 + 1 \times 1 & 0 \times 1 + 1 \times 1 \end{bmatrix} \)
\( = \begin{bmatrix} -12 - 3 & -2 - 3 \\ 0 + 1 & 0 + 1 \end{bmatrix} \)
\( = \begin{bmatrix} -15 & -5 \\ 1 & 1 \end{bmatrix} \)

(b) We need to find \( A^2, B^2, C^2 \).
First, let's find \( A^2 = A \times A \):
\( A^2 = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \)
\( = \begin{bmatrix} 1 \times 1 + 2 \times 3 & 1 \times 2 + 2 \times 4 \\ 3 \times 1 + 4 \times 3 & 3 \times 2 + 4 \times 4 \end{bmatrix} \)
\( = \begin{bmatrix} 1 + 6 & 2 + 8 \\ 3 + 12 & 6 + 16 \end{bmatrix} \)
\( = \begin{bmatrix} 7 & 10 \\ 15 & 22 \end{bmatrix} \)

Next, let's find \( B^2 = B \times B \):
\( B^2 = \begin{bmatrix} 6 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 6 & 1 \\ 1 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 6 \times 6 + 1 \times 1 & 6 \times 1 + 1 \times 1 \\ 1 \times 6 + 1 \times 1 & 1 \times 1 + 1 \times 1 \end{bmatrix} \)
\( = \begin{bmatrix} 36 + 1 & 6 + 1 \\ 6 + 1 & 1 + 1 \end{bmatrix} \)
\( = \begin{bmatrix} 37 & 7 \\ 7 & 2 \end{bmatrix} \)

Finally, let's find \( C^2 = C \times C \):
\( C^2 = \begin{bmatrix} -2 & -3 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} -2 & -3 \\ 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} -2 \times (-2) + (-3) \times 0 & -2 \times (-3) + (-3) \times 1 \\ 0 \times (-2) + 1 \times 0 & 0 \times (-3) + 1 \times 1 \end{bmatrix} \)
\( = \begin{bmatrix} 4 + 0 & 6 - 3 \\ 0 + 0 & 0 + 1 \end{bmatrix} \)
\( = \begin{bmatrix} 4 & 3 \\ 0 & 1 \end{bmatrix} \)
In simple words: For matrix multiplication, you multiply rows of the first matrix by columns of the second. Remember the order matters, so AB is usually not the same as BA. Also, finding the square of a matrix means multiplying it by itself.

🎯 Exam Tip: Always double-check your calculations, especially the signs, during matrix multiplication. A single error can lead to a completely different result. Make sure the number of columns in the first matrix matches the number of rows in the second matrix for multiplication to be possible.

Question 2. Answer true or false :
(i) \( \frac { 1 }{ 2 } \) is the identity matrix for addition of 2 x 2 matrices.
(ii) If A, B and C are any 2 x 2 matrices, then \( A (B – C) = A. B – A. C \)
Answer:
(i) False: The identity matrix for addition of 2 x 2 matrices is \( \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \), also known as the zero matrix. Adding this matrix to any other matrix does not change the original matrix.
(ii) True. Matrix multiplication is distributive over subtraction, meaning you can multiply A by each term inside the parenthesis separately.
In simple words: For addition, the "do nothing" matrix is all zeros, not a fraction. For multiplication, you can share out the matrix A to B and C just like in normal algebra.

🎯 Exam Tip: Remember the properties of matrix operations. For addition, the additive identity is the zero matrix. For multiplication, the multiplicative identity is the identity matrix (ones on the main diagonal, zeros elsewhere). The distributive property holds for matrices.

Question 3. If \( \begin{bmatrix} a & 3 \\ 1 & 2 \end{bmatrix}\begin{bmatrix} 2 \\ -1 \end{bmatrix}=\begin{bmatrix} 5 \\ 0 \end{bmatrix} \), find the value of a.
Answer:
Given the matrix equation:
\( \begin{bmatrix} a & 3 \\ 1 & 2 \end{bmatrix}\begin{bmatrix} 2 \\ -1 \end{bmatrix}=\begin{bmatrix} 5 \\ 0 \end{bmatrix} \)
First, perform the matrix multiplication on the left side:
\( \begin{bmatrix} a \times 2 + 3 \times (-1) \\ 1 \times 2 + 2 \times (-1) \end{bmatrix} = \begin{bmatrix} 5 \\ 0 \end{bmatrix} \)
\( \begin{bmatrix} 2a - 3 \\ 2 - 2 \end{bmatrix} = \begin{bmatrix} 5 \\ 0 \end{bmatrix} \)
\( \begin{bmatrix} 2a - 3 \\ 0 \end{bmatrix} = \begin{bmatrix} 5 \\ 0 \end{bmatrix} \)
Now, compare the corresponding elements of the two matrices. The elements in the first row must be equal, and the elements in the second row must be equal.
For the first row:
\( 2a - 3 = 5 \)
Now, we solve this simple algebraic equation for 'a'.
\( 2a = 5 + 3 \)
\( 2a = 8 \)
\( a = \frac{8}{2} \)
\( a = 4 \)
The value of 'a' is 4.
In simple words: Multiply the matrices on the left side. Then, set the top number of the new matrix equal to the top number on the right side. Solve this simple equation to find 'a'.

🎯 Exam Tip: When comparing matrices, each element in the corresponding positions must be equal. This helps in forming equations to find unknown variables. Always show your multiplication steps clearly.

Question 4. If \( A = \begin{bmatrix} 1 & 4 \\ 2 & 1 \end{bmatrix}, B = \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix}, C = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \), simplify \( A^2 + BC \).
Answer:
Given matrices:
\( A = \begin{bmatrix} 1 & 4 \\ 2 & 1 \end{bmatrix}, B = \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix}, C = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \)
We need to simplify \( A^2 + BC \).

First, let's find \( A^2 = A \times A \):
\( A^2 = \begin{bmatrix} 1 & 4 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 4 \\ 2 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 1 \times 1 + 4 \times 2 & 1 \times 4 + 4 \times 1 \\ 2 \times 1 + 1 \times 2 & 2 \times 4 + 1 \times 1 \end{bmatrix} \)
\( = \begin{bmatrix} 1 + 8 & 4 + 4 \\ 2 + 2 & 8 + 1 \end{bmatrix} \)
\( = \begin{bmatrix} 9 & 8 \\ 4 & 9 \end{bmatrix} \)

Next, let's find BC:
\( BC = \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \)
\( = \begin{bmatrix} -3 \times 1 + 2 \times 0 & -3 \times 0 + 2 \times 2 \\ 4 \times 1 + 0 \times 0 & 4 \times 0 + 0 \times 2 \end{bmatrix} \)
\( = \begin{bmatrix} -3 + 0 & 0 + 4 \\ 4 + 0 & 0 + 0 \end{bmatrix} \)
\( = \begin{bmatrix} -3 & 4 \\ 4 & 0 \end{bmatrix} \)

Finally, let's add \( A^2 \) and BC:
\( A^2 + BC = \begin{bmatrix} 9 & 8 \\ 4 & 9 \end{bmatrix} + \begin{bmatrix} -3 & 4 \\ 4 & 0 \end{bmatrix} \)
\( = \begin{bmatrix} 9 + (-3) & 8 + 4 \\ 4 + 4 & 9 + 0 \end{bmatrix} \)
\( = \begin{bmatrix} 9 - 3 & 12 \\ 8 & 9 \end{bmatrix} \)
\( = \begin{bmatrix} 6 & 12 \\ 8 & 9 \end{bmatrix} \)
In simple words: First, multiply matrix A by itself to get \( A^2 \). Then, multiply matrix B by matrix C to get BC. After that, add the resulting matrices \( A^2 \) and BC by adding their matching numbers.

🎯 Exam Tip: Remember to follow the order of operations: first calculate the powers (like \( A^2 \)), then multiplications (like BC), and finally additions or subtractions. Always ensure matrices have compatible dimensions for multiplication and addition.

Question 5. Find x and y if \( \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\begin{bmatrix} 3 \\ 4 \end{bmatrix}=\begin{bmatrix} x \\ y \end{bmatrix} \).
Answer:
Given the matrix equation:
\( \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\begin{bmatrix} 3 \\ 4 \end{bmatrix}=\begin{bmatrix} x \\ y \end{bmatrix} \)
First, perform the matrix multiplication on the left side:
\( \begin{bmatrix} 1 \times 3 + 0 \times 4 \\ 0 \times 3 + (-1) \times 4 \end{bmatrix} = \begin{bmatrix} x \\ y \end{bmatrix} \)
\( \begin{bmatrix} 3 + 0 \\ 0 - 4 \end{bmatrix} = \begin{bmatrix} x \\ y \end{bmatrix} \)
\( \begin{bmatrix} 3 \\ -4 \end{bmatrix} = \begin{bmatrix} x \\ y \end{bmatrix} \)
Now, compare the corresponding elements of the two matrices. This is how we find the values of x and y.
From the first row, \( x = 3 \).
From the second row, \( y = -4 \).
Thus, the values are \( x=3 \) and \( y=-4 \). This type of problem helps understand how matrices transform vectors.
In simple words: Multiply the two matrices on the left. The top number of the answer will be x, and the bottom number will be y.

🎯 Exam Tip: When multiplying a matrix by a column vector, the result is a column vector. Ensure you correctly perform the row-by-column multiplication for each element of the resulting vector.

Question 6. Let M \( \times \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 2 \end{bmatrix} \), where M is a matrix,
(i) State the order of the matrix M.
(ii) Find the matrix M.
Answer:
(i) Let the order of matrix M be \( m \times n \).
We have the equation: M \( \times \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 2 \end{bmatrix} \)
The order of \( \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} \) is \( 2 \times 2 \).
The order of \( \begin{bmatrix} 1 & 2 \end{bmatrix} \) is \( 1 \times 2 \).
For matrix multiplication to be possible, the number of columns in M must be equal to the number of rows in the second matrix. So, \( n = 2 \).
The order of the resulting matrix \( (M \times \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix}) \) is \( m \times 2 \).
Since this result is equal to \( \begin{bmatrix} 1 & 2 \end{bmatrix} \) which has order \( 1 \times 2 \), we must have \( m = 1 \).
Therefore, the order of matrix M is \( 1 \times 2 \). This means M is a row matrix.

(ii) Let matrix M be \( \begin{bmatrix} x & y \end{bmatrix} \).
So the equation becomes:
\( \begin{bmatrix} x & y \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 2 \end{bmatrix} \)
Perform the matrix multiplication on the left side:
\( \begin{bmatrix} x \times 1 + y \times 0 & x \times 1 + y \times 2 \end{bmatrix} = \begin{bmatrix} 1 & 2 \end{bmatrix} \)
\( \begin{bmatrix} x + 0 & x + 2y \end{bmatrix} = \begin{bmatrix} 1 & 2 \end{bmatrix} \)
\( \begin{bmatrix} x & x + 2y \end{bmatrix} = \begin{bmatrix} 1 & 2 \end{bmatrix} \)
Now, compare the corresponding elements:
\( x = 1 \)
\( x + 2y = 2 \)
Substitute the value of \( x=1 \) into the second equation:
\( 1 + 2y = 2 \)
\( 2y = 2 - 1 \)
\( 2y = 1 \)
\( y = \frac{1}{2} \)
Thus, the matrix M is \( \begin{bmatrix} 1 & \frac{1}{2} \end{bmatrix} \).
In simple words: First, figure out how big matrix M must be for the multiplication to work. Then, write M with unknown letters like x and y, multiply the matrices, and compare the numbers to solve for x and y.

🎯 Exam Tip: When finding an unknown matrix in a multiplication, start by determining its order based on the dimensions of the known matrices and the product matrix. This helps set up the correct number of variables for solving.

Question 7. Let \( A = \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix}, B = \begin{bmatrix} 0 & 2 \\ 1 & -1 \end{bmatrix}, C = \begin{bmatrix} -2 & 3 \\ 1 & -3 \end{bmatrix} \), find
(i) \( A^2 \)
(ii) BC
Answer:
Given matrices:
\( A = \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix}, B = \begin{bmatrix} 0 & 2 \\ 1 & -1 \end{bmatrix}, C = \begin{bmatrix} -2 & 3 \\ 1 & -3 \end{bmatrix} \)

(i) Find \( A^2 = A \times A \):
\( A^2 = \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix} \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix} \)
\( = \begin{bmatrix} 4 \times 4 + (-2) \times 6 & 4 \times (-2) + (-2) \times (-3) \\ 6 \times 4 + (-3) \times 6 & 6 \times (-2) + (-3) \times (-3) \end{bmatrix} \)
\( = \begin{bmatrix} 16 - 12 & -8 + 6 \\ 24 - 18 & -12 + 9 \end{bmatrix} \)
\( = \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix} \)
It's interesting to note that \( A^2 = A \) in this case, which means A is an idempotent matrix. This is a special property.

(ii) Find BC:
\( BC = \begin{bmatrix} 0 & 2 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} -2 & 3 \\ 1 & -3 \end{bmatrix} \)
\( = \begin{bmatrix} 0 \times (-2) + 2 \times 1 & 0 \times 3 + 2 \times (-3) \\ 1 \times (-2) + (-1) \times 1 & 1 \times 3 + (-1) \times (-3) \end{bmatrix} \)
\( = \begin{bmatrix} 0 + 2 & 0 - 6 \\ -2 - 1 & 3 + 3 \end{bmatrix} \)
\( = \begin{bmatrix} 2 & -6 \\ -3 & 6 \end{bmatrix} \)
In simple words: For \( A^2 \), multiply matrix A by itself. For BC, multiply matrix B by matrix C. Be very careful with negative numbers during multiplication.

🎯 Exam Tip: Pay close attention to the order of matrices in multiplication (e.g., BC is not necessarily CB). Also, ensure all terms in the dot product of rows and columns are correctly calculated, especially when dealing with negative values.

Question 8. If \( P = \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix}, Q = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \), then compute
(i) \( P^2 – Q^2 \)
(ii) \( (P + Q) (P – Q) \)
Answer:
Given matrices:
\( P = \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix}, Q = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \)

(i) Compute \( P^2 - Q^2 \):
First, find \( P^2 = P \times P \):
\( P^2 = \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} \)
\( = \begin{bmatrix} 1 \times 1 + 2 \times 2 & 1 \times 2 + 2 \times (-1) \\ 2 \times 1 + (-1) \times 2 & 2 \times 2 + (-1) \times (-1) \end{bmatrix} \)
\( = \begin{bmatrix} 1 + 4 & 2 - 2 \\ 2 - 2 & 4 + 1 \end{bmatrix} \)
\( = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} \)

Next, find \( Q^2 = Q \times Q \):
\( Q^2 = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 1 \times 1 + 0 \times 2 & 1 \times 0 + 0 \times 1 \\ 2 \times 1 + 1 \times 2 & 2 \times 0 + 1 \times 1 \end{bmatrix} \)
\( = \begin{bmatrix} 1 + 0 & 0 + 0 \\ 2 + 2 & 0 + 1 \end{bmatrix} \)
\( = \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix} \)

Finally, compute \( P^2 - Q^2 \):
\( P^2 - Q^2 = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 5 - 1 & 0 - 0 \\ 0 - 4 & 5 - 1 \end{bmatrix} \)
\( = \begin{bmatrix} 4 & 0 \\ -4 & 4 \end{bmatrix} \)

(ii) Compute \( (P + Q) (P - Q) \):
First, find \( P + Q \):
\( P + Q = \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 1 + 1 & 2 + 0 \\ 2 + 2 & -1 + 1 \end{bmatrix} \)
\( = \begin{bmatrix} 2 & 2 \\ 4 & 0 \end{bmatrix} \)

Next, find \( P - Q \):
\( P - Q = \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 1 - 1 & 2 - 0 \\ 2 - 2 & -1 - 1 \end{bmatrix} \)
\( = \begin{bmatrix} 0 & 2 \\ 0 & -2 \end{bmatrix} \)

Finally, multiply \( (P + Q) \) by \( (P - Q) \):
\( (P + Q) (P - Q) = \begin{bmatrix} 2 & 2 \\ 4 & 0 \end{bmatrix} \begin{bmatrix} 0 & 2 \\ 0 & -2 \end{bmatrix} \)
\( = \begin{bmatrix} 2 \times 0 + 2 \times 0 & 2 \times 2 + 2 \times (-2) \\ 4 \times 0 + 0 \times 0 & 4 \times 2 + 0 \times (-2) \end{bmatrix} \)
\( = \begin{bmatrix} 0 + 0 & 4 - 4 \\ 0 + 0 & 8 + 0 \end{bmatrix} \)
\( = \begin{bmatrix} 0 & 0 \\ 0 & 8 \end{bmatrix} \)
We can see that \( (P + Q) (P - Q) \ne P^2 - Q^2 \) for these matrices, unlike in scalar algebra. This highlights that matrix multiplication is generally not commutative.
In simple words: First, calculate the square of P and Q, then subtract them. Second, add P and Q, then subtract Q from P. Finally, multiply these two new matrices together. Notice that the answers for (i) and (ii) are different.

🎯 Exam Tip: Remember that the algebraic identity \( (A+B)(A-B) = A^2 - B^2 \) does NOT generally hold for matrices because matrix multiplication is not commutative (AB is usually not equal to BA). Always compute each term separately. Show all intermediate steps clearly for full marks.

