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

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

 

Question 1. Given matrix \( A = \begin{bmatrix} 3 \\ 2 \end{bmatrix}, B = \begin{bmatrix} -2 \\ -1 \end{bmatrix}, C = \begin{bmatrix} 1 \\ -3 \end{bmatrix} \), find the matrix X in each of the following cases:
(i) X = A + B + C
(ii) X = A + B -C
(iii) X - A = B
(iv) A + X = B + C
(v) 2X = A - C
Answer:
(i) To find X, we add matrices A, B, and C:
\( X = A + B + C \)
\( X = \begin{bmatrix} 3 \\ 2 \end{bmatrix} + \begin{bmatrix} -2 \\ -1 \end{bmatrix} + \begin{bmatrix} 1 \\ -3 \end{bmatrix} \)
\( X = \begin{bmatrix} 3-2+1 \\ 2-1-3 \end{bmatrix} \)
\( X = \begin{bmatrix} 2 \\ -2 \end{bmatrix} \)
(ii) To find X, we add A and B, then subtract C:
\( X = A + B - C \)
\( X = \begin{bmatrix} 3 \\ 2 \end{bmatrix} + \begin{bmatrix} -2 \\ -1 \end{bmatrix} - \begin{bmatrix} 1 \\ -3 \end{bmatrix} \)
\( X = \begin{bmatrix} 3-2-1 \\ 2-1-(-3) \end{bmatrix} \)
\( X = \begin{bmatrix} 3-2-1 \\ 2-1+3 \end{bmatrix} \)
\( X = \begin{bmatrix} 0 \\ 4 \end{bmatrix} \)
(iii) First, we need to rearrange the equation \( X - A = B \) to solve for X. We add A to both sides.
\( X = A + B \)
\( X = \begin{bmatrix} 3 \\ 2 \end{bmatrix} + \begin{bmatrix} -2 \\ -1 \end{bmatrix} \)
\( X = \begin{bmatrix} 3-2 \\ 2-1 \end{bmatrix} \)
\( X = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \)
(iv) First, we need to rearrange the equation \( A + X = B + C \) to solve for X. We subtract A from both sides.
\( X = B + C - A \)
\( X = \begin{bmatrix} -2 \\ -1 \end{bmatrix} + \begin{bmatrix} 1 \\ -3 \end{bmatrix} - \begin{bmatrix} 3 \\ 2 \end{bmatrix} \)
\( X = \begin{bmatrix} -2+1-3 \\ -1-3-2 \end{bmatrix} \)
\( X = \begin{bmatrix} -4 \\ -6 \end{bmatrix} \)
(v) First, we need to rearrange the equation \( 2X = A - C \) to solve for X. We divide both sides by 2, or multiply by \( \frac{1}{2} \).
\( 2X = \begin{bmatrix} 3 \\ 2 \end{bmatrix} - \begin{bmatrix} 1 \\ -3 \end{bmatrix} \)
\( 2X = \begin{bmatrix} 3-1 \\ 2-(-3) \end{bmatrix} \)
\( 2X = \begin{bmatrix} 2 \\ 2+3 \end{bmatrix} \)
\( 2X = \begin{bmatrix} 2 \\ 5 \end{bmatrix} \)
\( X = \frac{1}{2} \begin{bmatrix} 2 \\ 5 \end{bmatrix} \)
\( X = \begin{bmatrix} \frac{2}{2} \\ \frac{5}{2} \end{bmatrix} \)
\( X = \begin{bmatrix} 1 \\ \frac{5}{2} \end{bmatrix} \)
In simple words: To find matrix X, perform the given addition and subtraction operations with the matrices A, B, and C. Remember that matrix operations are done element by element, and scalar multiplication multiplies each element.

🎯 Exam Tip: Always pay close attention to the order of operations and signs, especially when subtracting matrices or terms. A small mistake in a sign can lead to an incorrect final matrix.

 

Question 2. If \( B = \begin{bmatrix} 2 & -3 \\ -4 & 5 \end{bmatrix} \) and \( C = \begin{bmatrix} 1 & -3 \\ -4 & 4 \end{bmatrix} \), find the matrix B – C.
Answer: To find the matrix \( B - C \), we subtract the corresponding elements of matrix C from matrix B.
Given: \( B = \begin{bmatrix} 2 & -3 \\ -4 & 5 \end{bmatrix} \) and \( C = \begin{bmatrix} 1 & -3 \\ -4 & 4 \end{bmatrix} \)
\( B - C = \begin{bmatrix} 2 & -3 \\ -4 & 5 \end{bmatrix} - \begin{bmatrix} 1 & -3 \\ -4 & 4 \end{bmatrix} \)
\( B - C = \begin{bmatrix} 2-1 & -3-(-3) \\ -4-(-4) & 5-4 \end{bmatrix} \)
\( B - C = \begin{bmatrix} 2-1 & -3+3 \\ -4+4 & 5-4 \end{bmatrix} \)
\( B - C = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
In simple words: To subtract matrices, just subtract the number in the same spot in the second matrix from the first matrix. This gives you a new matrix with the results.

🎯 Exam Tip: When subtracting negative numbers, remember that 'minus a minus' becomes a plus. This is a very common place for errors in matrix subtraction.

 

Question 3. If \( A = \begin{bmatrix} 2 & 0 \\ -3 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} 0 & 1 \\ -2 & 3 \end{bmatrix} \), find \( 2A – 3B \).
Answer: To find \( 2A - 3B \), we first multiply matrix A by 2 and matrix B by 3, then subtract the results.
First, find \( 2A \):
\( 2A = 2 \begin{bmatrix} 2 & 0 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} 2 \times 2 & 2 \times 0 \\ 2 \times (-3) & 2 \times 1 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ -6 & 2 \end{bmatrix} \)
Next, find \( 3B \):
\( 3B = 3 \begin{bmatrix} 0 & 1 \\ -2 & 3 \end{bmatrix} = \begin{bmatrix} 3 \times 0 & 3 \times 1 \\ 3 \times (-2) & 3 \times 3 \end{bmatrix} = \begin{bmatrix} 0 & 3 \\ -6 & 9 \end{bmatrix} \)
Now, subtract \( 3B \) from \( 2A \):
\( 2A - 3B = \begin{bmatrix} 4 & 0 \\ -6 & 2 \end{bmatrix} - \begin{bmatrix} 0 & 3 \\ -6 & 9 \end{bmatrix} \)
\( 2A - 3B = \begin{bmatrix} 4-0 & 0-3 \\ -6-(-6) & 2-9 \end{bmatrix} \)
\( 2A - 3B = \begin{bmatrix} 4 & -3 \\ -6+6 & -7 \end{bmatrix} \)
\( 2A - 3B = \begin{bmatrix} 4 & -3 \\ 0 & -7 \end{bmatrix} \)
In simple words: Multiply each matrix by its number first, and then subtract the resulting matrices element by element. This means you do all the multiplications before the subtraction.

🎯 Exam Tip: Remember that scalar multiplication (multiplying a matrix by a single number) means multiplying *every* element inside the matrix by that number. Don't forget any element!

 

Question 4. Given \( A = \begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix} \) and \( B = \begin{bmatrix} -4 & -1 \\ -3 & -2 \end{bmatrix} \).
(i) Find the matrix \( 2A + B \);
(ii) Find a matrix C such that \( C + B = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
Answer:
(i) To find \( 2A + B \), first multiply matrix A by 2, then add matrix B to the result.
\( 2A = 2 \begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} 2 \times 1 & 2 \times 4 \\ 2 \times 2 & 2 \times 3 \end{bmatrix} = \begin{bmatrix} 2 & 8 \\ 4 & 6 \end{bmatrix} \)
Now, add \( B \) to \( 2A \):
\( 2A + B = \begin{bmatrix} 2 & 8 \\ 4 & 6 \end{bmatrix} + \begin{bmatrix} -4 & -1 \\ -3 & -2 \end{bmatrix} \)
\( 2A + B = \begin{bmatrix} 2+(-4) & 8+(-1) \\ 4+(-3) & 6+(-2) \end{bmatrix} \)
\( 2A + B = \begin{bmatrix} 2-4 & 8-1 \\ 4-3 & 6-2 \end{bmatrix} \)
\( 2A + B = \begin{bmatrix} -2 & 7 \\ 1 & 4 \end{bmatrix} \)
(ii) We are given \( C + B = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \). We want to find C. This means C is the negative of B.
We can rearrange the equation to find C:
\( C = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} - B \)
So, C is the additive inverse of B.
\( C = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} - \begin{bmatrix} -4 & -1 \\ -3 & -2 \end{bmatrix} \)
\( C = \begin{bmatrix} 0-(-4) & 0-(-1) \\ 0-(-3) & 0-(-2) \end{bmatrix} \)
\( C = \begin{bmatrix} 4 & 1 \\ 3 & 2 \end{bmatrix} \)
In simple words: For the first part, multiply matrix A by 2, then add matrix B. For the second part, if C plus B equals a matrix of all zeros, then C must be the opposite of B, meaning all its numbers have opposite signs.

🎯 Exam Tip: A matrix with all zero elements is called a null matrix or zero matrix. When asked to find a matrix that, when added to another, results in a null matrix, you are essentially finding the additive inverse.

 

Question 5. If \( P = \begin{bmatrix} 4 & 6 \\ 2 & -8 \end{bmatrix} \) and \( Q = \begin{bmatrix} 2 & -3 \\ -1 & 1 \end{bmatrix} \), find \( P + 2Q \).
Answer: To find \( P + 2Q \), we first multiply matrix Q by 2, then add matrix P to the result.
First, find \( 2Q \):
\( 2Q = 2 \begin{bmatrix} 2 & -3 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 2 \times 2 & 2 \times (-3) \\ 2 \times (-1) & 2 \times 1 \end{bmatrix} = \begin{bmatrix} 4 & -6 \\ -2 & 2 \end{bmatrix} \)
Now, add \( P \) to \( 2Q \):
\( P + 2Q = \begin{bmatrix} 4 & 6 \\ 2 & -8 \end{bmatrix} + \begin{bmatrix} 4 & -6 \\ -2 & 2 \end{bmatrix} \)
\( P + 2Q = \begin{bmatrix} 4+4 & 6+(-6) \\ 2+(-2) & -8+2 \end{bmatrix} \)
\( P + 2Q = \begin{bmatrix} 4+4 & 6-6 \\ 2-2 & -8+2 \end{bmatrix} \)
\( P + 2Q = \begin{bmatrix} 8 & 0 \\ 0 & -6 \end{bmatrix} \)
In simple words: Multiply matrix Q by 2, which means multiplying every number inside Q by 2. Then, add this new matrix to matrix P by adding the numbers that are in the same position.