Question 9. If \( \begin{bmatrix} 3 & 4 \\ 2 & 5 \end{bmatrix}=\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \) write down the values of a, b, c and d.
Answer:
Given the matrix equation:
\( \begin{bmatrix} 3 & 4 \\ 2 & 5 \end{bmatrix}=\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \)
First, perform the matrix multiplication on the right side:
\( \begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} a \times 1 + b \times 0 & a \times 0 + b \times (-1) \\ c \times 1 + d \times 0 & c \times 0 + d \times (-1) \end{bmatrix} \)
\( = \begin{bmatrix} a + 0 & 0 - b \\ c + 0 & 0 - d \end{bmatrix} \)
\( = \begin{bmatrix} a & -b \\ c & -d \end{bmatrix} \)
So, the original equation becomes:
\( \begin{bmatrix} 3 & 4 \\ 2 & 5 \end{bmatrix} = \begin{bmatrix} a & -b \\ c & -d \end{bmatrix} \)
Now, compare the corresponding elements of the two matrices to find a, b, c, and d. Each matching position gives an equation.
\( a = 3 \)
\( -b = 4 \implies b = -4 \)
\( c = 2 \)
\( -d = 5 \implies d = -5 \)
Thus, the values are \( a=3, b=-4, c=2, d=-5 \). This method is useful for decomposing matrix transformations.
In simple words: Multiply the matrices on the right side first. Then, compare the numbers in the same spots on both sides of the equal sign to find a, b, c, and d.

🎯 Exam Tip: Be careful with signs, especially when variables like 'b' and 'd' appear as '-b' or '-d' after multiplication. Remember to solve for the positive variable. Comparing elements is a direct way to solve for unknowns in matrix equations.

Question 10. If \( A = \begin{bmatrix} 2 & 0 \\ -3 & 1 \end{bmatrix} \), and \( B = \begin{bmatrix} 0 & 1 \\ -2 & 3 \end{bmatrix} \), find (i) BA (ii) \( A^2 \).
Answer:
Given matrices:
\( A = \begin{bmatrix} 2 & 0 \\ -3 & 1 \end{bmatrix}, B = \begin{bmatrix} 0 & 1 \\ -2 & 3 \end{bmatrix} \)

(i) Find BA:
\( BA = \begin{bmatrix} 0 & 1 \\ -2 & 3 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ -3 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 0 \times 2 + 1 \times (-3) & 0 \times 0 + 1 \times 1 \\ -2 \times 2 + 3 \times (-3) & -2 \times 0 + 3 \times 1 \end{bmatrix} \)
\( = \begin{bmatrix} 0 - 3 & 0 + 1 \\ -4 - 9 & 0 + 3 \end{bmatrix} \)
\( = \begin{bmatrix} -3 & 1 \\ -13 & 3 \end{bmatrix} \)

(ii) Find \( A^2 \):
\( A^2 = A \times A = \begin{bmatrix} 2 & 0 \\ -3 & 1 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ -3 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 2 \times 2 + 0 \times (-3) & 2 \times 0 + 0 \times 1 \\ -3 \times 2 + 1 \times (-3) & -3 \times 0 + 1 \times 1 \end{bmatrix} \)
\( = \begin{bmatrix} 4 + 0 & 0 + 0 \\ -6 - 3 & 0 + 1 \end{bmatrix} \)
\( = \begin{bmatrix} 4 & 0 \\ -9 & 1 \end{bmatrix} \)
In simple words: For BA, multiply matrix B by matrix A. For \( A^2 \), multiply matrix A by itself. Always take care with negative numbers and make sure you are multiplying in the correct order.

🎯 Exam Tip: Matrix multiplication is not commutative, meaning BA is generally not equal to AB. Be sure to multiply in the exact order specified in the question. Any misordering will result in an incorrect answer.

Question 11. If \( A = \begin{bmatrix} 2 & x \\ 0 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} 4 & 36 \\ 0 & 1 \end{bmatrix} \), find the value of x, given that \( A^2 = B \).
Answer:
Given matrices:
\( A = \begin{bmatrix} 2 & x \\ 0 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} 4 & 36 \\ 0 & 1 \end{bmatrix} \)
We are given that \( A^2 = B \).

First, let's find \( A^2 = A \times A \):
\( A^2 = \begin{bmatrix} 2 & x \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & x \\ 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 2 \times 2 + x \times 0 & 2 \times x + x \times 1 \\ 0 \times 2 + 1 \times 0 & 0 \times x + 1 \times 1 \end{bmatrix} \)
\( = \begin{bmatrix} 4 + 0 & 2x + x \\ 0 + 0 & 0 + 1 \end{bmatrix} \)
\( = \begin{bmatrix} 4 & 3x \\ 0 & 1 \end{bmatrix} \)

Now, we use the condition \( A^2 = B \):
\( \begin{bmatrix} 4 & 3x \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 36 \\ 0 & 1 \end{bmatrix} \)
Comparing the corresponding elements of the two matrices, we can find the value of x. The other elements are already equal, which confirms our calculations.
From the top-right elements:
\( 3x = 36 \)
Now, solve for x:
\( x = \frac{36}{3} \)
\( x = 12 \)
So, the value of x is 12. This shows how matrix equality can be used to find unknown scalar values.
In simple words: Multiply matrix A by itself to find \( A^2 \). Then, set the numbers in \( A^2 \) equal to the numbers in matrix B. Solve the equation you get to find the value of x.

🎯 Exam Tip: When setting two matrices equal, every corresponding element must be equal. This means if the matrices have multiple unknown variables, you will get a system of equations. Make sure your matrix multiplication is perfect, as errors will lead to incorrect values for variables.

Question 12.
(i) Find p and q such that \( \begin{bmatrix} p & q \end{bmatrix} \begin{bmatrix} p \\ q \end{bmatrix} = \begin{bmatrix} 25 \end{bmatrix} \).
(ii) Find p if \( \begin{bmatrix} 1 & 3 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 2 \\ -1 \end{bmatrix}=\begin{bmatrix} p \\ q \end{bmatrix} \).
Answer:
(i) Given: \( \begin{bmatrix} p & q \end{bmatrix} \begin{bmatrix} p \\ q \end{bmatrix} = \begin{bmatrix} 25 \end{bmatrix} \)
Perform the matrix multiplication on the left side:
\( \begin{bmatrix} p \times p + q \times q \end{bmatrix} = \begin{bmatrix} 25 \end{bmatrix} \)
\( \begin{bmatrix} p^2 + q^2 \end{bmatrix} = \begin{bmatrix} 25 \end{bmatrix} \)
Comparing the elements, we get:
\( p^2 + q^2 = 25 \)
We need to find integers p and q whose squares add up to 25. These are Pythagorean triplets, or integer pairs.
Possible integer solutions for (p, q) include:
- If \( p=0 \), then \( q^2 = 25 \implies q = \pm 5 \). So, \( (0, \pm 5) \).
- If \( p=\pm 5 \), then \( q^2 = 0 \implies q = 0 \). So, \( (\pm 5, 0) \).
- If \( p=\pm 3 \), then \( 9 + q^2 = 25 \implies q^2 = 16 \implies q = \pm 4 \). So, \( (\pm 3, \pm 4) \).
- If \( p=\pm 4 \), then \( 16 + q^2 = 25 \implies q^2 = 9 \implies q = \pm 3 \). So, \( (\pm 4, \pm 3) \).

(ii) Given: \( \begin{bmatrix} 1 & 3 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 2 \\ -1 \end{bmatrix}=\begin{bmatrix} p \\ q \end{bmatrix} \)
Perform the matrix multiplication on the left side:
\( \begin{bmatrix} 1 \times 2 + 3 \times (-1) \\ 0 \times 2 + 0 \times (-1) \end{bmatrix} = \begin{bmatrix} p \\ q \end{bmatrix} \)
\( \begin{bmatrix} 2 - 3 \\ 0 + 0 \end{bmatrix} = \begin{bmatrix} p \\ q \end{bmatrix} \)
\( \begin{bmatrix} -1 \\ 0 \end{bmatrix} = \begin{bmatrix} p \\ q \end{bmatrix} \)
Comparing the corresponding elements, we get:
\( p = -1 \)
\( q = 0 \)
Thus, the value of p is -1. This process is fundamental to solving systems of linear equations using matrices.
In simple words: (i) Multiply the two matrices to get \( p^2 + q^2 \). Then, find pairs of numbers whose squares add up to 25. (ii) Multiply the matrices on the left. The top number of the new matrix will be 'p', and the bottom number will be 'q'.

🎯 Exam Tip: For part (i), remember that multiple integer pairs can satisfy \( p^2 + q^2 = 25 \). It's important to list all possible combinations, including positive and negative values. For part (ii), a careful single matrix multiplication is enough to find 'p' and 'q'.

Question 13. If A and B are any two 2 x 2 matrices such that AB = B and B is not a zero matrix, what can you say about the matrix A?
Answer:
Given that A and B are two 2 x 2 matrices.
We are told that \( AB = B \).
We are also told that B is not a zero matrix, meaning B is not \( \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \).
We want to determine what type of matrix A must be.
Consider the equation: \( AB = B \)
We can rewrite this as: \( AB - B = O \) (where O is the zero matrix)
Factor out B on the right side:
\( (A - I)B = O \) (where I is the 2 x 2 identity matrix \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \))
Since B is not a zero matrix, for \( (A - I)B = O \) to be true, the matrix \( (A - I) \) must be the zero matrix. If \( (A - I) \) were invertible, then \( B \) would have to be \( O \), which contradicts the given condition.
Therefore, \( A - I = O \)
\( A = I \)
So, matrix A must be the unit matrix or identity matrix, \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \). The identity matrix acts like the number '1' in scalar multiplication (1 * B = B).
In simple words: If multiplying matrix A by matrix B gives back matrix B, and B is not all zeros, then matrix A must be the "identity matrix" (a matrix with ones on the main diagonal and zeros elsewhere). It acts like multiplying by the number 1.

🎯 Exam Tip: This question tests your understanding of matrix properties, especially the role of the identity matrix. If \( AB = B \) and B is not the zero matrix, it implies A behaves like a multiplicative identity. This is a common theoretical question in matrix algebra.

Question 14. If \( A = \begin{bmatrix} 3 & 2 \\ 4 & -3 \end{bmatrix}, B = \begin{bmatrix} 2 & -3 \\ -4 & 5 \end{bmatrix} \) and \( C = \begin{bmatrix} 1 & -3 \\ -4 & 4 \end{bmatrix} \), find (i) BC (ii) \( A^2 + A \).
Answer:
Given matrices:
\( A = \begin{bmatrix} 3 & 2 \\ 4 & -3 \end{bmatrix}, B = \begin{bmatrix} 2 & -3 \\ -4 & 5 \end{bmatrix}, C = \begin{bmatrix} 1 & -3 \\ -4 & 4 \end{bmatrix} \)

(i) Find BC:
\( BC = \begin{bmatrix} 2 & -3 \\ -4 & 5 \end{bmatrix} \begin{bmatrix} 1 & -3 \\ -4 & 4 \end{bmatrix} \)
\( = \begin{bmatrix} 2 \times 1 + (-3) \times (-4) & 2 \times (-3) + (-3) \times 4 \\ -4 \times 1 + 5 \times (-4) & -4 \times (-3) + 5 \times 4 \end{bmatrix} \)
\( = \begin{bmatrix} 2 + 12 & -6 - 12 \\ -4 - 20 & 12 + 20 \end{bmatrix} \)
\( = \begin{bmatrix} 14 & -18 \\ -24 & 32 \end{bmatrix} \)

(ii) Find \( A^2 + A \):
First, find \( A^2 = A \times A \):
\( A^2 = \begin{bmatrix} 3 & 2 \\ 4 & -3 \end{bmatrix} \begin{bmatrix} 3 & 2 \\ 4 & -3 \end{bmatrix} \)
\( = \begin{bmatrix} 3 \times 3 + 2 \times 4 & 3 \times 2 + 2 \times (-3) \\ 4 \times 3 + (-3) \times 4 & 4 \times 2 + (-3) \times (-3) \end{bmatrix} \)
\( = \begin{bmatrix} 9 + 8 & 6 - 6 \\ 12 - 12 & 8 + 9 \end{bmatrix} \)
\( = \begin{bmatrix} 17 & 0 \\ 0 & 17 \end{bmatrix} \)
This result \( A^2 = 17I \) (where I is the identity matrix) is quite special for this matrix A.
Now, add \( A^2 \) and A:
\( A^2 + A = \begin{bmatrix} 17 & 0 \\ 0 & 17 \end{bmatrix} + \begin{bmatrix} 3 & 2 \\ 4 & -3 \end{bmatrix} \)
\( = \begin{bmatrix} 17 + 3 & 0 + 2 \\ 0 + 4 & 17 + (-3) \end{bmatrix} \)
\( = \begin{bmatrix} 20 & 2 \\ 4 & 14 \end{bmatrix} \)
In simple words: For BC, multiply matrix B by matrix C. For \( A^2+A \), first multiply A by itself to get \( A^2 \), then add the original matrix A to that result. Be extra careful with negative signs and make sure your calculations are accurate.

🎯 Exam Tip: Always perform matrix multiplication and addition/subtraction step-by-step. Break down complex expressions like \( A^2 + A \) into smaller, manageable calculations to avoid errors. Pay attention to signs when multiplying and adding.

Question 15. Given the matrix equations:
(i) \( \left[\begin{array}{ll} a-b & b-4 \\ b+4 & a-2 \end{array}\right]\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right]=\left[\begin{array}{cc} -2 & -2 \\ 14 & 0 \end{array}\right] \)
(ii) \( \left[\begin{array}{cc} 8 & -2 \\ 1 & 4 \end{array}\right] X=\left[\begin{array}{l} 12 \\ 10 \end{array}\right] \)
For part (i), write down the values of a, b, c and d. For part (ii), write down (a) the order of matrix X, (b) the matrix X.
Answer:
(i) We are given the equation:
\( \left[\begin{array}{ll} a-b & b-4 \\ b+4 & a-2 \end{array}\right]\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right]=\left[\begin{array}{cc} -2 & -2 \\ 14 & 0 \end{array}\right] \)
First, multiply the matrices on the left side:
\( \left[\begin{array}{ll} (a-b) \times 2+(b-4) \times 0 & (a-b) \times 0+(b-4) \times 2 \\ (b+4) \times 2+(a-2) \times 0 & (b+4) \times 0+(a-2) \times 2 \end{array}\right]=\left[\begin{array}{cc} -2 & -2 \\ 14 & 0 \end{array}\right] \)
This simplifies to:
\( \left[\begin{array}{cc} 2a-2b & 2b-8 \\ 2b+8 & 2a-4 \end{array}\right]=\left[\begin{array}{cc} -2 & -2 \\ 14 & 0 \end{array}\right] \)
Now, compare the corresponding elements of the matrices:
From \( 2b-8 = -2 \):
\( 2b = -2 + 8 \)
\( 2b = 6 \)
\( b = \frac{6}{2} \)
\( b = 3 \)
From \( 2a-4 = 0 \):
\( 2a = 4 \)
\( a = \frac{4}{2} \)
\( a = 2 \)
Thus, the values are \( a = 2 \) and \( b = 3 \). We don't have 'c' or 'd' explicitly in this part of the problem setup, but the structure matches typical matrix element naming, so we can assume they were meant to be \( a \) and \( b \).
(ii) Given \( \left[\begin{array}{cc} 8 & -2 \\ 1 & 4 \end{array}\right] X=\left[\begin{array}{l} 12 \\ 10 \end{array}\right] \)
(a) The first matrix is \( 2 \times 2 \) (2 rows, 2 columns). The result matrix is \( 2 \times 1 \) (2 rows, 1 column).
For matrix multiplication, if matrix A is \( m \times n \) and matrix X is \( p \times q \), then \( n \) must be equal to \( p \). The resulting matrix AX will be \( m \times q \).
Here, \( m=2, n=2 \). The result is \( 2 \times 1 \), so \( m=2 \) and \( q=1 \).
Since \( n=p \), \( p=2 \).
So, matrix X must have 2 rows and 1 column.
The order of matrix X is \( 2 \times 1 \).
(b) Let matrix \( X = \left[\begin{array}{l} a \\ b \end{array}\right] \).
Then the equation becomes:
\( \left[\begin{array}{cc} 8 & -2 \\ 1 & 4 \end{array}\right]\left[\begin{array}{l} a \\ b \end{array}\right]=\left[\begin{array}{l} 12 \\ 10 \end{array}\right] \)
Multiply the matrices on the left:
\( \left[\begin{array}{c} 8a-2b \\ a+4b \end{array}\right]=\left[\begin{array}{c} 12 \\ 10 \end{array}\right] \)
Compare the corresponding elements to form a system of linear equations:
1) \( 8a - 2b = 12 \)
2) \( a + 4b = 10 \)
From equation (1), divide by 2: \( 4a - b = 6 \implies b = 4a - 6 \).
Substitute this into equation (2):
\( a + 4(4a - 6) = 10 \)
\( a + 16a - 24 = 10 \)
\( 17a - 24 = 10 \)
\( 17a = 10 + 24 \)
\( 17a = 34 \)
\( a = \frac{34}{17} \)
\( a = 2 \)
Now substitute \( a=2 \) back into \( b = 4a - 6 \):
\( b = 4(2) - 6 \)
\( b = 8 - 6 \)
\( b = 2 \)
Therefore, matrix \( X = \left[\begin{array}{l} 2 \\ 2 \end{array}\right] \)
In simple words: For the first part, we multiplied the two matrices and then matched the elements on both sides to find the values of 'a' and 'b'. For the second part, we first figured out how big matrix X should be, then we set it up with unknown values 'a' and 'b'. Multiplying the matrices gave us two simple equations, which we solved to find 'a' and 'b' for matrix X.

🎯 Exam Tip: When dealing with matrix equations, remember to perform matrix operations (like multiplication) first before equating corresponding elements. Always double-check the order of matrices for multiplication and the resulting order.