🎯 Exam Tip: Remember that matrix addition and subtraction are only possible if the matrices have the same dimensions. In this case, both P and 2Q are 2x2 matrices, so addition is valid.

 

Question 6. If \( 2\begin{bmatrix} 3 & 4 \\ 5 & x \end{bmatrix} + \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix} \), find the values of x and y.
Answer: To find the values of x and y, we first perform the matrix operations on the left side of the equation, and then compare the elements of the resulting matrix with the elements of the matrix on the right side.
Multiply the first matrix by 2:
\( \begin{bmatrix} 2 \times 3 & 2 \times 4 \\ 2 \times 5 & 2 \times x \end{bmatrix} + \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix} \)
\( \begin{bmatrix} 6 & 8 \\ 10 & 2x \end{bmatrix} + \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix} \)
Now, add the two matrices on the left:
\( \begin{bmatrix} 6+1 & 8+y \\ 10+0 & 2x+1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix} \)
\( \begin{bmatrix} 7 & 8+y \\ 10 & 2x+1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix} \)
By comparing the corresponding elements, we can set up two equations:
For the top right element: \( 8 + y = 0 \)
\( y = 0 - 8 \)
\( y = -8 \)
For the bottom right element: \( 2x + 1 = 5 \)
\( 2x = 5 - 1 \)
\( 2x = 4 \)
\( x = \frac{4}{2} \)
\( x = 2 \)
So, the values are \( x = 2 \) and \( y = -8 \).
In simple words: First, multiply the first matrix by 2. Then, add the two matrices together. After that, match the numbers in the same positions on both sides of the equal sign to find the unknown values of x and y.

🎯 Exam Tip: For matrix equality, every corresponding element must be equal. This gives you a system of equations, one for each position in the matrix. Be careful with calculations when solving these equations.

 

Question 7. If \( \begin{bmatrix} a & 3 \\ 4 & 2 \end{bmatrix} + \begin{bmatrix} 2 & b \\ 1 & -2 \end{bmatrix} - \begin{bmatrix} 1 & 1 \\ -2 & c \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 7 & 3 \end{bmatrix} \), find the value of a, b and c.
Answer: To find the values of a, b, and c, we first perform the matrix operations on the left side of the equation, then compare the elements of the resulting matrix with the elements of the matrix on the right side.
Combine the matrices on the left side:
\( \begin{bmatrix} a+2-1 & 3+b-1 \\ 4+1-(-2) & 2-2-c \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 7 & 3 \end{bmatrix} \)
\( \begin{bmatrix} a+1 & b+2 \\ 4+1+2 & 2-2-c \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 7 & 3 \end{bmatrix} \)
\( \begin{bmatrix} a+1 & b+2 \\ 7 & -c \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 7 & 3 \end{bmatrix} \)
Now, compare the corresponding elements to form equations:
For the top left element: \( a + 1 = 5 \)
\( a = 5 - 1 \)
\( a = 4 \)
For the top right element: \( b + 2 = 0 \)
\( b = -2 \)
For the bottom right element: \( -c = 3 \)
\( c = -3 \)
So, the values are \( a = 4, b = -2, c = -3 \).
In simple words: First, do all the adding and subtracting of the matrices on the left side by working on each position. Then, set the final numbers in each position equal to the numbers in the same position on the right side to figure out a, b, and c.

🎯 Exam Tip: Be careful with the signs, especially when subtracting negative numbers (e.g., \( -(-2) \) becomes \( +2 \)). A common error is mixing up a positive and negative sign.

 

Question 8. If \( \begin{bmatrix} 2 & -1 \\ 2 & 0 \end{bmatrix} + 2A = \begin{bmatrix} -3 & 5 \\ 4 & 3 \end{bmatrix} \), find A.
Answer: To find matrix A, we first need to isolate \( 2A \) by subtracting the first matrix from both sides, then multiply the result by \( \frac{1}{2} \).
Let the first matrix be M.
\( M + 2A = \begin{bmatrix} -3 & 5 \\ 4 & 3 \end{bmatrix} \)
\( 2A = \begin{bmatrix} -3 & 5 \\ 4 & 3 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ 2 & 0 \end{bmatrix} \)
\( 2A = \begin{bmatrix} -3-2 & 5-(-1) \\ 4-2 & 3-0 \end{bmatrix} \)
\( 2A = \begin{bmatrix} -5 & 5+1 \\ 2 & 3 \end{bmatrix} \)
\( 2A = \begin{bmatrix} -5 & 6 \\ 2 & 3 \end{bmatrix} \)
Now, multiply both sides by \( \frac{1}{2} \) to find A:
\( A = \frac{1}{2} \begin{bmatrix} -5 & 6 \\ 2 & 3 \end{bmatrix} \)
\( A = \begin{bmatrix} \frac{-5}{2} & \frac{6}{2} \\ \frac{2}{2} & \frac{3}{2} \end{bmatrix} \)
\( A = \begin{bmatrix} -\frac{5}{2} & 3 \\ 1 & \frac{3}{2} \end{bmatrix} \)
In simple words: First, move the matrix on the left side to the right side by subtracting it. This will leave \( 2A \) by itself. Then, divide every number in the resulting matrix by 2 to find A.

🎯 Exam Tip: Remember that when moving a matrix from one side of an equation to the other, you change its sign. This is just like solving equations with regular numbers.

 

Question 9. If \( 2\begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} + 3\begin{bmatrix} 1 & 3 \\ y & 2 \end{bmatrix} = \begin{bmatrix} z & -7 \\ 15 & 8 \end{bmatrix} \), find the values of x, y and z.
Answer: To find the values of x, y, and z, we first perform the matrix operations on the left side, and then compare the elements of the resulting matrix with the elements of the matrix on the right side.
First, perform the scalar multiplications:
\( \begin{bmatrix} 2 \times 3 & 2 \times x \\ 2 \times 0 & 2 \times 1 \end{bmatrix} + \begin{bmatrix} 3 \times 1 & 3 \times 3 \\ 3 \times y & 3 \times 2 \end{bmatrix} = \begin{bmatrix} z & -7 \\ 15 & 8 \end{bmatrix} \)
\( \begin{bmatrix} 6 & 2x \\ 0 & 2 \end{bmatrix} + \begin{bmatrix} 3 & 9 \\ 3y & 6 \end{bmatrix} = \begin{bmatrix} z & -7 \\ 15 & 8 \end{bmatrix} \)
Next, add the two matrices on the left:
\( \begin{bmatrix} 6+3 & 2x+9 \\ 0+3y & 2+6 \end{bmatrix} = \begin{bmatrix} z & -7 \\ 15 & 8 \end{bmatrix} \)
\( \begin{bmatrix} 9 & 2x+9 \\ 3y & 8 \end{bmatrix} = \begin{bmatrix} z & -7 \\ 15 & 8 \end{bmatrix} \)
Now, compare the corresponding elements to set up equations:
For the top left element: \( 9 = z \)
\( z = 9 \)
For the top right element: \( 2x + 9 = -7 \)
\( 2x = -7 - 9 \)
\( 2x = -16 \)
\( x = \frac{-16}{2} \)
\( x = -8 \)
For the bottom left element: \( 3y = 15 \)
\( y = \frac{15}{3} \)
\( y = 5 \)
So, the values are \( x = -8, y = 5, z = 9 \).
In simple words: Multiply each matrix by its number. Then, add the matrices on the left. Finally, compare the numbers in the same spots in the matrices on both sides to find x, y, and z.

🎯 Exam Tip: This type of question tests your ability to perform both scalar multiplication and matrix addition, as well as solving linear equations. Work carefully through each step to avoid errors.

 

Question 10. Given that \( M = \begin{bmatrix} 2 & 0 \\ 1 & 2 \end{bmatrix} \) and \( N = \begin{bmatrix} 2 & 0 \\ -1 & 2 \end{bmatrix} \), find \( M + 2N \).
Answer: To find \( M + 2N \), we first multiply matrix N by 2, then add matrix M to the result.
First, find \( 2N \):
\( 2N = 2 \begin{bmatrix} 2 & 0 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 2 \times 2 & 2 \times 0 \\ 2 \times (-1) & 2 \times 2 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ -2 & 4 \end{bmatrix} \)
Now, add \( M \) to \( 2N \):
\( M + 2N = \begin{bmatrix} 2 & 0 \\ 1 & 2 \end{bmatrix} + \begin{bmatrix} 4 & 0 \\ -2 & 4 \end{bmatrix} \)
\( M + 2N = \begin{bmatrix} 2+4 & 0+0 \\ 1+(-2) & 2+4 \end{bmatrix} \)
\( M + 2N = \begin{bmatrix} 6 & 0 \\ 1-2 & 6 \end{bmatrix} \)
\( M + 2N = \begin{bmatrix} 6 & 0 \\ -1 & 6 \end{bmatrix} \)
In simple words: Multiply matrix N by the number 2, which means multiplying every element inside N by 2. Then, add this new matrix to matrix M by adding the numbers that are in the same position.

🎯 Exam Tip: Always double-check your arithmetic, especially when dealing with negative numbers, to ensure accurate addition and subtraction within the matrices.