Question 16. If \( A = \left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right] \), show that \( A^2 - 4A + 5I = 0 \), where I is the unit matrix.
Answer:
Given matrix \( A = \left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right] \).
First, calculate \( A^2 = A \times A \):
\( A^2 = \left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right] \left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right] \)
\( A^2 = \left[\begin{array}{cc} (1 \times 1) + (-1 \times 2) & (1 \times -1) + (-1 \times 3) \\ (2 \times 1) + (3 \times 2) & (2 \times -1) + (3 \times 3) \end{array}\right] \)
\( A^2 = \left[\begin{array}{cc} 1 - 2 & -1 - 3 \\ 2 + 6 & -2 + 9 \end{array}\right] \)
\( A^2 = \left[\begin{array}{cc} -1 & -4 \\ 8 & 7 \end{array}\right] \)
Next, calculate \( 4A \):
\( 4A = 4 \left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right] \)
\( 4A = \left[\begin{array}{cc} 4 \times 1 & 4 \times -1 \\ 4 \times 2 & 4 \times 3 \end{array}\right] \)
\( 4A = \left[\begin{array}{cc} 4 & -4 \\ 8 & 12 \end{array}\right] \)
The unit matrix I for a \( 2 \times 2 \) matrix is \( I = \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right] \).
Calculate \( 5I \):
\( 5I = 5 \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right] \)
\( 5I = \left[\begin{array}{cc} 5 & 0 \\ 0 & 5 \end{array}\right] \)
Now, substitute these into the expression \( A^2 - 4A + 5I \):
\( A^2 - 4A + 5I = \left[\begin{array}{cc} -1 & -4 \\ 8 & 7 \end{array}\right] - \left[\begin{array}{cc} 4 & -4 \\ 8 & 12 \end{array}\right] + \left[\begin{array}{cc} 5 & 0 \\ 0 & 5 \end{array}\right] \)
First, perform the subtraction:
\( A^2 - 4A = \left[\begin{array}{cc} -1 - 4 & -4 - (-4) \\ 8 - 8 & 7 - 12 \end{array}\right] \)
\( A^2 - 4A = \left[\begin{array}{cc} -5 & 0 \\ 0 & -5 \end{array}\right] \)
Now, add \( 5I \):
\( A^2 - 4A + 5I = \left[\begin{array}{cc} -5 & 0 \\ 0 & -5 \end{array}\right] + \left[\begin{array}{cc} 5 & 0 \\ 0 & 5 \end{array}\right] \)
\( A^2 - 4A + 5I = \left[\begin{array}{cc} -5+5 & 0+0 \\ 0+0 & -5+5 \end{array}\right] \)
\( A^2 - 4A + 5I = \left[\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}\right] \)
This is the zero matrix, which is denoted by 0.
Therefore, \( A^2 - 4A + 5I = 0 \). The unit matrix plays a crucial role in matrix algebra, similar to the number 1 in scalar arithmetic.
In simple words: We first calculated A multiplied by itself (A squared), then we found 4 times A, and 5 times the special unit matrix. After that, we put all these results together by subtracting and adding them. The final answer was a matrix with all zeros, which means we proved the given statement.

🎯 Exam Tip: Remember to calculate each term (A², 4A, 5I) separately and accurately before combining them. Pay close attention to signs during subtraction, especially with negative numbers.

Question 17. If \( A = \left[\begin{array}{ll} 1 & 4 \\ 1 & 0 \end{array}\right] \), \( B = \left[\begin{array}{cc} 2 & 1 \\ 3 & -1 \end{array}\right] \) and \( C = \left[\begin{array}{ll} 2 & 3 \\ 0 & 5 \end{array}\right] \), compute (AB) C and (CB) A.
Answer:
Given matrices:
\( A = \left[\begin{array}{ll} 1 & 4 \\ 1 & 0 \end{array}\right] \)
\( B = \left[\begin{array}{cc} 2 & 1 \\ 3 & -1 \end{array}\right] \)
\( C = \left[\begin{array}{ll} 2 & 3 \\ 0 & 5 \end{array}\right] \)

**Part 1: Compute (AB) C**
First, calculate \( AB \):
\( AB = \left[\begin{array}{ll} 1 & 4 \\ 1 & 0 \end{array}\right] \left[\begin{array}{cc} 2 & 1 \\ 3 & -1 \end{array}\right] \)
\( AB = \left[\begin{array}{ll} (1 \times 2) + (4 \times 3) & (1 \times 1) + (4 \times -1) \\ (1 \times 2) + (0 \times 3) & (1 \times 1) + (0 \times -1) \end{array}\right] \)
\( AB = \left[\begin{array}{ll} 2 + 12 & 1 - 4 \\ 2 + 0 & 1 + 0 \end{array}\right] \)
\( AB = \left[\begin{array}{cc} 14 & -3 \\ 2 & 1 \end{array}\right] \)
Next, calculate \( (AB) C \):
\( (AB) C = \left[\begin{array}{cc} 14 & -3 \\ 2 & 1 \end{array}\right] \left[\begin{array}{ll} 2 & 3 \\ 0 & 5 \end{array}\right] \)
\( (AB) C = \left[\begin{array}{ll} (14 \times 2) + (-3 \times 0) & (14 \times 3) + (-3 \times 5) \\ (2 \times 2) + (1 \times 0) & (2 \times 3) + (1 \times 5) \end{array}\right] \)
\( (AB) C = \left[\begin{array}{ll} 28 + 0 & 42 - 15 \\ 4 + 0 & 6 + 5 \end{array}\right] \)
\( (AB) C = \left[\begin{array}{cc} 28 & 27 \\ 4 & 11 \end{array}\right] \)

**Part 2: Compute (CB) A**
First, calculate \( CB \):
\( CB = \left[\begin{array}{ll} 2 & 3 \\ 0 & 5 \end{array}\right] \left[\begin{array}{cc} 2 & 1 \\ 3 & -1 \end{array}\right] \)
\( CB = \left[\begin{array}{ll} (2 \times 2) + (3 \times 3) & (2 \times 1) + (3 \times -1) \\ (0 \times 2) + (5 \times 3) & (0 \times 1) + (5 \times -1) \end{array}\right] \)
\( CB = \left[\begin{array}{ll} 4 + 9 & 2 - 3 \\ 0 + 15 & 0 - 5 \end{array}\right] \)
\( CB = \left[\begin{array}{cc} 13 & -1 \\ 15 & -5 \end{array}\right] \)
Next, calculate \( (CB) A \):
\( (CB) A = \left[\begin{array}{cc} 13 & -1 \\ 15 & -5 \end{array}\right] \left[\begin{array}{ll} 1 & 4 \\ 1 & 0 \end{array}\right] \)
\( (CB) A = \left[\begin{array}{ll} (13 \times 1) + (-1 \times 1) & (13 \times 4) + (-1 \times 0) \\ (15 \times 1) + (-5 \times 1) & (15 \times 4) + (-5 \times 0) \end{array}\right] \)
\( (CB) A = \left[\begin{array}{ll} 13 - 1 & 52 + 0 \\ 15 - 5 & 60 + 0 \end{array}\right] \)
\( (CB) A = \left[\begin{array}{cc} 12 & 52 \\ 10 & 60 \end{array}\right] \)
From the results, we can see that \( (AB)C \neq (CB)A \), which is a general property of matrix multiplication (it is not commutative).
In simple words: We calculated two different matrix multiplications. First, we multiplied A by B, and then took that result and multiplied it by C. Second, we multiplied C by B, and then multiplied that result by A. We found that the final answers were different matrices, showing that the order of multiplication matters for matrices.

🎯 Exam Tip: Matrix multiplication is associative (A(BC) = (AB)C) but not commutative (AB ≠ BA generally). Always perform operations inside parentheses first and follow the exact order given in the question.

Question 18.
(i) Write as a single matrix: \( \left[\begin{array}{cc} 6 & -2 \\ 1 & 2 \end{array}\right]\left[\begin{array}{ll} 5 & 3 \\ 2 & 4 \end{array}\right] \)
(ii) Find x and y if \( \left[\begin{array}{cc} 2 & 3 \\ -1 & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} 7 \\ -2 \end{array}\right] \).
Answer:
(i) To write the expression as a single matrix, we perform the matrix multiplication:
\( \left[\begin{array}{cc} 6 & -2 \\ 1 & 2 \end{array}\right]\left[\begin{array}{ll} 5 & 3 \\ 2 & 4 \end{array}\right] \)
\( = \left[\begin{array}{ll} (6 \times 5) + (-2 \times 2) & (6 \times 3) + (-2 \times 4) \\ (1 \times 5) + (2 \times 2) & (1 \times 3) + (2 \times 4) \end{array}\right] \)
\( = \left[\begin{array}{ll} 30 - 4 & 18 - 8 \\ 5 + 4 & 3 + 8 \end{array}\right] \)
\( = \left[\begin{array}{cc} 26 & 10 \\ 9 & 11 \end{array}\right] \)

(ii) We are given the matrix equation:
\( \left[\begin{array}{cc} 2 & 3 \\ -1 & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} 7 \\ -2 \end{array}\right] \)
First, perform the matrix multiplication on the left side:
\( \left[\begin{array}{c} (2 \times x) + (3 \times y) \\ (-1 \times x) + (0 \times y) \end{array}\right]=\left[\begin{array}{c} 7 \\ -2 \end{array}\right] \)
\( \implies \left[\begin{array}{c} 2x+3y \\ -x \end{array}\right]=\left[\begin{array}{c} 7 \\ -2 \end{array}\right] \)
Now, compare the corresponding elements:
From the second row: \( -x = -2 \)
\( \implies x = 2 \)
From the first row: \( 2x + 3y = 7 \)
Substitute \( x=2 \) into this equation:
\( 2(2) + 3y = 7 \)
\( 4 + 3y = 7 \)
\( 3y = 7 - 4 \)
\( 3y = 3 \)
\( y = \frac{3}{3} \)
\( \implies y = 1 \)
So, the values are \( x = 2 \) and \( y = 1 \). This involves a simple system of equations.
In simple words: For the first part, we just multiplied two matrices to get one single matrix. For the second part, we multiplied a matrix by a column of variables to get another column. Then, we matched the numbers in the resulting column to solve for the unknown values 'x' and 'y' using basic algebra.

🎯 Exam Tip: For matrix multiplication, carefully follow the row-by-column rule. When comparing matrices, ensure you equate elements in the exact same positions to avoid errors in forming equations.

Question 19. If \( A = \left[\begin{array}{ll} 3 & 1 \\ 2 & 1 \end{array}\right] \) and \( B = \left[\begin{array}{cc} 1 & -2 \\ 5 & 3 \end{array}\right] \), verify if \( (A - B)^2 = A^2 - 2AB + B^2 \).
Answer:
Given matrices:
\( A = \left[\begin{array}{ll} 3 & 1 \\ 2 & 1 \end{array}\right] \)
\( B = \left[\begin{array}{cc} 1 & -2 \\ 5 & 3 \end{array}\right] \)

**Step 1: Calculate the Left Hand Side (LHS) - \( (A - B)^2 \)**
First, find \( A - B \):
\( A - B = \left[\begin{array}{ll} 3 & 1 \\ 2 & 1 \end{array}\right] - \left[\begin{array}{cc} 1 & -2 \\ 5 & 3 \end{array}\right] \)
\( A - B = \left[\begin{array}{ll} 3 - 1 & 1 - (-2) \\ 2 - 5 & 1 - 3 \end{array}\right] \)
\( A - B = \left[\begin{array}{cc} 2 & 3 \\ -3 & -2 \end{array}\right] \)
Now, calculate \( (A - B)^2 = (A - B) \times (A - B) \):
\( (A - B)^2 = \left[\begin{array}{cc} 2 & 3 \\ -3 & -2 \end{array}\right] \left[\begin{array}{cc} 2 & 3 \\ -3 & -2 \end{array}\right] \)
\( (A - B)^2 = \left[\begin{array}{ll} (2 \times 2) + (3 \times -3) & (2 \times 3) + (3 \times -2) \\ (-3 \times 2) + (-2 \times -3) & (-3 \times 3) + (-2 \times -2) \end{array}\right] \)
\( (A - B)^2 = \left[\begin{array}{ll} 4 - 9 & 6 - 6 \\ -6 + 6 & -9 + 4 \end{array}\right] \)
\( (A - B)^2 = \left[\begin{array}{cc} -5 & 0 \\ 0 & -5 \end{array}\right] \)

**Step 2: Calculate the Right Hand Side (RHS) - \( A^2 - 2AB + B^2 \)**
First, calculate \( A^2 \):
\( A^2 = A \times A = \left[\begin{array}{ll} 3 & 1 \\ 2 & 1 \end{array}\right] \left[\begin{array}{ll} 3 & 1 \\ 2 & 1 \end{array}\right] \)
\( A^2 = \left[\begin{array}{ll} (3 \times 3) + (1 \times 2) & (3 \times 1) + (1 \times 1) \\ (2 \times 3) + (1 \times 2) & (2 \times 1) + (1 \times 1) \end{array}\right] \)
\( A^2 = \left[\begin{array}{ll} 9 + 2 & 3 + 1 \\ 6 + 2 & 2 + 1 \end{array}\right] \)
\( A^2 = \left[\begin{array}{cc} 11 & 4 \\ 8 & 3 \end{array}\right] \)
Next, calculate \( AB \):
\( AB = \left[\begin{array}{ll} 3 & 1 \\ 2 & 1 \end{array}\right] \left[\begin{array}{cc} 1 & -2 \\ 5 & 3 \end{array}\right] \)
\( AB = \left[\begin{array}{ll} (3 \times 1) + (1 \times 5) & (3 \times -2) + (1 \times 3) \\ (2 \times 1) + (1 \times 5) & (2 \times -2) + (1 \times 3) \end{array}\right] \)
\( AB = \left[\begin{array}{ll} 3 + 5 & -6 + 3 \\ 2 + 5 & -4 + 3 \end{array}\right] \)
\( AB = \left[\begin{array}{cc} 8 & -3 \\ 7 & -1 \end{array}\right] \)
Then, calculate \( 2AB \):
\( 2AB = 2 \left[\begin{array}{cc} 8 & -3 \\ 7 & -1 \end{array}\right] = \left[\begin{array}{cc} 16 & -6 \\ 14 & -2 \end{array}\right] \)
Now, calculate \( B^2 \):
\( B^2 = B \times B = \left[\begin{array}{cc} 1 & -2 \\ 5 & 3 \end{array}\right] \left[\begin{array}{cc} 1 & -2 \\ 5 & 3 \end{array}\right] \)
\( B^2 = \left[\begin{array}{ll} (1 \times 1) + (-2 \times 5) & (1 \times -2) + (-2 \times 3) \\ (5 \times 1) + (3 \times 5) & (5 \times -2) + (3 \times 3) \end{array}\right] \)
\( B^2 = \left[\begin{array}{ll} 1 - 10 & -2 - 6 \\ 5 + 15 & -10 + 9 \end{array}\right] \)
\( B^2 = \left[\begin{array}{cc} -9 & -8 \\ 20 & -1 \end{array}\right] \)
Finally, combine \( A^2 - 2AB + B^2 \):
\( A^2 - 2AB + B^2 = \left[\begin{array}{cc} 11 & 4 \\ 8 & 3 \end{array}\right] - \left[\begin{array}{cc} 16 & -6 \\ 14 & -2 \end{array}\right] + \left[\begin{array}{cc} -9 & -8 \\ 20 & -1 \end{array}\right] \)
First, perform the subtraction:
\( \left[\begin{array}{cc} 11 - 16 & 4 - (-6) \\ 8 - 14 & 3 - (-2) \end{array}\right] = \left[\begin{array}{cc} -5 & 10 \\ -6 & 5 \end{array}\right] \)
Then add \( B^2 \):
\( \left[\begin{array}{cc} -5 & 10 \\ -6 & 5 \end{array}\right] + \left[\begin{array}{cc} -9 & -8 \\ 20 & -1 \end{array}\right] = \left[\begin{array}{cc} -5 - 9 & 10 - 8 \\ -6 + 20 & 5 - 1 \end{array}\right] \)
\( = \left[\begin{array}{cc} -14 & 2 \\ 14 & 4 \end{array}\right] \)
Comparing the LHS and RHS: \( \left[\begin{array}{cc} -5 & 0 \\ 0 & -5 \end{array}\right] \neq \left[\begin{array}{cc} -14 & 2 \\ 14 & 4 \end{array}\right] \)
Therefore, \( (A - B)^2 \neq A^2 - 2AB + B^2 \) in matrix algebra. This shows that the standard algebraic identity for squares of binomials does not directly apply to matrices in the same way because matrix multiplication is not commutative.
In simple words: We checked if a common algebra rule, \( (A-B)^2 = A^2 - 2AB + B^2 \), works for matrices. We calculated the left side by first subtracting A from B and then multiplying the result by itself. Then, we calculated the right side by finding \( A^2 \), \( B^2 \), and \( 2AB \), and putting them together. We found that the two sides were not equal. This means that the normal algebra rule doesn't always apply to matrices because the order of multiplying matrices matters.

🎯 Exam Tip: Always remember that standard algebraic identities like \( (A \pm B)^2 = A^2 \pm 2AB + B^2 \) do NOT generally hold true for matrices because \( AB \neq BA \). Instead, you must expand it as \( (A-B)(A-B) = A^2 - AB - BA + B^2 \).

Question 20. Given that \( A = \left[\begin{array}{ll} 3 & 0 \\ 0 & 4 \end{array}\right] \) and \( B = \left[\begin{array}{ll} a & b \\ 0 & c \end{array}\right] \) and that \( AB = A + B \), find the values of a, b and c.
Answer:
Given matrices:
\( A = \left[\begin{array}{ll} 3 & 0 \\ 0 & 4 \end{array}\right] \)
\( B = \left[\begin{array}{ll} a & b \\ 0 & c \end{array}\right] \)
We are given the condition \( AB = A + B \).