 

Question 11.
(a) If \( A = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}, B = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}, C = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \), find \( A - B + C \).
(b) "If A and B are two ‘2 x 2' matrices such that A – B = A + B, then B is a zero matrix." Is this statement true or false? You do not have to give any reason or proof.
Answer:
(a) To find \( A - B + C \), we perform the subtraction and addition operations in order from left to right.
\( A - B + C = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} - \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
Combine the elements:
\( A - B + C = \begin{bmatrix} 3-1+1 & 0-3+0 \\ 0-0+0 & 3-1+1 \end{bmatrix} \)
\( A - B + C = \begin{bmatrix} 3 & -3 \\ 0 & 3 \end{bmatrix} \)
(b) The statement is true. Let's look at the equation given: \( A - B = A + B \).
If we subtract matrix A from both sides, we get:
\( -B = B \)
This equation can only be true if B is a matrix where all its elements are zero. For example, if B was \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \), then \( -B \) would be \( \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \), which is not equal to B. So, B must be a zero matrix.
In simple words: For part (a), just add and subtract the numbers in the same spots in the matrices. For part (b), the statement is true. If subtracting B from A gives the same result as adding B to A, then B must be a matrix where all its numbers are zero.

🎯 Exam Tip: For matrix algebra questions like part (b), treat the matrices as variables in an algebraic equation. Operations like adding or subtracting a matrix from both sides follow the same rules as with scalar numbers. A key property to remember is that \( -B = B \) implies B is a zero matrix.

 

Question 12. If \( P = \begin{bmatrix} -3 & 1 \\ 2 & 5 \end{bmatrix}, Q = \begin{bmatrix} 1 & 6 \\ -4 & 0 \end{bmatrix} \) and \( R = \begin{bmatrix} 4 & -1 \\ 2 & 3 \end{bmatrix} \). Find the value of :
(i) \( 2P + 3Q – R \)
(ii) \( 4P – 2Q + 3R \)
Answer:
(i) To find \( 2P + 3Q - R \), we first perform scalar multiplications, then add and subtract the resulting matrices.
First, find \( 2P \) and \( 3Q \):
\( 2P = 2 \begin{bmatrix} -3 & 1 \\ 2 & 5 \end{bmatrix} = \begin{bmatrix} -6 & 2 \\ 4 & 10 \end{bmatrix} \)
\( 3Q = 3 \begin{bmatrix} 1 & 6 \\ -4 & 0 \end{bmatrix} = \begin{bmatrix} 3 & 18 \\ -12 & 0 \end{bmatrix} \)
Now, perform the addition and subtraction:
\( 2P + 3Q - R = \begin{bmatrix} -6 & 2 \\ 4 & 10 \end{bmatrix} + \begin{bmatrix} 3 & 18 \\ -12 & 0 \end{bmatrix} - \begin{bmatrix} 4 & -1 \\ 2 & 3 \end{bmatrix} \)
\( = \begin{bmatrix} -6+3-4 & 2+18-(-1) \\ 4+(-12)-2 & 10+0-3 \end{bmatrix} \)
\( = \begin{bmatrix} -3-4 & 20+1 \\ 4-12-2 & 7 \end{bmatrix} \)
\( = \begin{bmatrix} -7 & 21 \\ -8-2 & 7 \end{bmatrix} \)
\( = \begin{bmatrix} -7 & 21 \\ -10 & 7 \end{bmatrix} \)
(ii) To find \( 4P - 2Q + 3R \), we first perform scalar multiplications, then add and subtract the resulting matrices.
First, find \( 4P \), \( 2Q \) and \( 3R \):
\( 4P = 4 \begin{bmatrix} -3 & 1 \\ 2 & 5 \end{bmatrix} = \begin{bmatrix} -12 & 4 \\ 8 & 20 \end{bmatrix} \)
\( 2Q = 2 \begin{bmatrix} 1 & 6 \\ -4 & 0 \end{bmatrix} = \begin{bmatrix} 2 & 12 \\ -8 & 0 \end{bmatrix} \)
\( 3R = 3 \begin{bmatrix} 4 & -1 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} 12 & -3 \\ 6 & 9 \end{bmatrix} \)
Now, perform the addition and subtraction:
\( 4P - 2Q + 3R = \begin{bmatrix} -12 & 4 \\ 8 & 20 \end{bmatrix} - \begin{bmatrix} 2 & 12 \\ -8 & 0 \end{bmatrix} + \begin{bmatrix} 12 & -3 \\ 6 & 9 \end{bmatrix} \)
\( = \begin{bmatrix} -12-2+12 & 4-12+(-3) \\ 8-(-8)+6 & 20-0+9 \end{bmatrix} \)
\( = \begin{bmatrix} -2 & 4-12-3 \\ 8+8+6 & 29 \end{bmatrix} \)
\( = \begin{bmatrix} -2 & -8-3 \\ 16+6 & 29 \end{bmatrix} \)
\( = \begin{bmatrix} -2 & -11 \\ 22 & 29 \end{bmatrix} \)
In simple words: For both parts, first multiply each matrix by its number. Then, add or subtract the new matrices, making sure to pay close attention to all the plus and minus signs for each number in its correct spot.

🎯 Exam Tip: When performing multiple matrix operations, it's best to calculate each scalar multiplication separately before combining them through addition or subtraction. This reduces the chance of making a calculation error.

 

Question 13. Compute: \( \begin{bmatrix} \cos^2\theta & 0 \\ \cot^2\theta & 1 \end{bmatrix} + \begin{bmatrix} \sin^2\theta & 1 \\ -\operatorname{cosec}^2\theta & 0 \end{bmatrix} + \begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix} \)
Answer: To compute the sum, we add the corresponding elements of the three matrices.
\( \begin{bmatrix} \cos^2\theta & 0 \\ \cot^2\theta & 1 \end{bmatrix} + \begin{bmatrix} \sin^2\theta & 1 \\ -\operatorname{cosec}^2\theta & 0 \end{bmatrix} + \begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix} \)
Add the elements in the top-left position:
\( \cos^2\theta + \sin^2\theta + 0 \)
Add the elements in the top-right position:
\( 0 + 1 + (-1) \)
Add the elements in the bottom-left position:
\( \cot^2\theta + (-\operatorname{cosec}^2\theta) + (-1) \)
Add the elements in the bottom-right position:
\( 1 + 0 + 0 \)

Combining these, the resulting matrix is:
\( \begin{bmatrix} \cos^2\theta + \sin^2\theta + 0 & 0+1-1 \\ \cot^2\theta - \operatorname{cosec}^2\theta - 1 & 1+0+0 \end{bmatrix} \)
Now, use the trigonometric identities:
\( \cos^2\theta + \sin^2\theta = 1 \)
\( 1 + \cot^2\theta = \operatorname{cosec}^2\theta \implies \cot^2\theta - \operatorname{cosec}^2\theta = -1 \)
Substitute these identities into the matrix:
\( \begin{bmatrix} 1 & 0 \\ -1 - 1 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix} \)
In simple words: Add the numbers in the same position from all three matrices. Use basic trigonometry rules like \( \sin^2\theta + \cos^2\theta = 1 \) and \( \cot^2\theta - \operatorname{cosec}^2\theta = -1 \) to simplify the final numbers.

🎯 Exam Tip: This question combines matrix addition with trigonometric identities. Make sure you recall the fundamental identities: \( \sin^2\theta + \cos^2\theta = 1 \) and \( \cot^2\theta - \operatorname{cosec}^2\theta = -1 \). Errors often occur if these identities are forgotten or misapplied.

 

Question 14. If \( A = \begin{bmatrix} x & y \\ z & w \end{bmatrix}, B = \begin{bmatrix} x & -y \\ -z & w \end{bmatrix} \) and \( C = \begin{bmatrix} 0 & y \\ 2z & 0 \end{bmatrix} \). Show that \( (A + B) + C = A + (B + C) \).
Answer: We need to show that matrix addition is associative, meaning \( (A + B) + C = A + (B + C) \). We will calculate both sides of the equation separately and show they are equal.
**Left Hand Side (L.H.S.): \( (A + B) + C \)**
First, calculate \( A + B \):
\( A + B = \begin{bmatrix} x & y \\ z & w \end{bmatrix} + \begin{bmatrix} x & -y \\ -z & w \end{bmatrix} \)
\( = \begin{bmatrix} x+x & y+(-y) \\ z+(-z) & w+w \end{bmatrix} \)
\( = \begin{bmatrix} 2x & 0 \\ 0 & 2w \end{bmatrix} \)
Now, add C to \( (A+B) \):
\( (A + B) + C = \begin{bmatrix} 2x & 0 \\ 0 & 2w \end{bmatrix} + \begin{bmatrix} 0 & y \\ 2z & 0 \end{bmatrix} \)
\( = \begin{bmatrix} 2x+0 & 0+y \\ 0+2z & 2w+0 \end{bmatrix} \)
\( = \begin{bmatrix} 2x & y \\ 2z & 2w \end{bmatrix} \)

**Right Hand Side (R.H.S.): \( A + (B + C) \)**
First, calculate \( B + C \):
\( B + C = \begin{bmatrix} x & -y \\ -z & w \end{bmatrix} + \begin{bmatrix} 0 & y \\ 2z & 0 \end{bmatrix} \)
\( = \begin{bmatrix} x+0 & -y+y \\ -z+2z & w+0 \end{bmatrix} \)
\( = \begin{bmatrix} x & 0 \\ z & w \end{bmatrix} \)
Now, add A to \( (B+C) \):
\( A + (B + C) = \begin{bmatrix} x & y \\ z & w \end{bmatrix} + \begin{bmatrix} x & 0 \\ z & w \end{bmatrix} \)
\( = \begin{bmatrix} x+x & y+0 \\ z+z & w+w \end{bmatrix} \)
\( = \begin{bmatrix} 2x & y \\ 2z & 2w \end{bmatrix} \)
Since L.H.S. \( = \begin{bmatrix} 2x & y \\ 2z & 2w \end{bmatrix} \) and R.H.S. \( = \begin{bmatrix} 2x & y \\ 2z & 2w \end{bmatrix} \), we have shown that L.H.S. = R.H.S.
Thus, \( (A + B) + C = A + (B + C) \).
In simple words: To show the equation is true, calculate the left side by adding A and B first, then add C to that. Then, calculate the right side by adding B and C first, then add A to that. If both results are the same matrix, then you've shown the equation is correct.