First, calculate \( AB \):
\( AB = \left[\begin{array}{ll} 3 & 0 \\ 0 & 4 \end{array}\right] \left[\begin{array}{ll} a & b \\ 0 & c \end{array}\right] \)
\( AB = \left[\begin{array}{ll} (3 \times a) + (0 \times 0) & (3 \times b) + (0 \times c) \\ (0 \times a) + (4 \times 0) & (0 \times b) + (4 \times c) \end{array}\right] \)
\( AB = \left[\begin{array}{cc} 3a & 3b \\ 0 & 4c \end{array}\right] \)

Next, calculate \( A + B \):
\( A + B = \left[\begin{array}{ll} 3 & 0 \\ 0 & 4 \end{array}\right] + \left[\begin{array}{ll} a & b \\ 0 & c \end{array}\right] \)
\( A + B = \left[\begin{array}{cc} 3+a & 0+b \\ 0+0 & 4+c \end{array}\right] \)
\( A + B = \left[\begin{array}{cc} 3+a & b \\ 0 & 4+c \end{array}\right] \)

Now, equate the corresponding elements of \( AB \) and \( A + B \):
Since \( AB = A + B \), we have:
\( \left[\begin{array}{cc} 3a & 3b \\ 0 & 4c \end{array}\right] = \left[\begin{array}{cc} 3+a & b \\ 0 & 4+c \end{array}\right] \)
Comparing the elements:
1) \( 3a = 3+a \)
\( 3a - a = 3 \)
\( 2a = 3 \)
\( a = \frac{3}{2} \)
2) \( 3b = b \)
\( 3b - b = 0 \)
\( 2b = 0 \)
\( b = 0 \)
3) \( 4c = 4+c \)
\( 4c - c = 4 \)
\( 3c = 4 \)
\( c = \frac{4}{3} \)
Therefore, the values are \( a = \frac{3}{2}, b = 0, \) and \( c = \frac{4}{3} \). Matrix equations often lead to systems of equations that can be solved algebraically.
In simple words: We were given two matrices and a rule that said multiplying them (A times B) gives the same result as adding them (A plus B). We first did the multiplication and the addition to get two new matrices. Then, we matched the numbers in the same positions in these two result matrices to create simple equations. Solving these equations helped us find the values for 'a', 'b', and 'c'.

🎯 Exam Tip: Break down matrix equations into individual steps: calculate each side of the equation separately, then equate corresponding elements to form linear equations. This systematic approach minimizes errors.

Question 21.
(i) Solve the matrix equation \( \left[\begin{array}{ll} 2 & 1 \\ 5 & 0 \end{array}\right] – 3X = \left[\begin{array}{cc} -7 & 4 \\ 2 & 6 \end{array}\right] \).
(ii) If \( A = \left[\begin{array}{ll} 2 & 1 \\ 1 & 2 \end{array}\right] \), \( B = \left[\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right] \), compute A (BA).
Answer:
(i) We need to solve for matrix X in the equation:
\( \left[\begin{array}{ll} 2 & 1 \\ 5 & 0 \end{array}\right] – 3X = \left[\begin{array}{cc} -7 & 4 \\ 2 & 6 \end{array}\right] \)
First, rearrange the equation to isolate \( 3X \):
\( 3X = \left[\begin{array}{ll} 2 & 1 \\ 5 & 0 \end{array}\right] - \left[\begin{array}{cc} -7 & 4 \\ 2 & 6 \end{array}\right] \)
Perform the matrix subtraction:
\( 3X = \left[\begin{array}{ll} 2 - (-7) & 1 - 4 \\ 5 - 2 & 0 - 6 \end{array}\right] \)
\( 3X = \left[\begin{array}{cc} 2 + 7 & -3 \\ 3 & -6 \end{array}\right] \)
\( 3X = \left[\begin{array}{cc} 9 & -3 \\ 3 & -6 \end{array}\right] \)
Now, divide each element by 3 to find X:
\( X = \frac{1}{3} \left[\begin{array}{cc} 9 & -3 \\ 3 & -6 \end{array}\right] \)
\( X = \left[\begin{array}{cc} \frac{9}{3} & \frac{-3}{3} \\ \frac{3}{3} & \frac{-6}{3} \end{array}\right] \)
\( X = \left[\begin{array}{cc} 3 & -1 \\ 1 & -2 \end{array}\right] \)

(ii) Given matrices:
\( A = \left[\begin{array}{ll} 2 & 1 \\ 1 & 2 \end{array}\right] \)
\( B = \left[\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right] \)
We need to compute \( A(BA) \).
First, calculate the product \( BA \):
\( BA = \left[\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right] \left[\begin{array}{ll} 2 & 1 \\ 1 & 2 \end{array}\right] \)
\( BA = \left[\begin{array}{ll} (1 \times 2) + (2 \times 1) & (1 \times 1) + (2 \times 2) \\ (2 \times 2) + (1 \times 1) & (2 \times 1) + (1 \times 2) \end{array}\right] \)
\( BA = \left[\begin{array}{ll} 2 + 2 & 1 + 4 \\ 4 + 1 & 2 + 2 \end{array}\right] \)
\( BA = \left[\begin{array}{ll} 4 & 5 \\ 5 & 4 \end{array}\right] \)
Next, calculate \( A(BA) \):
\( A(BA) = \left[\begin{array}{ll} 2 & 1 \\ 1 & 2 \end{array}\right] \left[\begin{array}{ll} 4 & 5 \\ 5 & 4 \end{array}\right] \)
\( A(BA) = \left[\begin{array}{ll} (2 \times 4) + (1 \times 5) & (2 \times 5) + (1 \times 4) \\ (1 \times 4) + (2 \times 5) & (1 \times 5) + (2 \times 4) \end{array}\right] \)
\( A(BA) = \left[\begin{array}{ll} 8 + 5 & 10 + 4 \\ 4 + 10 & 5 + 8 \end{array}\right] \)
\( A(BA) = \left[\begin{array}{cc} 13 & 14 \\ 14 & 13 \end{array}\right] \)
In simple words: For the first problem, we moved the terms around to get 3X by itself, then subtracted the matrices. Finally, we divided every number in the resulting matrix by 3 to find X. For the second problem, we first multiplied B by A. Then, we took that result and multiplied it by A again, following the correct order of operations, to get the final matrix.

🎯 Exam Tip: When solving matrix equations, treat them like algebraic equations but remember matrix operation rules. For expressions like A(BA), calculate the innermost product (BA) first, then multiply the result by A.

Question 22. If \( A = \left[\begin{array}{ll} 9 & 1 \\ 5 & 3 \end{array}\right] \) and \( B = \left[\begin{array}{cc} 1 & 5 \\ 7 & -11 \end{array}\right] \), find the matrix X such that \( 3A + 5B – 2X = 0 \).
Answer:
Given matrices:
\( A = \left[\begin{array}{ll} 9 & 1 \\ 5 & 3 \end{array}\right] \)
\( B = \left[\begin{array}{cc} 1 & 5 \\ 7 & -11 \end{array}\right] \)
We need to find matrix X such that \( 3A + 5B – 2X = 0 \).
Rearrange the equation to solve for \( 2X \):
\( 2X = 3A + 5B \)
First, calculate \( 3A \):
\( 3A = 3 \left[\begin{array}{ll} 9 & 1 \\ 5 & 3 \end{array}\right] = \left[\begin{array}{ll} 3 \times 9 & 3 \times 1 \\ 3 \times 5 & 3 \times 3 \end{array}\right] = \left[\begin{array}{cc} 27 & 3 \\ 15 & 9 \end{array}\right] \)
Next, calculate \( 5B \):
\( 5B = 5 \left[\begin{array}{cc} 1 & 5 \\ 7 & -11 \end{array}\right] = \left[\begin{array}{cc} 5 \times 1 & 5 \times 5 \\ 5 \times 7 & 5 \times -11 \end{array}\right] = \left[\begin{array}{cc} 5 & 25 \\ 35 & -55 \end{array}\right] \)
Now, calculate \( 3A + 5B \):
\( 3A + 5B = \left[\begin{array}{cc} 27 & 3 \\ 15 & 9 \end{array}\right] + \left[\begin{array}{cc} 5 & 25 \\ 35 & -55 \end{array}\right] \)
\( 3A + 5B = \left[\begin{array}{cc} 27 + 5 & 3 + 25 \\ 15 + 35 & 9 + (-55) \end{array}\right] \)
\( 3A + 5B = \left[\begin{array}{cc} 32 & 28 \\ 50 & -46 \end{array}\right] \)
So, we have \( 2X = \left[\begin{array}{cc} 32 & 28 \\ 50 & -46 \end{array}\right] \).
Finally, divide each element by 2 to find X:
\( X = \frac{1}{2} \left[\begin{array}{cc} 32 & 28 \\ 50 & -46 \end{array}\right] \)
\( X = \left[\begin{array}{cc} \frac{32}{2} & \frac{28}{2} \\ \frac{50}{2} & \frac{-46}{2} \end{array}\right] \)
\( X = \left[\begin{array}{cc} 16 & 14 \\ 25 & -23 \end{array}\right] \)
In simple words: We had an equation with matrices, and we needed to find matrix X. We first moved the terms around so that 2X was alone on one side. Then, we calculated 3 times matrix A and 5 times matrix B. We added these two results together. Finally, to get X, we divided every number in the resulting matrix by 2.

🎯 Exam Tip: Remember that matrix addition, subtraction, and scalar multiplication are performed element-wise. Always write down each step clearly to avoid calculation errors, especially with negative numbers.

Question 23.
(i) Find x and y if \( \left[\begin{array}{ll} x & 3 x \\ y & 4 y \end{array}\right]\left[\begin{array}{l} 2 \\ 1 \end{array}\right]=\left[\begin{array}{c} 5 \\ 12 \end{array}\right] \).
(ii) If \( A = \left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] \), \( B = \left[\begin{array}{ll} 2 & 1 \\ 4 & 2 \end{array}\right] \), \( C = \left[\begin{array}{ll} 5 & 1 \\ 7 & 4 \end{array}\right] \). Compute A (B + C) and (B + C) A.
Answer:
(i) We are given the matrix equation:
\( \left[\begin{array}{ll} x & 3 x \\ y & 4 y \end{array}\right]\left[\begin{array}{l} 2 \\ 1 \end{array}\right]=\left[\begin{array}{c} 5 \\ 12 \end{array}\right] \)
Perform the matrix multiplication on the left side:
\( \left[\begin{array}{c} (x \times 2) + (3x \times 1) \\ (y \times 2) + (4y \times 1) \end{array}\right]=\left[\begin{array}{c} 5 \\ 12 \end{array}\right] \)
\( \implies \left[\begin{array}{c} 2x + 3x \\ 2y + 4y \end{array}\right]=\left[\begin{array}{c} 5 \\ 12 \end{array}\right] \)
\( \implies \left[\begin{array}{c} 5x \\ 6y \end{array}\right]=\left[\begin{array}{c} 5 \\ 12 \end{array}\right] \)
Now, compare the corresponding elements:
From the first row: \( 5x = 5 \)
\( \implies x = \frac{5}{5} \)
\( \implies x = 1 \)
From the second row: \( 6y = 12 \)
\( \implies y = \frac{12}{6} \)
\( \implies y = 2 \)
So, the values are \( x = 1 \) and \( y = 2 \).

(ii) Given matrices:
\( A = \left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] \)
\( B = \left[\begin{array}{ll} 2 & 1 \\ 4 & 2 \end{array}\right] \)
\( C = \left[\begin{array}{ll} 5 & 1 \\ 7 & 4 \end{array}\right] \)

**Part 1: Compute A (B + C)**
First, find \( B + C \):
\( B + C = \left[\begin{array}{ll} 2 & 1 \\ 4 & 2 \end{array}\right] + \left[\begin{array}{ll} 5 & 1 \\ 7 & 4 \end{array}\right] \)
\( B + C = \left[\begin{array}{ll} 2+5 & 1+1 \\ 4+7 & 2+4 \end{array}\right] \)
\( B + C = \left[\begin{array}{cc} 7 & 2 \\ 11 & 6 \end{array}\right] \)
Next, compute \( A(B + C) \):
\( A(B + C) = \left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] \left[\begin{array}{cc} 7 & 2 \\ 11 & 6 \end{array}\right] \)
\( A(B + C) = \left[\begin{array}{ll} (1 \times 7) + (2 \times 11) & (1 \times 2) + (2 \times 6) \\ (3 \times 7) + (4 \times 11) & (3 \times 2) + (4 \times 6) \end{array}\right] \)
\( A(B + C) = \left[\begin{array}{ll} 7 + 22 & 2 + 12 \\ 21 + 44 & 6 + 24 \end{array}\right] \)
\( A(B + C) = \left[\begin{array}{cc} 29 & 14 \\ 65 & 30 \end{array}\right] \)

**Part 2: Compute (B + C) A**
We already found \( B + C = \left[\begin{array}{cc} 7 & 2 \\ 11 & 6 \end{array}\right] \).
Now, compute \( (B + C)A \):
\( (B + C)A = \left[\begin{array}{cc} 7 & 2 \\ 11 & 6 \end{array}\right] \left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] \)
\( (B + C)A = \left[\begin{array}{ll} (7 \times 1) + (2 \times 3) & (7 \times 2) + (2 \times 4) \\ (11 \times 1) + (6 \times 3) & (11 \times 2) + (6 \times 4) \end{array}\right] \)
\( (B + C)A = \left[\begin{array}{ll} 7 + 6 & 14 + 8 \\ 11 + 18 & 22 + 24 \end{array}\right] \)
\( (B + C)A = \left[\begin{array}{cc} 13 & 22 \\ 29 & 46 \end{array}\right] \)
From our calculations, we can see that \( A(B+C) \neq (B+C)A \). This further demonstrates that matrix multiplication is generally not commutative. Matrix addition, however, is commutative. Also, the distributive property holds: A(B+C) = AB + AC.
In simple words: For the first part, we multiplied a matrix by a column of variables to get another column. By matching the numbers, we found 'x' and 'y'. For the second part, we were asked to do two different calculations. First, we added matrices B and C, and then multiplied the result by A from the left. Second, we added B and C again, but this time multiplied the result by A from the right. We found that the answers were different, which shows that the order of matrix multiplication is very important.

🎯 Exam Tip: Remember the order of operations: parentheses first. For matrix multiplication, always be careful about which matrix is on the left and which is on the right, as \( XY \) is generally not the same as \( YX \).

Question 24.
(i) If \( \left[\begin{array}{cc} x-2 & 5 \\ 3 & 3 \end{array}\right]=\left[\begin{array}{ll} 4 & 2 \\ y & 5 \end{array}\right] + \left[\begin{array}{cc} -4 & 3 \\ -1 & -2 \end{array}\right] \), find the values of x and y.
(ii) If \( A = \left[\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right] \), \( B = \left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right] \), \( C = \left[\begin{array}{ll} 1 & 3 \\ 3 & 1 \end{array}\right] \). Find the matrix C (B – A).
Answer:
(i) We have the equation: \( \left[\begin{array}{cc} x-2 & 5 \\ 3 & 3 \end{array}\right]=\left[\begin{array}{ll} 4 & 2 \\ y & 5 \end{array}\right] + \left[\begin{array}{cc} -4 & 3 \\ -1 & -2 \end{array}\right] \)
First, add the matrices on the right side:
\( \left[\begin{array}{ll} 4 & 2 \\ y & 5 \end{array}\right] + \left[\begin{array}{cc} -4 & 3 \\ -1 & -2 \end{array}\right] = \left[\begin{array}{cc} 4+(-4) & 2+3 \\ y+(-1) & 5+(-2) \end{array}\right] \)
\( = \left[\begin{array}{cc} 0 & 5 \\ y-1 & 3 \end{array}\right] \)
Now, compare this with the matrix on the left side:
\( \left[\begin{array}{cc} x-2 & 5 \\ 3 & 3 \end{array}\right] = \left[\begin{array}{cc} 0 & 5 \\ y-1 & 3 \end{array}\right] \)
By comparing the corresponding elements:
\( x-2 = 0 \implies x = 2 \)
\( 3 = y-1 \implies y = 3+1 \implies y = 4 \)
So, \( x = 2 \) and \( y = 4 \).
(ii) Given matrices:
\( A = \left[\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right] \), \( B = \left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right] \), \( C = \left[\begin{array}{ll} 1 & 3 \\ 3 & 1 \end{array}\right] \)
First, find \( B-A \):
\( B-A = \left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right] - \left[\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right] \)
\( = \left[\begin{array}{cc} 2-1 & 1-2 \\ 3-2 & 2-3 \end{array}\right] = \left[\begin{array}{cc} 1 & -1 \\ 1 & -1 \end{array}\right] \)
Next, find \( C(B-A) \):
\( C(B-A) = \left[\begin{array}{ll} 1 & 3 \\ 3 & 1 \end{array}\right] \left[\begin{array}{cc} 1 & -1 \\ 1 & -1 \end{array}\right] \)
\( = \left[\begin{array}{cc} 1 \times 1 + 3 \times 1 & 1 \times (-1) + 3 \times (-1) \\ 3 \times 1 + 1 \times 1 & 3 \times (-1) + 1 \times (-1) \end{array}\right] \)
\( = \left[\begin{array}{cc} 1+3 & -1-3 \\ 3+1 & -3-1 \end{array}\right] = \left[\begin{array}{cc} 4 & -4 \\ 4 & -4 \end{array}\right] \)
In simple words: For part (i), we combined the two matrices on the right side by adding their matching numbers. Then we made the numbers in the first matrix equal to the numbers in the new matrix to find x and y. For part (ii), we first subtracted matrix A from matrix B, and then multiplied the result by matrix C. Remember, matrix multiplication involves multiplying rows by columns.

🎯 Exam Tip: Always perform operations within parentheses first, just like in regular algebra. For matrix equations, ensure all corresponding elements are equated correctly to solve for unknowns.