🎯 Exam Tip: Proving matrix properties involves calculating each side of the equation separately and showing that they result in the same matrix. This confirms the property holds true for the given matrices.

 

Question 15. If \( A = \begin{bmatrix} 8 & 1 \\ 5 & 2 \end{bmatrix}, B = \begin{bmatrix} -4 & 6 \\ -2 & 12 \end{bmatrix} \), find the matrix C such that \( 2A + 3B + 4C \) is a null matrix.
Answer: We are given that \( 2A + 3B + 4C \) is a null matrix. A null matrix is a matrix where all elements are zero. For 2x2 matrices, the null matrix is \( \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \).
So, the equation is: \( 2A + 3B + 4C = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \).
We need to find C. First, calculate \( 2A \) and \( 3B \).
\( 2A = 2 \begin{bmatrix} 8 & 1 \\ 5 & 2 \end{bmatrix} = \begin{bmatrix} 16 & 2 \\ 10 & 4 \end{bmatrix} \)
\( 3B = 3 \begin{bmatrix} -4 & 6 \\ -2 & 12 \end{bmatrix} = \begin{bmatrix} -12 & 18 \\ -6 & 36 \end{bmatrix} \)
Now, substitute these into the main equation:
\( \begin{bmatrix} 16 & 2 \\ 10 & 4 \end{bmatrix} + \begin{bmatrix} -12 & 18 \\ -6 & 36 \end{bmatrix} + 4C = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
Add the two matrices on the left:
\( \begin{bmatrix} 16+(-12) & 2+18 \\ 10+(-6) & 4+36 \end{bmatrix} + 4C = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
\( \begin{bmatrix} 16-12 & 20 \\ 10-6 & 40 \end{bmatrix} + 4C = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
\( \begin{bmatrix} 4 & 20 \\ 4 & 40 \end{bmatrix} + 4C = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
Now, subtract \( \begin{bmatrix} 4 & 20 \\ 4 & 40 \end{bmatrix} \) from both sides to isolate \( 4C \):
\( 4C = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} - \begin{bmatrix} 4 & 20 \\ 4 & 40 \end{bmatrix} \)
\( 4C = \begin{bmatrix} 0-4 & 0-20 \\ 0-4 & 0-40 \end{bmatrix} \)
\( 4C = \begin{bmatrix} -4 & -20 \\ -4 & -40 \end{bmatrix} \)
Finally, divide all elements by 4 (or multiply by \( \frac{1}{4} \)) to find C:
\( C = \frac{1}{4} \begin{bmatrix} -4 & -20 \\ -4 & -40 \end{bmatrix} \)
\( C = \begin{bmatrix} \frac{-4}{4} & \frac{-20}{4} \\ \frac{-4}{4} & \frac{-40}{4} \end{bmatrix} \)
\( C = \begin{bmatrix} -1 & -5 \\ -1 & -10 \end{bmatrix} \)
In simple words: First, multiply A by 2 and B by 3. Add these two new matrices together. Then, subtract this total from the zero matrix to find \( 4C \). Lastly, divide all numbers in \( 4C \) by 4 to get matrix C.

🎯 Exam Tip: When an equation states that an expression equals a "null matrix," remember to set the expression equal to a matrix of all zeros of the appropriate dimensions. Then, use standard matrix operations to solve for the unknown matrix.

 

Question 16. If \( P = \begin{bmatrix} 6 & -2 \\ 4 & -6 \end{bmatrix}, Q = \begin{bmatrix} 5 & 3 \\ 2 & 0 \end{bmatrix} \), find X such that \( 3P – 2Q + 3X = 0 \).
Answer: We are given the equation \( 3P - 2Q + 3X = 0 \), where 0 represents a null matrix (a matrix of all zeros). We need to solve for X.
First, rearrange the equation to isolate \( 3X \):
\( 3X = -3P + 2Q \)
Alternatively, \( 3X = 2Q - 3P \)
Now, calculate \( 2Q \) and \( 3P \):
\( 2Q = 2 \begin{bmatrix} 5 & 3 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} 10 & 6 \\ 4 & 0 \end{bmatrix} \)
\( 3P = 3 \begin{bmatrix} 6 & -2 \\ 4 & -6 \end{bmatrix} = \begin{bmatrix} 18 & -6 \\ 12 & -18 \end{bmatrix} \)
Next, calculate \( 2Q - 3P \):
\( 3X = \begin{bmatrix} 10 & 6 \\ 4 & 0 \end{bmatrix} - \begin{bmatrix} 18 & -6 \\ 12 & -18 \end{bmatrix} \)
\( 3X = \begin{bmatrix} 10-18 & 6-(-6) \\ 4-12 & 0-(-18) \end{bmatrix} \)
\( 3X = \begin{bmatrix} -8 & 6+6 \\ -8 & 0+18 \end{bmatrix} \)
\( 3X = \begin{bmatrix} -8 & 12 \\ -8 & 18 \end{bmatrix} \)
Finally, divide all elements by 3 (or multiply by \( \frac{1}{3} \)) to find X:
\( X = \frac{1}{3} \begin{bmatrix} -8 & 12 \\ -8 & 18 \end{bmatrix} \)
\( X = \begin{bmatrix} \frac{-8}{3} & \frac{12}{3} \\ \frac{-8}{3} & \frac{18}{3} \end{bmatrix} \)
\( X = \begin{bmatrix} -\frac{8}{3} & 4 \\ -\frac{8}{3} & 6 \end{bmatrix} \)
In simple words: First, move the terms involving P and Q to the other side of the equation. Multiply Q by 2 and P by 3. Then, subtract the multiplied P from the multiplied Q. Finally, divide every number in the resulting matrix by 3 to find matrix X.