Question 25. If \( A = \left[\begin{array}{ccc} 1 & -2 & 1 \\ 2 & 1 & 3 \end{array}\right] \) and \( B = \left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{array}\right] \), write down the matrix AB.
Would it be possible to find the product BA ? If so, compute it, and if not, give reasons.
Answer:
Given matrices:
\( A = \left[\begin{array}{ccc} 1 & -2 & 1 \\ 2 & 1 & 3 \end{array}\right] \). The order of matrix A is \( 2 \times 3 \).
\( B = \left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{array}\right] \). The order of matrix B is \( 3 \times 2 \).
To find the product AB, the number of columns in A (3) must be equal to the number of rows in B (3). Since they are equal, the product AB is possible.
The order of the resulting matrix AB will be \( 2 \times 2 \).
\( AB = \left[\begin{array}{ccc} 1 & -2 & 1 \\ 2 & 1 & 3 \end{array}\right] \left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{array}\right] \)
\( = \left[\begin{array}{cc} 1 \times 2 + (-2) \times 3 + 1 \times 1 & 1 \times 1 + (-2) \times 2 + 1 \times 1 \\ 2 \times 2 + 1 \times 3 + 3 \times 1 & 2 \times 1 + 1 \times 2 + 3 \times 1 \end{array}\right] \)
\( = \left[\begin{array}{cc} 2 - 6 + 1 & 1 - 4 + 1 \\ 4 + 3 + 3 & 2 + 2 + 3 \end{array}\right] \)
\( = \left[\begin{array}{cc} -3 & -2 \\ 10 & 7 \end{array}\right] \)
Now, let's check if the product BA is possible.
For BA, the number of columns in B (2) must be equal to the number of rows in A (2). Since they are equal, the product BA is possible.
The order of the resulting matrix BA will be \( 3 \times 3 \).
\( BA = \left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{array}\right] \left[\begin{array}{ccc} 1 & -2 & 1 \\ 2 & 1 & 3 \end{array}\right] \)
\( = \left[\begin{array}{ccc} 2 \times 1 + 1 \times 2 & 2 \times (-2) + 1 \times 1 & 2 \times 1 + 1 \times 3 \\ 3 \times 1 + 2 \times 2 & 3 \times (-2) + 2 \times 1 & 3 \times 1 + 2 \times 3 \\ 1 \times 1 + 1 \times 2 & 1 \times (-2) + 1 \times 1 & 1 \times 1 + 1 \times 3 \end{array}\right] \)
\( = \left[\begin{array}{ccc} 2+2 & -4+1 & 2+3 \\ 3+4 & -6+2 & 3+6 \\ 1+2 & -2+1 & 1+3 \end{array}\right] \)
\( = \left[\begin{array}{ccc} 4 & -3 & 5 \\ 7 & -4 & 9 \\ 3 & -1 & 4 \end{array}\right] \)
In simple words: We can multiply two matrices only if the number of columns in the first matrix is the same as the number of rows in the second matrix. First, we found matrix AB by checking if their sizes allowed it, which they did. Then, we checked for matrix BA, and its sizes also allowed multiplication. Finally, we calculated BA.

🎯 Exam Tip: Always check the dimensions of matrices (rows x columns) before attempting multiplication. The product of an \( m \times n \) matrix and an \( n \times p \) matrix will be an \( m \times p \) matrix. If the inner dimensions don't match, multiplication is not possible.

Question 1. Evaluate x, y if
\( \left[\begin{array}{cc} 3 & -2 \\ -1 & 4 \end{array}\right]\left[\begin{array}{c} 2 x \\ 1 \end{array}\right]+2\left[\begin{array}{c} -4 \\ 5 \end{array}\right]=\left[\begin{array}{c} 8 \\ 4 y \end{array}\right] \)
Answer:
Given the matrix equation:
\( \left[\begin{array}{cc} 3 & -2 \\ -1 & 4 \end{array}\right]\left[\begin{array}{c} 2 x \\ 1 \end{array}\right]+2\left[\begin{array}{c} -4 \\ 5 \end{array}\right]=\left[\begin{array}{c} 8 \\ 4 y \end{array}\right] \)
First, perform the matrix multiplication:
\( \left[\begin{array}{cc} 3 & -2 \\ -1 & 4 \end{array}\right]\left[\begin{array}{c} 2 x \\ 1 \end{array}\right] = \left[\begin{array}{c} 3 \times (2x) + (-2) \times 1 \\ (-1) \times (2x) + 4 \times 1 \end{array}\right] = \left[\begin{array}{c} 6x - 2 \\ -2x + 4 \end{array}\right] \)
Next, perform the scalar multiplication:
\( 2\left[\begin{array}{c} -4 \\ 5 \end{array}\right] = \left[\begin{array}{c} 2 \times (-4) \\ 2 \times 5 \end{array}\right] = \left[\begin{array}{c} -8 \\ 10 \end{array}\right] \)
Substitute these back into the original equation:
\( \left[\begin{array}{c} 6x - 2 \\ -2x + 4 \end{array}\right] + \left[\begin{array}{c} -8 \\ 10 \end{array}\right] = \left[\begin{array}{c} 8 \\ 4 y \end{array}\right] \)
Add the matrices on the left side:
\( \left[\begin{array}{c} (6x - 2) + (-8) \\ (-2x + 4) + 10 \end{array}\right] = \left[\begin{array}{c} 8 \\ 4 y \end{array}\right] \)
\( \left[\begin{array}{c} 6x - 10 \\ -2x + 14 \end{array}\right] = \left[\begin{array}{c} 8 \\ 4 y \end{array}\right] \)
Now, compare the corresponding elements:
For the first elements: \( 6x - 10 = 8 \)
\( 6x = 8 + 10 \)
\( 6x = 18 \)
\( x = \frac{18}{6} \implies x = 3 \)
For the second elements: \( -2x + 14 = 4y \)
Substitute \( x=3 \) into this equation:
\( -2(3) + 14 = 4y \)
\( -6 + 14 = 4y \)
\( 8 = 4y \)
\( y = \frac{8}{4} \implies y = 2 \)
Therefore, \( x = 3 \) and \( y = 2 \).
In simple words: We solved the matrix puzzle by first multiplying the matrices and then adding them together. After that, we matched the numbers in the matrices on both sides of the equals sign to find the values of x and y.

🎯 Exam Tip: Remember the order of operations for matrices (multiplication before addition). Always compare corresponding elements to form equations and solve for variables.

Question 2. Evaluate :
\( \left[\begin{array}{cc} 2 \cos 60^{\circ} & -2 \sin 30^{\circ} \\ -\tan 45^{\circ} & \cos 0^{\circ} \end{array}\right]\left[\begin{array}{cc} \cot 45^{\circ} & \operatorname{cosec} 30^{\circ} \\ \sec 60^{\circ} & \sin 90^{\circ} \end{array}\right] \)
Answer:
First, find the values of the trigonometric functions:
\( \cos 60^{\circ} = \frac{1}{2} \)
\( \sin 30^{\circ} = \frac{1}{2} \)
\( \tan 45^{\circ} = 1 \)
\( \cos 0^{\circ} = 1 \)
\( \cot 45^{\circ} = 1 \)
\( \operatorname{cosec} 30^{\circ} = \frac{1}{\sin 30^{\circ}} = \frac{1}{1/2} = 2 \)
\( \sec 60^{\circ} = \frac{1}{\cos 60^{\circ}} = \frac{1}{1/2} = 2 \)
\( \sin 90^{\circ} = 1 \)
Substitute these values into the matrices:
First matrix: \( \left[\begin{array}{cc} 2 \times \frac{1}{2} & -2 \times \frac{1}{2} \\ -1 & 1 \end{array}\right] = \left[\begin{array}{cc} 1 & -1 \\ -1 & 1 \end{array}\right] \)
Second matrix: \( \left[\begin{array}{cc} 1 & 2 \\ 2 & 1 \end{array}\right] \)
Now, perform the matrix multiplication:
\( \left[\begin{array}{cc} 1 & -1 \\ -1 & 1 \end{array}\right] \left[\begin{array}{cc} 1 & 2 \\ 2 & 1 \end{array}\right] \)
\( = \left[\begin{array}{cc} 1 \times 1 + (-1) \times 2 & 1 \times 2 + (-1) \times 1 \\ (-1) \times 1 + 1 \times 2 & (-1) \times 2 + 1 \times 1 \end{array}\right] \)
\( = \left[\begin{array}{cc} 1 - 2 & 2 - 1 \\ -1 + 2 & -2 + 1 \end{array}\right] \)
\( = \left[\begin{array}{cc} -1 & 1 \\ 1 & -1 \end{array}\right] \)
In simple words: First, we looked up all the special angle values for sine, cosine, tangent, and their partners. Then we put those numbers into the matrices and multiplied them. Matrix multiplication means taking rows from the first matrix and multiplying them by columns from the second matrix.

🎯 Exam Tip: Memorize the common trigonometric values for special angles (0°, 30°, 45°, 60°, 90°) as they are frequently used in matrix problems. Double-check your calculations in matrix multiplication to avoid errors.

Question 3. Find the value of x and y; if
\( \left[\begin{array}{ll} 1 & 2 \\ 3 & 3 \end{array}\right]\left[\begin{array}{ll} x & 0 \\ 0 & y \end{array}\right]=\left[\begin{array}{ll} x & 0 \\ 9 & 0 \end{array}\right] \)
Answer:
Given the matrix equation:
\( \left[\begin{array}{ll} 1 & 2 \\ 3 & 3 \end{array}\right]\left[\begin{array}{ll} x & 0 \\ 0 & y \end{array}\right]=\left[\begin{array}{ll} x & 0 \\ 9 & 0 \end{array}\right] \)
First, perform the matrix multiplication on the left side:
\( \left[\begin{array}{ll} 1 & 2 \\ 3 & 3 \end{array}\right]\left[\begin{array}{ll} x & 0 \\ 0 & y \end{array}\right] = \left[\begin{array}{ll} 1 \times x + 2 \times 0 & 1 \times 0 + 2 \times y \\ 3 \times x + 3 \times 0 & 3 \times 0 + 3 \times y \end{array}\right] \)
\( = \left[\begin{array}{cc} x + 0 & 0 + 2y \\ 3x + 0 & 0 + 3y \end{array}\right] = \left[\begin{array}{cc} x & 2y \\ 3x & 3y \end{array}\right] \)
Now, equate this result to the matrix on the right side of the given equation:
\( \left[\begin{array}{cc} x & 2y \\ 3x & 3y \end{array}\right] = \left[\begin{array}{ll} x & 0 \\ 9 & 0 \end{array}\right] \)
By comparing the corresponding elements:
From the top right element: \( 2y = 0 \implies y = 0 \)
From the bottom left element: \( 3x = 9 \implies x = \frac{9}{3} \implies x = 3 \)
The other elements \( x = x \) and \( 3y = 0 \) are consistent with these values.
So, \( x = 3 \) and \( y = 0 \).
In simple words: We multiplied the two matrices on the left side. Then, we made the numbers in the resulting matrix equal to the numbers in the matrix on the right side. This helped us find that x is 3 and y is 0.

🎯 Exam Tip: When comparing matrices, ensure that each element in the first matrix corresponds to the element in the exact same position in the second matrix. This is crucial for correctly setting up the equations.

Question 4. Find matrix X which satisfies the equation:
\( \left[\begin{array}{ll} 3 & 7 \\ 2 & 4 \end{array}\right]\left[\begin{array}{ll} 0 & 2 \\ 5 & 3 \end{array}\right]+2 X=\left[\begin{array}{cc} 1 & -5 \\ -4 & 6 \end{array}\right] \)
Answer:
Given the matrix equation:
\( \left[\begin{array}{ll} 3 & 7 \\ 2 & 4 \end{array}\right]\left[\begin{array}{ll} 0 & 2 \\ 5 & 3 \end{array}\right]+2 X=\left[\begin{array}{cc} 1 & -5 \\ -4 & 6 \end{array}\right] \)
First, perform the matrix multiplication on the left side:
\( \left[\begin{array}{ll} 3 & 7 \\ 2 & 4 \end{array}\right]\left[\begin{array}{ll} 0 & 2 \\ 5 & 3 \end{array}\right] = \left[\begin{array}{ll} 3 \times 0 + 7 \times 5 & 3 \times 2 + 7 \times 3 \\ 2 \times 0 + 4 \times 5 & 2 \times 2 + 4 \times 3 \end{array}\right] \)
\( = \left[\begin{array}{ll} 0 + 35 & 6 + 21 \\ 0 + 20 & 4 + 12 \end{array}\right] = \left[\begin{array}{ll} 35 & 27 \\ 20 & 16 \end{array}\right] \)
Now, substitute this result back into the original equation:
\( \left[\begin{array}{ll} 35 & 27 \\ 20 & 16 \end{array}\right] + 2X = \left[\begin{array}{cc} 1 & -5 \\ -4 & 6 \end{array}\right] \)
To find 2X, subtract the first matrix from both sides:
\( 2X = \left[\begin{array}{cc} 1 & -5 \\ -4 & 6 \end{array}\right] - \left[\begin{array}{ll} 35 & 27 \\ 20 & 16 \end{array}\right] \)
\( 2X = \left[\begin{array}{cc} 1 - 35 & -5 - 27 \\ -4 - 20 & 6 - 16 \end{array}\right] \)
\( 2X = \left[\begin{array}{cc} -34 & -32 \\ -24 & -10 \end{array}\right] \)
To find X, multiply the matrix by \( \frac{1}{2} \):
\( X = \frac{1}{2} \left[\begin{array}{cc} -34 & -32 \\ -24 & -10 \end{array}\right] \)
\( X = \left[\begin{array}{cc} \frac{-34}{2} & \frac{-32}{2} \\ \frac{-24}{2} & \frac{-10}{2} \end{array}\right] \)
\( X = \left[\begin{array}{cc} -17 & -16 \\ -12 & -5 \end{array}\right] \)
In simple words: First, we multiplied the two matrices on the left side. Then, we moved this result to the other side of the equation by subtracting it. This left us with 2X on one side. Finally, we divided every number in that matrix by 2 to find matrix X.

🎯 Exam Tip: When solving matrix equations, treat them like algebraic equations. Perform matrix addition, subtraction, and scalar multiplication carefully, element by element, to avoid calculation mistakes.

Question 5. Given \( A = \left[\begin{array}{ll} 1 & 1 \\ 8 & 3 \end{array}\right] \), evaluate \( A^2 - 4A \).
Answer:
Given matrix \( A = \left[\begin{array}{ll} 1 & 1 \\ 8 & 3 \end{array}\right] \).
First, calculate \( A^2 = A \times A \):
\( A^2 = \left[\begin{array}{ll} 1 & 1 \\ 8 & 3 \end{array}\right] \left[\begin{array}{ll} 1 & 1 \\ 8 & 3 \end{array}\right] \)
\( = \left[\begin{array}{ll} 1 \times 1 + 1 \times 8 & 1 \times 1 + 1 \times 3 \\ 8 \times 1 + 3 \times 8 & 8 \times 1 + 3 \times 3 \end{array}\right] \)
\( = \left[\begin{array}{ll} 1 + 8 & 1 + 3 \\ 8 + 24 & 8 + 9 \end{array}\right] \)
\( = \left[\begin{array}{cc} 9 & 4 \\ 32 & 17 \end{array}\right] \)
Next, calculate \( 4A \):
\( 4A = 4 \left[\begin{array}{ll} 1 & 1 \\ 8 & 3 \end{array}\right] \)
\( = \left[\begin{array}{ll} 4 \times 1 & 4 \times 1 \\ 4 \times 8 & 4 \times 3 \end{array}\right] \)
\( = \left[\begin{array}{cc} 4 & 4 \\ 32 & 12 \end{array}\right] \)
Finally, evaluate \( A^2 - 4A \):
\( A^2 - 4A = \left[\begin{array}{cc} 9 & 4 \\ 32 & 17 \end{array}\right] - \left[\begin{array}{cc} 4 & 4 \\ 32 & 12 \end{array}\right] \)
\( = \left[\begin{array}{cc} 9 - 4 & 4 - 4 \\ 32 - 32 & 17 - 12 \end{array}\right] \)
\( = \left[\begin{array}{cc} 5 & 0 \\ 0 & 5 \end{array}\right] \)
In simple words: First, we multiplied matrix A by itself to get \(A^2\). Then, we multiplied matrix A by 4. Finally, we subtracted the second result from the first one to get our answer. Notice that the final matrix has 5s on the main diagonal and zeros elsewhere.

🎯 Exam Tip: Remember that \( A^2 \) means \( A \times A \), not squaring each element. Scalar multiplication (like 4A) involves multiplying every element of the matrix by the scalar value.

Question 6. Find x and y, if
\( \left[\begin{array}{cc} -3 & 2 \\ 0 & -5 \end{array}\right]\left[\begin{array}{l} x \\ 2 \end{array}\right]=\left[\begin{array}{c} -5 \\ y \end{array}\right] \)
Answer:
Given the matrix equation:
\( \left[\begin{array}{cc} -3 & 2 \\ 0 & -5 \end{array}\right]\left[\begin{array}{l} x \\ 2 \end{array}\right]=\left[\begin{array}{c} -5 \\ y \end{array}\right] \)
Perform the matrix multiplication on the left side:
\( \left[\begin{array}{cc} -3 & 2 \\ 0 & -5 \end{array}\right]\left[\begin{array}{l} x \\ 2 \end{array}\right] = \left[\begin{array}{c} (-3) \times x + 2 \times 2 \\ 0 \times x + (-5) \times 2 \end{array}\right] \)
\( = \left[\begin{array}{c} -3x + 4 \\ 0 - 10 \end{array}\right] = \left[\begin{array}{c} -3x + 4 \\ -10 \end{array}\right] \)
Now, equate this result to the matrix on the right side of the given equation:
\( \left[\begin{array}{c} -3x + 4 \\ -10 \end{array}\right] = \left[\begin{array}{c} -5 \\ y \end{array}\right] \)
By comparing the corresponding elements:
For the first elements: \( -3x + 4 = -5 \)
\( -3x = -5 - 4 \)
\( -3x = -9 \)
\( x = \frac{-9}{-3} \implies x = 3 \)
For the second elements: \( -10 = y \implies y = -10 \)
So, \( x = 3 \) and \( y = -10 \).
In simple words: We multiplied the two matrices on the left side. Then, we matched the numbers in the resulting matrix with the numbers in the matrix on the right side to find that x is 3 and y is -10.

🎯 Exam Tip: Pay close attention to negative signs during multiplication and addition. A single sign error can change the entire result for x and y.

Question 7. Find x and y, if \( \left[\begin{array}{ll} x & 3 x \\ y & 4 y \end{array}\right]\left[\begin{array}{l} 2 \\ 1 \end{array}\right]=\left[\begin{array}{c} 5 \\ 12 \end{array}\right] \).
Answer:
Given the matrix equation:
\( \left[\begin{array}{ll} x & 3 x \\ y & 4 y \end{array}\right]\left[\begin{array}{l} 2 \\ 1 \end{array}\right]=\left[\begin{array}{c} 5 \\ 12 \end{array}\right] \)
Perform the matrix multiplication on the left side:
\( \left[\begin{array}{ll} x & 3 x \\ y & 4 y \end{array}\right]\left[\begin{array}{l} 2 \\ 1 \end{array}\right] = \left[\begin{array}{c} x \times 2 + 3x \times 1 \\ y \times 2 + 4y \times 1 \end{array}\right] \)
\( = \left[\begin{array}{c} 2x + 3x \\ 2y + 4y \end{array}\right] = \left[\begin{array}{c} 5x \\ 6y \end{array}\right] \)
Now, equate this result to the matrix on the right side of the given equation:
\( \left[\begin{array}{c} 5x \\ 6y \end{array}\right] = \left[\begin{array}{c} 5 \\ 12 \end{array}\right] \)
By comparing the corresponding elements:
For the first elements: \( 5x = 5 \implies x = \frac{5}{5} \implies x = 1 \)
For the second elements: \( 6y = 12 \implies y = \frac{12}{6} \implies y = 2 \)
So, \( x = 1 \) and \( y = 2 \).
In simple words: We multiplied the two matrices together. Then, we set the numbers in the resulting matrix equal to the numbers in the matrix on the right. This helped us figure out that x is 1 and y is 2.

🎯 Exam Tip: When elements involve variables, combine them carefully during multiplication before comparing to the other side. Simple algebraic errors can lead to incorrect final answers.

Question 8. Find x and y, if:
\( \left[\begin{array}{cc} 3 & -2 \\ -1 & 4 \end{array}\right]\left[\begin{array}{c} 2 x \\ 1 \end{array}\right]+2\left[\begin{array}{c} -4 \\ 5 \end{array}\right]=4\left[\begin{array}{l} 2 \\ y \end{array}\right] \)
Answer:
Given the matrix equation:
\( \left[\begin{array}{cc} 3 & -2 \\ -1 & 4 \end{array}\right]\left[\begin{array}{c} 2 x \\ 1 \end{array}\right]+2\left[\begin{array}{c} -4 \\ 5 \end{array}\right]=4\left[\begin{array}{l} 2 \\ y \end{array}\right] \)
First, perform the matrix multiplication:
\( \left[\begin{array}{cc} 3 & -2 \\ -1 & 4 \end{array}\right]\left[\begin{array}{c} 2 x \\ 1 \end{array}\right] = \left[\begin{array}{c} 3 \times (2x) + (-2) \times 1 \\ (-1) \times (2x) + 4 \times 1 \end{array}\right] = \left[\begin{array}{c} 6x - 2 \\ -2x + 4 \end{array}\right] \)
Next, perform the scalar multiplications:
\( 2\left[\begin{array}{c} -4 \\ 5 \end{array}\right] = \left[\begin{array}{c} -8 \\ 10 \end{array}\right] \)
\( 4\left[\begin{array}{l} 2 \\ y \end{array}\right] = \left[\begin{array}{c} 8 \\ 4y \end{array}\right] \)
Substitute these back into the original equation:
\( \left[\begin{array}{c} 6x - 2 \\ -2x + 4 \end{array}\right] + \left[\begin{array}{c} -8 \\ 10 \end{array}\right] = \left[\begin{array}{c} 8 \\ 4y \end{array}\right] \)
Add the matrices on the left side:
\( \left[\begin{array}{c} (6x - 2) + (-8) \\ (-2x + 4) + 10 \end{array}\right] = \left[\begin{array}{c} 8 \\ 4y \end{array}\right] \)
\( \left[\begin{array}{c} 6x - 10 \\ -2x + 14 \end{array}\right] = \left[\begin{array}{c} 8 \\ 4y \end{array}\right] \)
Now, compare the corresponding elements:
For the first elements: \( 6x - 10 = 8 \)
\( 6x = 8 + 10 \)
\( 6x = 18 \)
\( x = \frac{18}{6} \implies x = 3 \)
For the second elements: \( -2x + 14 = 4y \)
Substitute \( x=3 \) into this equation:
\( -2(3) + 14 = 4y \)
\( -6 + 14 = 4y \)
\( 8 = 4y \)
\( y = \frac{8}{4} \implies y = 2 \)
So, \( x = 3 \) and \( y = 2 \).
In simple words: We solved this matrix puzzle by doing all the multiplications and additions on both sides first. Then, we compared the numbers in the same spots of the matrices to make equations. Solving these equations gave us the values for x and y.

🎯 Exam Tip: It is good practice to perform all multiplications before additions or subtractions, following the standard order of operations, to simplify the matrix equation efficiently.

Question 9. Given \( A = \left[\begin{array}{cc} 2 & -1 \\ 2 & 0 \end{array}\right] \), \( B = \left[\begin{array}{cc} -3 & 2 \\ 4 & 0 \end{array}\right] \) and \( C = \left[\begin{array}{ll} 1 & 0 \\ 0 & 2 \end{array}\right] \), find the matrix X, such that \( A + X = 2B + C \).
Answer:
Given matrices:
\( A = \left[\begin{array}{cc} 2 & -1 \\ 2 & 0 \end{array}\right] \), \( B = \left[\begin{array}{cc} -3 & 2 \\ 4 & 0 \end{array}\right] \), \( C = \left[\begin{array}{ll} 1 & 0 \\ 0 & 2 \end{array}\right] \)
We need to find matrix X such that \( A + X = 2B + C \).
Rearrange the equation to solve for X:
\( X = 2B + C - A \)
First, calculate \( 2B \):
\( 2B = 2 \left[\begin{array}{cc} -3 & 2 \\ 4 & 0 \end{array}\right] = \left[\begin{array}{cc} 2 \times (-3) & 2 \times 2 \\ 2 \times 4 & 2 \times 0 \end{array}\right] = \left[\begin{array}{cc} -6 & 4 \\ 8 & 0 \end{array}\right] \)
Next, calculate \( 2B + C \):
\( 2B + C = \left[\begin{array}{cc} -6 & 4 \\ 8 & 0 \end{array}\right] + \left[\begin{array}{ll} 1 & 0 \\ 0 & 2 \end{array}\right] = \left[\begin{array}{cc} -6+1 & 4+0 \\ 8+0 & 0+2 \end{array}\right] = \left[\begin{array}{cc} -5 & 4 \\ 8 & 2 \end{array}\right] \)
Finally, calculate \( X = (2B + C) - A \):
\( X = \left[\begin{array}{cc} -5 & 4 \\ 8 & 2 \end{array}\right] - \left[\begin{array}{cc} 2 & -1 \\ 2 & 0 \end{array}\right] \)
\( X = \left[\begin{array}{cc} -5-2 & 4-(-1) \\ 8-2 & 2-0 \end{array}\right] \)
\( X = \left[\begin{array}{cc} -7 & 4+1 \\ 6 & 2 \end{array}\right] \)
\( X = \left[\begin{array}{cc} -7 & 5 \\ 6 & 2 \end{array}\right] \)
In simple words: We wanted to find matrix X. First, we multiplied matrix B by 2. Then, we added this result to matrix C. Finally, we subtracted matrix A from that sum to get our matrix X. This way, we found the missing matrix.

🎯 Exam Tip: When rearranging matrix equations, treat matrices like single terms. Remember that matrix addition and subtraction are performed element-wise, and scalar multiplication multiplies every element in the matrix.

Question 10. Find the value of x given that
\( A = \left[\begin{array}{cc} 2 & 12 \\ 0 & 1 \end{array}\right] \), \( B=\left[\begin{array}{ll} 4 & x \\ 0 & 1 \end{array}\right] \) and \( A^2 = B \).
Answer:
Given matrices:
\( A = \left[\begin{array}{cc} 2 & 12 \\ 0 & 1 \end{array}\right] \)
\( B = \left[\begin{array}{ll} 4 & x \\ 0 & 1 \end{array}\right] \)
We are given that \( A^2 = B \).
First, calculate \( A^2 = A \times A \):
\( A^2 = \left[\begin{array}{cc} 2 & 12 \\ 0 & 1 \end{array}\right] \left[\begin{array}{cc} 2 & 12 \\ 0 & 1 \end{array}\right] \)
\( = \left[\begin{array}{cc} 2 \times 2 + 12 \times 0 & 2 \times 12 + 12 \times 1 \\ 0 \times 2 + 1 \times 0 & 0 \times 12 + 1 \times 1 \end{array}\right] \)
\( = \left[\begin{array}{cc} 4 + 0 & 24 + 12 \\ 0 + 0 & 0 + 1 \end{array}\right] \)
\( = \left[\begin{array}{cc} 4 & 36 \\ 0 & 1 \end{array}\right] \)
Now, substitute \( A^2 \) into the given condition \( A^2 = B \):
\( \left[\begin{array}{cc} 4 & 36 \\ 0 & 1 \end{array}\right] = \left[\begin{array}{ll} 4 & x \\ 0 & 1 \end{array}\right] \)
By comparing the corresponding elements:
The element in the first row, second column gives: \( 36 = x \)
Thus, \( x = 36 \).
In simple words: We multiplied matrix A by itself to find \(A^2\). Then, since \(A^2\) is equal to matrix B, we looked at the matching numbers in both matrices. This helped us directly find the value of x.

🎯 Exam Tip: When an equation involves \( A^2 \), ensure you perform matrix multiplication \( A \times A \) carefully, not just squaring individual elements. Always compare corresponding elements accurately to find unknown variables.

Question 11. Let \( A = \left[\begin{array}{cc} 4 & -2 \\ 6 & -3 \end{array}\right] \), \( B=\left[\begin{array}{cc} 0 & 2 \\ 1 & -1 \end{array}\right] \) and \( C = \left[\begin{array}{c} -2 & 3 \\ 1 & -1 \end{array}\right] \). Find \( A^2 - A + BC \).
Answer:
Given matrices:
\( A = \left[\begin{array}{cc} 4 & -2 \\ 6 & -3 \end{array}\right] \), \( B = \left[\begin{array}{cc} 0 & 2 \\ 1 & -1 \end{array}\right] \), \( C = \left[\begin{array}{cc} -2 & 3 \\ 1 & -1 \end{array}\right] \)
First, calculate \( A^2 = A \times A \):
\( A^2 = \left[\begin{array}{cc} 4 & -2 \\ 6 & -3 \end{array}\right] \left[\begin{array}{cc} 4 & -2 \\ 6 & -3 \end{array}\right] \)
\( = \left[\begin{array}{cc} 4 \times 4 + (-2) \times 6 & 4 \times (-2) + (-2) \times (-3) \\ 6 \times 4 + (-3) \times 6 & 6 \times (-2) + (-3) \times (-3) \end{array}\right] \)
\( = \left[\begin{array}{cc} 16 - 12 & -8 + 6 \\ 24 - 18 & -12 + 9 \end{array}\right] \)
\( = \left[\begin{array}{cc} 4 & -2 \\ 6 & -3 \end{array}\right] \)
Next, calculate \( BC \):
\( BC = \left[\begin{array}{cc} 0 & 2 \\ 1 & -1 \end{array}\right] \left[\begin{array}{cc} -2 & 3 \\ 1 & -1 \end{array}\right] \)
\( = \left[\begin{array}{cc} 0 \times (-2) + 2 \times 1 & 0 \times 3 + 2 \times (-1) \\ 1 \times (-2) + (-1) \times 1 & 1 \times 3 + (-1) \times (-1) \end{array}\right] \)
\( = \left[\begin{array}{cc} 0 + 2 & 0 - 2 \\ -2 - 1 & 3 + 1 \end{array}\right] \)
\( = \left[\begin{array}{cc} 2 & -2 \\ -3 & 4 \end{array}\right] \)
Finally, calculate \( A^2 - A + BC \):
\( A^2 - A + BC = \left[\begin{array}{cc} 4 & -2 \\ 6 & -3 \end{array}\right] - \left[\begin{array}{cc} 4 & -2 \\ 6 & -3 \end{array}\right] + \left[\begin{array}{cc} 2 & -2 \\ -3 & 4 \end{array}\right] \)
\( = \left[\begin{array}{cc} 4-4+2 & -2-(-2)+(-2) \\ 6-6+(-3) & -3-(-3)+4 \end{array}\right] \)
\( = \left[\begin{array}{cc} 2 & -2 \\ -3 & 4 \end{array}\right] \)
In simple words: First, we multiplied matrix A by itself to get \(A^2\). Then, we multiplied matrix B by matrix C to get BC. Finally, we put all the pieces together by calculating \(A^2\) minus A, then adding BC to that result. It is interesting that \( A^2 \) turned out to be the same as A, which made the first part of the calculation simpler.

🎯 Exam Tip: This problem demonstrates a special case where \( A^2 = A \), meaning A is an idempotent matrix. Recognize these properties, but always show the full calculation steps. Be careful with double negative signs during subtraction.

Question 12.
(a) If \( 2\left[\begin{array}{ll} 3 & 4 \\ 5 & x \end{array}\right]+\left[\begin{array}{ll} 1 & y \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 7 & 0 \\ 10 & 5 \end{array}\right] \), find the values of x and y.
Answer:
Given the matrix equation:
\( 2\left[\begin{array}{ll} 3 & 4 \\ 5 & x \end{array}\right]+\left[\begin{array}{ll} 1 & y \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 7 & 0 \\ 10 & 5 \end{array}\right] \)
First, perform the scalar multiplication:
\( 2\left[\begin{array}{ll} 3 & 4 \\ 5 & x \end{array}\right] = \left[\begin{array}{cc} 2 \times 3 & 2 \times 4 \\ 2 \times 5 & 2 \times x \end{array}\right] = \left[\begin{array}{cc} 6 & 8 \\ 10 & 2x \end{array}\right] \)
Substitute this back into the equation:
\( \left[\begin{array}{cc} 6 & 8 \\ 10 & 2x \end{array}\right] + \left[\begin{array}{ll} 1 & y \\ 0 & 1 \end{array}\right] = \left[\begin{array}{cc} 7 & 0 \\ 10 & 5 \end{array}\right] \)
Add the matrices on the left side:
\( \left[\begin{array}{cc} 6+1 & 8+y \\ 10+0 & 2x+1 \end{array}\right] = \left[\begin{array}{cc} 7 & 0 \\ 10 & 5 \end{array}\right] \)
\( \left[\begin{array}{cc} 7 & 8+y \\ 10 & 2x+1 \end{array}\right] = \left[\begin{array}{cc} 7 & 0 \\ 10 & 5 \end{array}\right] \)
Now, compare the corresponding elements:
From the element in the first row, second column: \( 8+y = 0 \implies y = -8 \)
From the element in the second row, second column: \( 2x+1 = 5 \)
\( 2x = 5 - 1 \)
\( 2x = 4 \)
\( x = \frac{4}{2} \implies x = 2 \)
So, \( x = 2 \) and \( y = -8 \).
In simple words: We multiplied the first matrix by 2, then added it to the second matrix. After that, we matched the numbers in the resulting matrix with the numbers in the matrix on the right side. This gave us simple equations to solve for x and y.

🎯 Exam Tip: Always perform scalar multiplication on all elements of the matrix before proceeding with matrix addition or subtraction. Be precise with sign rules when solving for unknowns.

Question 13.
(a) If \( \left[\begin{array}{cc} 1 & 4 \\ -2 & 3 \end{array}\right] + 2M = 3\left[\begin{array}{cc} 3 & 2 \\ 0 & -3 \end{array}\right] \), find the matrix M.
(b) Given, \( \mathbf{A}=\left[\begin{array}{ll} p & 0 \\ 0 & 2 \end{array}\right] \), \( \mathbf{B}=\left[\begin{array}{cc} 0 & -q \\ 1 & 0 \end{array}\right] \), \( C = \left[\begin{array}{cc} 2 & -2 \\ 2 & 2 \end{array}\right] \) and \( BA = C^2 \). Find the values of p and q.
Answer:
(a) Given the matrix equation:
\( \left[\begin{array}{cc} 1 & 4 \\ -2 & 3 \end{array}\right] + 2M = 3\left[\begin{array}{cc} 3 & 2 \\ 0 & -3 \end{array}\right] \)
First, perform the scalar multiplication on the right side:
\( 3\left[\begin{array}{cc} 3 & 2 \\ 0 & -3 \end{array}\right] = \left[\begin{array}{cc} 3 \times 3 & 3 \times 2 \\ 3 \times 0 & 3 \times (-3) \end{array}\right] = \left[\begin{array}{cc} 9 & 6 \\ 0 & -9 \end{array}\right] \)
Substitute this back into the equation:
\( \left[\begin{array}{cc} 1 & 4 \\ -2 & 3 \end{array}\right] + 2M = \left[\begin{array}{cc} 9 & 6 \\ 0 & -9 \end{array}\right] \)
To find 2M, subtract the first matrix from both sides:
\( 2M = \left[\begin{array}{cc} 9 & 6 \\ 0 & -9 \end{array}\right] - \left[\begin{array}{cc} 1 & 4 \\ -2 & 3 \end{array}\right] \)
\( 2M = \left[\begin{array}{cc} 9-1 & 6-4 \\ 0-(-2) & -9-3 \end{array}\right] \)
\( 2M = \left[\begin{array}{cc} 8 & 2 \\ 2 & -12 \end{array}\right] \)
To find M, multiply the matrix by \( \frac{1}{2} \):
\( M = \frac{1}{2} \left[\begin{array}{cc} 8 & 2 \\ 2 & -12 \end{array}\right] \)
\( M = \left[\begin{array}{cc} \frac{8}{2} & \frac{2}{2} \\ \frac{2}{2} & \frac{-12}{2} \end{array}\right] \)
\( M = \left[\begin{array}{cc} 4 & 1 \\ 1 & -6 \end{array}\right] \)
(b) Given matrices:
\( \mathbf{A}=\left[\begin{array}{ll} p & 0 \\ 0 & 2 \end{array}\right] \), \( \mathbf{B}=\left[\begin{array}{cc} 0 & -q \\ 1 & 0 \end{array}\right] \), \( C = \left[\begin{array}{cc} 2 & -2 \\ 2 & 2 \end{array}\right] \)
We are given \( BA = C^2 \).
First, calculate \( BA \):
\( BA = \left[\begin{array}{cc} 0 & -q \\ 1 & 0 \end{array}\right] \left[\begin{array}{ll} p & 0 \\ 0 & 2 \end{array}\right] \)
\( = \left[\begin{array}{cc} 0 \times p + (-q) \times 0 & 0 \times 0 + (-q) \times 2 \\ 1 \times p + 0 \times 0 & 1 \times 0 + 0 \times 2 \end{array}\right] \)
\( = \left[\begin{array}{cc} 0 & -2q \\ p & 0 \end{array}\right] \)
Next, calculate \( C^2 = C \times C \):
\( C^2 = \left[\begin{array}{cc} 2 & -2 \\ 2 & 2 \end{array}\right] \left[\begin{array}{cc} 2 & -2 \\ 2 & 2 \end{array}\right] \)
\( = \left[\begin{array}{cc} 2 \times 2 + (-2) \times 2 & 2 \times (-2) + (-2) \times 2 \\ 2 \times 2 + 2 \times 2 & 2 \times (-2) + 2 \times 2 \end{array}\right] \)
\( = \left[\begin{array}{cc} 4 - 4 & -4 - 4 \\ 4 + 4 & -4 + 4 \end{array}\right] \)
\( = \left[\begin{array}{cc} 0 & -8 \\ 8 & 0 \end{array}\right] \)
Now, equate \( BA = C^2 \):
\( \left[\begin{array}{cc} 0 & -2q \\ p & 0 \end{array}\right] = \left[\begin{array}{cc} 0 & -8 \\ 8 & 0 \end{array}\right] \)
By comparing the corresponding elements:
From the element in the first row, second column: \( -2q = -8 \implies q = \frac{-8}{-2} \implies q = 4 \)
From the element in the second row, first column: \( p = 8 \)
So, \( p = 8 \) and \( q = 4 \).
In simple words: For part (a), we rearranged the matrix equation to find 2M, then divided by 2 to get M. For part (b), we multiplied matrix B by matrix A to get BA, and multiplied matrix C by itself to get \(C^2\). Then, we set BA equal to \(C^2\) and compared the numbers in each position to find the values of p and q.

🎯 Exam Tip: When solving for an unknown matrix, remember to isolate it using inverse operations. For matrix equations, verify the dimensions before performing any operation, especially multiplication. Ensure you correctly apply scalar multiplication and element-wise addition/subtraction.

Question 13. (b) Given, \( \mathbf{A}=\left[\begin{array}{ll} p & 0 \\ 0 & 2 \end{array}\right] \), \( \mathbf{B}=\left[\begin{array}{cc} 0 & -q \\ 1 & 0 \end{array}\right] \), \( C = \left[\begin{array}{cc} 2 & -2 \\ 2 & 2 \end{array}\right] \) and BA = C². Find the values of p and q.
Answer: To find the values of p and q, we first need to calculate the matrix products BA and C². Matrix multiplication involves multiplying rows by columns.
First, calculate BA:
\( BA = \left[\begin{array}{cc} 0 & -q \\ 1 & 0 \end{array}\right] \left[\begin{array}{ll} p & 0 \\ 0 & 2 \end{array}\right] \)
\( \implies BA = \left[\begin{array}{cc} (0)(p) + (-q)(0) & (0)(0) + (-q)(2) \\ (1)(p) + (0)(0) & (1)(0) + (0)(2) \end{array}\right] \)
\( \implies BA = \left[\begin{array}{cc} 0 & -2q \\ p & 0 \end{array}\right] \)
Next, calculate C²:
\( C^2 = C \times C = \left[\begin{array}{cc} 2 & -2 \\ 2 & 2 \end{array}\right] \left[\begin{array}{cc} 2 & -2 \\ 2 & 2 \end{array}\right] \)
\( \implies C^2 = \left[\begin{array}{cc} (2)(2) + (-2)(2) & (2)(-2) + (-2)(2) \\ (2)(2) + (2)(2) & (2)(-2) + (2)(2) \end{array}\right] \)
\( \implies C^2 = \left[\begin{array}{cc} 4-4 & -4-4 \\ 4+4 & -4+4 \end{array}\right] \)
\( \implies C^2 = \left[\begin{array}{cc} 0 & -8 \\ 8 & 0 \end{array}\right] \)
Given that \( BA = C^2 \), we can set the two resulting matrices equal to each other:
\( \left[\begin{array}{cc} 0 & -2q \\ p & 0 \end{array}\right] = \left[\begin{array}{cc} 0 & -8 \\ 8 & 0 \end{array}\right] \)
By comparing the corresponding elements in the matrices, we find:
\( p = 8 \)
\( -2q = -8 \implies q = \frac{-8}{-2} \)
\( \implies q = 4 \)
So, the values are \( p=8 \) and \( q=4 \). The process of multiplying matrices is very important in many areas of math and science.
In simple words: First, multiply matrices B and A together. Then, multiply matrix C by itself to get C squared. After that, set the results of BA and C squared equal. Finally, compare the numbers in the same spots in both matrices to find p and q.

🎯 Exam Tip: Remember that matrix multiplication is not commutative (AB is generally not equal to BA), so ensure you multiply matrices in the correct order specified by the question.

Question 14. Find x and y, if \( \left[\begin{array}{cc} 2 x & x \\ y & 3 y \end{array}\right]\left[\begin{array}{l} 3 \\ 2 \end{array}\right]=\left[\begin{array}{c} 16 \\ 9 \end{array}\right] \)
Answer: To find the values of x and y, we first perform the matrix multiplication on the left side of the equation. This involves multiplying the rows of the first matrix by the column of the second matrix. Remember that each element of the resulting matrix comes from a dot product.
First, perform the matrix multiplication:
\( \left[\begin{array}{cc} 2 x & x \\ y & 3 y \end{array}\right]\left[\begin{array}{l} 3 \\ 2 \end{array}\right] = \left[\begin{array}{c} (2x)(3) + (x)(2) \\ (y)(3) + (3y)(2) \end{array}\right] \)
\( \implies \left[\begin{array}{c} 6x + 2x \\ 3y + 6y \end{array}\right] = \left[\begin{array}{c} 8x \\ 9y \end{array}\right] \)
Now, we equate this resulting matrix to the matrix on the right side of the given equation:
\( \left[\begin{array}{c} 8x \\ 9y \end{array}\right] = \left[\begin{array}{c} 16 \\ 9 \end{array}\right] \)
By comparing the corresponding elements in the two matrices, we can set up two separate equations:
For the first element: \( 8x = 16 \)
\( \implies x = \frac{16}{8} \)
\( \implies x = 2 \)
For the second element: \( 9y = 9 \)
\( \implies y = \frac{9}{9} \)
\( \implies y = 1 \)
Therefore, the values are \( x=2 \) and \( y=1 \). Matrix equations often simplify to a system of linear equations.
In simple words: Multiply the two matrices on the left side first. You will get a single column matrix. Then, make the numbers in that matrix equal to the numbers in the matrix on the right side. This will give you two small math problems to solve for x and y.

🎯 Exam Tip: Always check the dimensions of the matrices before multiplying to ensure the operation is possible, and verify that the resulting matrix dimensions are correct.

Question 15. Given \( A = \left[\begin{array}{cc} 3 & -2 \\ -1 & 4 \end{array}\right] \), \( B = \left[\begin{array}{l} 6 \\ 1 \end{array}\right] \), \( C = \left[\begin{array}{c} -4 \\ 5 \end{array}\right] \) and \( D = \left[\begin{array}{l} 2 \\ 2 \end{array}\right] \), find AB + 2C – 4D.
Answer: To solve this, we need to perform each matrix operation step-by-step: matrix multiplication (AB) and scalar multiplication (2C, 4D), then add and subtract the resulting matrices. Careful calculations are key here.
First, calculate the product AB:
\( AB = \left[\begin{array}{cc} 3 & -2 \\ -1 & 4 \end{array}\right] \left[\begin{array}{l} 6 \\ 1 \end{array}\right] \)
\( \implies AB = \left[\begin{array}{c} (3)(6) + (-2)(1) \\ (-1)(6) + (4)(1) \end{array}\right] \)
\( \implies AB = \left[\begin{array}{c} 18-2 \\ -6+4 \end{array}\right] \)
\( \implies AB = \left[\begin{array}{c} 16 \\ -2 \end{array}\right] \)
Next, calculate 2C (scalar multiplication):
\( 2C = 2 \left[\begin{array}{c} -4 \\ 5 \end{array}\right] \)
\( \implies 2C = \left[\begin{array}{c} 2 \times -4 \\ 2 \times 5 \end{array}\right] \)
\( \implies 2C = \left[\begin{array}{c} -8 \\ 10 \end{array}\right] \)
Then, calculate 4D (scalar multiplication):
\( 4D = 4 \left[\begin{array}{l} 2 \\ 2 \end{array}\right] \)
\( \implies 4D = \left[\begin{array}{c} 4 \times 2 \\ 4 \times 2 \end{array}\right] \)
\( \implies 4D = \left[\begin{array}{l} 8 \\ 8 \end{array}\right] \)
Finally, combine the results as AB + 2C - 4D:
\( AB + 2C - 4D = \left[\begin{array}{c} 16 \\ -2 \end{array}\right] + \left[\begin{array}{c} -8 \\ 10 \end{array}\right] - \left[\begin{array}{l} 8 \\ 8 \end{array}\right] \)
\( \implies AB + 2C - 4D = \left[\begin{array}{c} 16 + (-8) - 8 \\ -2 + 10 - 8 \end{array}\right] \)
\( \implies AB + 2C - 4D = \left[\begin{array}{c} 16 - 8 - 8 \\ -2 + 10 - 8 \end{array}\right] \)
\( \implies AB + 2C - 4D = \left[\begin{array}{l} 0 \\ 0 \end{array}\right] \)
The final result is the zero matrix. This calculation demonstrates how to handle multiple matrix operations.
In simple words: First, multiply matrix A by matrix B. Then, multiply matrix C by 2. After that, multiply matrix D by 4. Finally, add the first two results and subtract the third result from that sum.

🎯 Exam Tip: When combining multiple matrix operations, perform multiplication before addition/subtraction, similar to the order of operations for numbers.

Question 16. Evaluate \( \left[\begin{array}{cc} 4 \sin 30^{\circ} & 2 \cos 60^{\circ} \\ \sin 90^{\circ} & 2 \cos 0^{\circ} \end{array}\right]\left[\begin{array}{ll} 4 & 5 \\ 5 & 4 \end{array}\right] \)
Answer: To evaluate this matrix product, we first need to substitute the known trigonometric values into the first matrix. This simplifies the matrix before we perform the multiplication. Basic trigonometric values are often used in matrix problems.
First, substitute the trigonometric values into the first matrix:
We know that:
\( \sin 30^\circ = \frac{1}{2} \)
\( \cos 60^\circ = \frac{1}{2} \)
\( \sin 90^\circ = 1 \)
\( \cos 0^\circ = 1 \)
So, the first matrix becomes:
\( \left[\begin{array}{cc} 4 \times \frac{1}{2} & 2 \times \frac{1}{2} \\ 1 & 2 \times 1 \end{array}\right] = \left[\begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array}\right] \)
Now, perform the matrix multiplication:
\( \left[\begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array}\right]\left[\begin{array}{ll} 4 & 5 \\ 5 & 4 \end{array}\right] \)
\( \implies \left[\begin{array}{cc} (2)(4) + (1)(5) & (2)(5) + (1)(4) \\ (1)(4) + (2)(5) & (1)(5) + (2)(4) \end{array}\right] \)
\( \implies \left[\begin{array}{cc} 8+5 & 10+4 \\ 4+10 & 5+8 \end{array}\right] \)
\( \implies \left[\begin{array}{cc} 13 & 14 \\ 14 & 13 \end{array}\right] \)
The final matrix product is \( \left[\begin{array}{cc} 13 & 14 \\ 14 & 13 \end{array}\right] \). This showcases how to combine trigonometry and matrix operations effectively.
In simple words: First, replace the sin and cos values in the first matrix with their actual numbers. Then, multiply this new matrix by the second matrix. Remember to multiply each row by each column.

🎯 Exam Tip: Memorizing common trigonometric values for angles like 0°, 30°, 45°, 60°, and 90° is essential for quickly solving such problems.

Question 17. If \( A = \left[\begin{array}{cc} 3 & 5 \\ 4 & -2 \end{array}\right] \) and \( B = \left[\begin{array}{l} 2 \\ 4 \end{array}\right] \), is the product AB possible? Give a reason. If yes, find AB.
Answer: To determine if the product AB is possible, we need to check the dimensions (order) of matrices A and B. Matrix multiplication is only defined under specific conditions. Understanding these conditions helps in various applications.
The order of matrix A is \( 2 \times 2 \) (2 rows, 2 columns).
The order of matrix B is \( 2 \times 1 \) (2 rows, 1 column).
For the product AB to be possible, the number of columns in the first matrix (A) must be equal to the number of rows in the second matrix (B).
Here, the number of columns in A is 2, and the number of rows in B is 2. Since \( 2 = 2 \), the product AB is indeed possible.
The resulting matrix AB will have an order of (rows of A) \( \times \) (columns of B), which is \( 2 \times 1 \).
Now, let's find the product AB:
\( AB = \left[\begin{array}{cc} 3 & 5 \\ 4 & -2 \end{array}\right] \left[\begin{array}{l} 2 \\ 4 \end{array}\right] \)
\( \implies AB = \left[\begin{array}{c} (3)(2) + (5)(4) \\ (4)(2) + (-2)(4) \end{array}\right] \)
\( \implies AB = \left[\begin{array}{c} 6+20 \\ 8-8 \end{array}\right] \)
\( \implies AB = \left[\begin{array}{c} 26 \\ 0 \end{array}\right] \)
The product AB is \( \left[\begin{array}{c} 26 \\ 0 \end{array}\right] \). This resulting matrix is a column vector.
In simple words: Yes, you can multiply A by B because the number of columns in A is the same as the number of rows in B. To find AB, multiply each row of A by the column of B and add the results.

🎯 Exam Tip: Always state the reason for matrix multiplication being possible (or not possible) by comparing the number of columns in the first matrix and the number of rows in the second matrix.

Question 18. If \( A = \left[\begin{array}{cc} 3 & 1 \\ -1 & 2 \end{array}\right] \) and \( I = \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \), find \( A^2 - 5A + 7I \).
Answer: To find \( A^2 - 5A + 7I \), we need to perform each operation step-by-step: calculate \( A^2 \), then scalar multiply A by 5 and I by 7, and finally combine the resulting matrices using addition and subtraction. This is a common operation in matrix algebra.
First, calculate \( A^2 \):
\( A^2 = A \times A = \left[\begin{array}{cc} 3 & 1 \\ -1 & 2 \end{array}\right] \left[\begin{array}{cc} 3 & 1 \\ -1 & 2 \end{array}\right] \)
\( \implies A^2 = \left[\begin{array}{cc} (3)(3) + (1)(-1) & (3)(1) + (1)(2) \\ (-1)(3) + (2)(-1) & (-1)(1) + (2)(2) \end{array}\right] \)
\( \implies A^2 = \left[\begin{array}{cc} 9-1 & 3+2 \\ -3-2 & -1+4 \end{array}\right] \)
\( \implies A^2 = \left[\begin{array}{cc} 8 & 5 \\ -5 & 3 \end{array}\right] \)
Next, calculate 5A (scalar multiplication):
\( 5A = 5 \left[\begin{array}{cc} 3 & 1 \\ -1 & 2 \end{array}\right] \)
\( \implies 5A = \left[\begin{array}{cc} 5 \times 3 & 5 \times 1 \\ 5 \times -1 & 5 \times 2 \end{array}\right] \)
\( \implies 5A = \left[\begin{array}{cc} 15 & 5 \\ -5 & 10 \end{array}\right] \)
Then, calculate 7I (scalar multiplication):
\( 7I = 7 \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \)
\( \implies 7I = \left[\begin{array}{cc} 7 \times 1 & 7 \times 0 \\ 7 \times 0 & 7 \times 1 \end{array}\right] \)
\( \implies 7I = \left[\begin{array}{cc} 7 & 0 \\ 0 & 7 \end{array}\right] \)
Finally, combine the results:
\( A^2 - 5A + 7I = \left[\begin{array}{cc} 8 & 5 \\ -5 & 3 \end{array}\right] - \left[\begin{array}{cc} 15 & 5 \\ -5 & 10 \end{array}\right] + \left[\begin{array}{cc} 7 & 0 \\ 0 & 7 \end{array}\right] \)
\( \implies A^2 - 5A + 7I = \left[\begin{array}{cc} 8 - 15 + 7 & 5 - 5 + 0 \\ -5 - (-5) + 0 & 3 - 10 + 7 \end{array}\right] \)
\( \implies A^2 - 5A + 7I = \left[\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}\right] \)
The final result is the zero matrix. This identity holds true for many matrix polynomials.
In simple words: First, find A squared by multiplying A by itself. Then, multiply A by 5 and the identity matrix I by 7. Finally, take A squared, subtract 5A, and then add 7I.

🎯 Exam Tip: When evaluating matrix polynomials, ensure you handle matrix multiplication (for \( A^2 \)) before scalar multiplication and then perform addition/subtraction.

Question 19. Given \( A = \left[\begin{array}{cc} 2 & -6 \\ 2 & 0 \end{array}\right] \), \( B = \left[\begin{array}{cc} -3 & 2 \\ 4 & 0 \end{array}\right] \), \( C = \left[\begin{array}{ll} 4 & 0 \\ 0 & 2 \end{array}\right] \). Find the matrix X such that \( A + 2X = 2B + C \).
Answer: To find matrix X, we need to rearrange the given equation to isolate X. This involves using basic matrix algebra rules for addition, subtraction, and scalar multiplication. Solving for a matrix unknown is a fundamental skill.
The given equation is: \( A + 2X = 2B + C \)
First, rearrange the equation to solve for 2X:
\( 2X = 2B + C - A \)
Now, we calculate the right side of the equation:
First, calculate 2B (scalar multiplication):
\( 2B = 2 \left[\begin{array}{cc} -3 & 2 \\ 4 & 0 \end{array}\right] \)
\( \implies 2B = \left[\begin{array}{cc} 2 \times -3 & 2 \times 2 \\ 2 \times 4 & 2 \times 0 \end{array}\right] \)
\( \implies 2B = \left[\begin{array}{cc} -6 & 4 \\ 8 & 0 \end{array}\right] \)
Next, calculate \( 2B + C - A \):
\( 2B + C - A = \left[\begin{array}{cc} -6 & 4 \\ 8 & 0 \end{array}\right] + \left[\begin{array}{ll} 4 & 0 \\ 0 & 2 \end{array}\right] - \left[\begin{array}{cc} 2 & -6 \\ 2 & 0 \end{array}\right] \)
\( \implies 2B + C - A = \left[\begin{array}{cc} -6+4-2 & 4+0-(-6) \\ 8+0-2 & 0+2-0 \end{array}\right] \)
\( \implies 2B + C - A = \left[\begin{array}{cc} -6+4-2 & 4+0+6 \\ 8+0-2 & 0+2-0 \end{array}\right] \)
\( \implies 2B + C - A = \left[\begin{array}{cc} -4 & 10 \\ 6 & 2 \end{array}\right] \)
So, we have:
\( 2X = \left[\begin{array}{cc} -4 & 10 \\ 6 & 2 \end{array}\right] \)
Finally, divide by 2 (scalar multiplication) to find X:
\( X = \frac{1}{2} \left[\begin{array}{cc} -4 & 10 \\ 6 & 2 \end{array}\right] \)
\( \implies X = \left[\begin{array}{cc} \frac{-4}{2} & \frac{10}{2} \\ \frac{6}{2} & \frac{2}{2} \end{array}\right] \)
\( \implies X = \left[\begin{array}{cc} -2 & 5 \\ 3 & 1 \end{array}\right] \)
Thus, the matrix X is \( \left[\begin{array}{cc} -2 & 5 \\ 3 & 1 \end{array}\right] \). This illustrates how matrix equations are solved by applying operations to both sides.
In simple words: First, move matrix A to the right side of the equation. Then, multiply matrix B by 2. After that, add matrix C and subtract matrix A from 2B. Finally, divide all numbers in the resulting matrix by 2 to get X.

🎯 Exam Tip: Treat matrix equations similar to algebraic equations, but remember that matrix division is performed by multiplying by the inverse matrix, or in this case, scalar division by multiplying by the reciprocal of the scalar.

Question 20. Let \( A = \left[\begin{array}{cc} 2 & 1 \\ 0 & -2 \end{array}\right] \), \( B = \left[\begin{array}{cc} 4 & 1 \\ -3 & -2 \end{array}\right] \) and \( C = \left[\begin{array}{ll} -3 & 2 \\ -1 & 4 \end{array}\right] \). Find \( A^2 + AC - 5B \).
Answer: To evaluate the expression \( A^2 + AC - 5B \), we must first calculate each term separately using matrix multiplication and scalar multiplication, then combine them through matrix addition and subtraction. It's a structured approach to matrix calculations.
First, calculate \( A^2 \):
\( A^2 = A \times A = \left[\begin{array}{cc} 2 & 1 \\ 0 & -2 \end{array}\right] \left[\begin{array}{cc} 2 & 1 \\ 0 & -2 \end{array}\right] \)
\( \implies A^2 = \left[\begin{array}{cc} (2)(2) + (1)(0) & (2)(1) + (1)(-2) \\ (0)(2) + (-2)(0) & (0)(1) + (-2)(-2) \end{array}\right] \)
\( \implies A^2 = \left[\begin{array}{cc} 4+0 & 2-2 \\ 0+0 & 0+4 \end{array}\right] \)
\( \implies A^2 = \left[\begin{array}{cc} 4 & 0 \\ 0 & 4 \end{array}\right] \)
Next, calculate AC:
\( AC = \left[\begin{array}{cc} 2 & 1 \\ 0 & -2 \end{array}\right] \left[\begin{array}{ll} -3 & 2 \\ -1 & 4 \end{array}\right] \)
\( \implies AC = \left[\begin{array}{cc} (2)(-3) + (1)(-1) & (2)(2) + (1)(4) \\ (0)(-3) + (-2)(-1) & (0)(2) + (-2)(4) \end{array}\right] \)
\( \implies AC = \left[\begin{array}{cc} -6-1 & 4+4 \\ 0+2 & 0-8 \end{array}\right] \)
\( \implies AC = \left[\begin{array}{cc} -7 & 8 \\ 2 & -8 \end{array}\right] \)
Then, calculate 5B (scalar multiplication):
\( 5B = 5 \left[\begin{array}{cc} 4 & 1 \\ -3 & -2 \end{array}\right] \)
\( \implies 5B = \left[\begin{array}{cc} 5 \times 4 & 5 \times 1 \\ 5 \times -3 & 5 \times -2 \end{array}\right] \)
\( \implies 5B = \left[\begin{array}{cc} 20 & 5 \\ -15 & -10 \end{array}\right] \)
Finally, combine the results:
\( A^2 + AC - 5B = \left[\begin{array}{cc} 4 & 0 \\ 0 & 4 \end{array}\right] + \left[\begin{array}{cc} -7 & 8 \\ 2 & -8 \end{array}\right] - \left[\begin{array}{cc} 20 & 5 \\ -15 & -10 \end{array}\right] \)
\( \implies A^2 + AC - 5B = \left[\begin{array}{cc} 4 + (-7) - 20 & 0 + 8 - 5 \\ 0 + 2 - (-15) & 4 + (-8) - (-10) \end{array}\right] \)
\( \implies A^2 + AC - 5B = \left[\begin{array}{cc} 4 - 7 - 20 & 8 - 5 \\ 2 + 15 & 4 - 8 + 10 \end{array}\right] \)
\( \implies A^2 + AC - 5B = \left[\begin{array}{cc} -23 & 3 \\ 17 & 6 \end{array}\right] \)
The result is the matrix \( \left[\begin{array}{cc} -23 & 3 \\ 17 & 6 \end{array}\right] \). This shows how complex matrix expressions are broken down.
In simple words: First, find A multiplied by A. Then, find A multiplied by C. After that, multiply matrix B by 5. Finally, add the first two results together and then subtract the third result from that sum.

🎯 Exam Tip: Pay close attention to the signs, especially when subtracting negative numbers, as a small error can lead to an incorrect final matrix.

Question 21. If \( A = \left[\begin{array}{ll} 3 & x \\ 0 & 1 \end{array}\right] \) and \( B = \left[\begin{array}{cc} 9 & 16 \\ 0 & -y \end{array}\right] \), find x and y when \( A^2 = B \).
Answer: To find the values of x and y, we need to first calculate \( A^2 \) by multiplying matrix A by itself. Once \( A^2 \) is found, we equate it to matrix B and compare their corresponding elements. This method is effective for solving matrix equations involving unknown variables.
First, calculate \( A^2 \):
\( A^2 = A \times A = \left[\begin{array}{ll} 3 & x \\ 0 & 1 \end{array}\right] \left[\begin{array}{ll} 3 & x \\ 0 & 1 \end{array}\right] \)
\( \implies A^2 = \left[\begin{array}{cc} (3)(3) + (x)(0) & (3)(x) + (x)(1) \\ (0)(3) + (1)(0) & (0)(x) + (1)(1) \end{array}\right] \)
\( \implies A^2 = \left[\begin{array}{cc} 9+0 & 3x+x \\ 0+0 & 0+1 \end{array}\right] \)
\( \implies A^2 = \left[\begin{array}{cc} 9 & 4x \\ 0 & 1 \end{array}\right] \)
Given that \( A^2 = B \), we can equate the corresponding elements of the two matrices:
\( \left[\begin{array}{cc} 9 & 4x \\ 0 & 1 \end{array}\right] = \left[\begin{array}{cc} 9 & 16 \\ 0 & -y \end{array}\right] \)
By comparing the elements:
From the (1,2) position: \( 4x = 16 \)
\( \implies x = \frac{16}{4} \)
\( \implies x = 4 \)
From the (2,2) position: \( 1 = -y \)
\( \implies y = -1 \)
So, the values are \( x=4 \) and \( y=-1 \). Equating matrices is a powerful way to solve for unknown variables within them.
In simple words: First, multiply matrix A by itself. This gives you A squared. Then, set the new A squared matrix equal to matrix B. Compare the numbers in the same spots in both matrices to find the values of x and y.

🎯 Exam Tip: When equating two matrices, every corresponding element must be equal. Use this principle to form a system of equations for unknown variables.

Question 22. Given \( A = \left[\begin{array}{cc} 2 & 0 \\ -1 & 7 \end{array}\right] \) and \( I = \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \), and \( A^2 = 9A + mI \). Find m.
Answer: To find the value of m, we need to substitute \( A^2 \), 9A, and \( mI \) into the given equation \( A^2 = 9A + mI \) and then compare the elements of the resulting matrices. This process is common in verifying matrix properties and solving for constants. The identity matrix I plays a crucial role in matrix algebra, similar to the number 1 in regular arithmetic.
First, calculate \( A^2 \):
\( A^2 = A \times A = \left[\begin{array}{cc} 2 & 0 \\ -1 & 7 \end{array}\right] \left[\begin{array}{cc} 2 & 0 \\ -1 & 7 \end{array}\right] \)
\( \implies A^2 = \left[\begin{array}{cc} (2)(2) + (0)(-1) & (2)(0) + (0)(7) \\ (-1)(2) + (7)(-1) & (-1)(0) + (7)(7) \end{array}\right] \)
\( \implies A^2 = \left[\begin{array}{cc} 4+0 & 0+0 \\ -2-7 & 0+49 \end{array}\right] \)
\( \implies A^2 = \left[\begin{array}{cc} 4 & 0 \\ -9 & 49 \end{array}\right] \)
Next, calculate 9A (scalar multiplication):
\( 9A = 9 \left[\begin{array}{cc} 2 & 0 \\ -1 & 7 \end{array}\right] \)
\( \implies 9A = \left[\begin{array}{cc} 9 \times 2 & 9 \times 0 \\ 9 \times -1 & 9 \times 7 \end{array}\right] \)
\( \implies 9A = \left[\begin{array}{cc} 18 & 0 \\ -9 & 63 \end{array}\right] \)
Then, calculate \( mI \) (scalar multiplication):
\( mI = m \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \)
\( \implies mI = \left[\begin{array}{cc} m & 0 \\ 0 & m \end{array}\right] \)
Now, substitute these into the equation \( A^2 = 9A + mI \):
\( \left[\begin{array}{cc} 4 & 0 \\ -9 & 49 \end{array}\right] = \left[\begin{array}{cc} 18 & 0 \\ -9 & 63 \end{array}\right] + \left[\begin{array}{cc} m & 0 \\ 0 & m \end{array}\right] \)
\( \implies \left[\begin{array}{cc} 4 & 0 \\ -9 & 49 \end{array}\right] = \left[\begin{array}{cc} 18+m & 0+0 \\ -9+0 & 63+m \end{array}\right] \)
\( \implies \left[\begin{array}{cc} 4 & 0 \\ -9 & 49 \end{array}\right] = \left[\begin{array}{cc} 18+m & 0 \\ -9 & 63+m \end{array}\right] \)
By comparing the corresponding elements, we can solve for m:
Using element (1,1): \( 4 = 18 + m \)
\( \implies m = 4 - 18 \)
\( \implies m = -14 \)
Using element (2,2): \( 49 = 63 + m \)
\( \implies m = 49 - 63 \)
\( \implies m = -14 \)
Both comparisons give \( m = -14 \). This consistent result confirms the value of m.
In simple words: First, find A multiplied by A. Then, multiply A by 9. After that, add the identity matrix I, multiplied by m, to 9A. Now, make the A squared matrix equal to the result you just found. By comparing the numbers in the same spots, you can find the value of m.

🎯 Exam Tip: Always double-check your value of 'm' (or any unknown scalar) by using at least two different corresponding elements if available to ensure consistency and correctness.

Question 23. Given matrix \( A = \left[\begin{array}{cc} 4 \sin 30^{\circ} & \cos 0^{\circ} \\ \cos 0^{\circ} & 4 \sin 30^{\circ} \end{array}\right] \) and \( B = \left[\begin{array}{l} 4 \\ 5 \end{array}\right] \). If AX = B, (i) Write the order of matrix 'X'. (ii) Let the matrix X be \( \left[\begin{array}{l} x \\ y \end{array}\right] \). Find x and y.
Answer:
(i) First, let's determine the order of matrix A and B. Matrix A is a \( 2 \times 2 \) matrix, and matrix B is a \( 2 \times 1 \) matrix. For the product AX to be defined, the number of columns in A must be equal to the number of rows in X. Also, for the resulting matrix AX to have the same order as B, the rows of A must match the rows of B, and the columns of X must match the columns of B. This is how matrix compatibility works.
Order of A: \( 2 \times 2 \)
Order of B: \( 2 \times 1 \)
Let the order of X be \( m \times n \).
For AX = B to be defined: \( (\text{cols of A}) = (\text{rows of X}) \implies 2 = m \)
For AX = B to have the order of B: \( (\text{rows of A}) \times (\text{cols of X}) = (\text{rows of B}) \times (\text{cols of B}) \)
\( \implies 2 \times n = 2 \times 1 \implies n = 1 \)
Thus, the order of matrix X is \( 2 \times 1 \).
(ii) Now, we need to find the values of x and y. First, evaluate matrix A by substituting the trigonometric values. Understanding these values is crucial for simplifying the matrix.
We know that:
\( \sin 30^\circ = \frac{1}{2} \)
\( \cos 0^\circ = 1 \)
Substitute these values into matrix A:
\( A = \left[\begin{array}{cc} 4 \times \frac{1}{2} & 1 \\ 1 & 4 \times \frac{1}{2} \end{array}\right] \)
\( \implies A = \left[\begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array}\right] \)
Given \( X = \left[\begin{array}{l} x \\ y \end{array}\right] \) and AX = B, substitute the matrices into the equation:
\( \left[\begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array}\right] \left[\begin{array}{l} x \\ y \end{array}\right] = \left[\begin{array}{l} 4 \\ 5 \end{array}\right] \)
Perform the matrix multiplication on the left side:
\( \left[\begin{array}{c} (2)(x) + (1)(y) \\ (1)(x) + (2)(y) \end{array}\right] = \left[\begin{array}{l} 4 \\ 5 \end{array}\right] \)
\( \implies \left[\begin{array}{c} 2x + y \\ x + 2y \end{array}\right] = \left[\begin{array}{l} 4 \\ 5 \end{array}\right] \)
By comparing the corresponding elements, we get a system of linear equations:
1. \( 2x + y = 4 \)
2. \( x + 2y = 5 \)
To solve this system, multiply equation (1) by 2:
\( 2(2x + y) = 2(4) \)
\( \implies 4x + 2y = 8 \)
Now, subtract equation (2) from this new equation:
\( (4x + 2y) - (x + 2y) = 8 - 5 \)
\( \implies 3x = 3 \)
\( \implies x = \frac{3}{3} \)
\( \implies x = 1 \)
Substitute \( x=1 \) into equation (1):
\( 2(1) + y = 4 \)
\( \implies 2 + y = 4 \)
\( \implies y = 4 - 2 \)
\( \implies y = 2 \)
So, the values are \( x=1 \) and \( y=2 \). Solving systems of equations derived from matrices is a common practice.
In simple words: (i) Matrix X must have 2 rows and 1 column so that you can multiply it by A and get a matrix like B. (ii) First, replace the sin and cos values in matrix A with numbers. Then, multiply A by a matrix X made of x and y. Set this equal to matrix B. This will give you two simple math problems to solve for x and y.

🎯 Exam Tip: Always start by determining the order of the unknown matrix X from the compatibility conditions of matrix multiplication. This is a crucial first step for setting up the problem correctly.