🎯 Exam Tip: When isolating an unknown matrix in an equation, remember to treat the "0" on the right side as a null matrix of the same dimensions as the other matrices. Rearrange terms by adding or subtracting matrices just like with numbers.

 

Question 17. Find X and Y if \( X + Y = \begin{bmatrix} 9 & 7 \\ 2 & 11 \end{bmatrix} \) and \( X - Y = \begin{bmatrix} -5 & -6 \\ 4 & 3 \end{bmatrix} \).
Answer: We have a system of two matrix equations with two unknown matrices, X and Y.
Let the equations be:
(1) \( X + Y = \begin{bmatrix} 9 & 7 \\ 2 & 11 \end{bmatrix} \)
(2) \( X - Y = \begin{bmatrix} -5 & -6 \\ 4 & 3 \end{bmatrix} \)

**To find X:** Add equation (1) and equation (2). When we add, \( +Y \) and \( -Y \) will cancel out.
\( (X + Y) + (X - Y) = \begin{bmatrix} 9 & 7 \\ 2 & 11 \end{bmatrix} + \begin{bmatrix} -5 & -6 \\ 4 & 3 \end{bmatrix} \)
\( 2X = \begin{bmatrix} 9+(-5) & 7+(-6) \\ 2+4 & 11+3 \end{bmatrix} \)
\( 2X = \begin{bmatrix} 9-5 & 7-6 \\ 6 & 14 \end{bmatrix} \)
\( 2X = \begin{bmatrix} 4 & 1 \\ 6 & 14 \end{bmatrix} \)
Now, multiply by \( \frac{1}{2} \) to find X:
\( X = \frac{1}{2} \begin{bmatrix} 4 & 1 \\ 6 & 14 \end{bmatrix} \)
\( X = \begin{bmatrix} \frac{4}{2} & \frac{1}{2} \\ \frac{6}{2} & \frac{14}{2} \end{bmatrix} \)
\( X = \begin{bmatrix} 2 & \frac{1}{2} \\ 3 & 7 \end{bmatrix} \)

**To find Y:** Subtract equation (2) from equation (1). When we subtract, \( X \) will cancel out.
\( (X + Y) - (X - Y) = \begin{bmatrix} 9 & 7 \\ 2 & 11 \end{bmatrix} - \begin{bmatrix} -5 & -6 \\ 4 & 3 \end{bmatrix} \)
\( X + Y - X + Y = \begin{bmatrix} 9-(-5) & 7-(-6) \\ 2-4 & 11-3 \end{bmatrix} \)
\( 2Y = \begin{bmatrix} 9+5 & 7+6 \\ -2 & 8 \end{bmatrix} \)
\( 2Y = \begin{bmatrix} 14 & 13 \\ -2 & 8 \end{bmatrix} \)
Now, multiply by \( \frac{1}{2} \) to find Y:
\( Y = \frac{1}{2} \begin{bmatrix} 14 & 13 \\ -2 & 8 \end{bmatrix} \)
\( Y = \begin{bmatrix} \frac{14}{2} & \frac{13}{2} \\ \frac{-2}{2} & \frac{8}{2} \end{bmatrix} \)
\( Y = \begin{bmatrix} 7 & \frac{13}{2} \\ -1 & 4 \end{bmatrix} \)
In simple words: Treat these matrix equations like regular algebra problems. Add the two equations together to find X, then subtract them to find Y. Just make sure to perform all the matrix additions and subtractions correctly, element by element.

🎯 Exam Tip: This method is similar to solving simultaneous linear equations. Adding the two matrix equations eliminates one variable (Y), and subtracting them eliminates the other (X), allowing you to solve for each matrix separately.

 

Question 18. Solve for A and B, when \( 2A + B = \begin{bmatrix} 3 & -4 \\ 2 & 7 \end{bmatrix} \) and \( A – 2B = \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix} \).
Answer: We have a system of two matrix equations with two unknown matrices, A and B.
Let the equations be:
(i) \( 2A + B = \begin{bmatrix} 3 & -4 \\ 2 & 7 \end{bmatrix} \)
(ii) \( A - 2B = \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix} \)
We can use elimination method similar to solving linear equations. Multiply equation (i) by 2:
(iii) \( 2(2A + B) = 2 \begin{bmatrix} 3 & -4 \\ 2 & 7 \end{bmatrix} \)
\( 4A + 2B = \begin{bmatrix} 6 & -8 \\ 4 & 14 \end{bmatrix} \)
Now, add equation (ii) and equation (iii):
\( (A - 2B) + (4A + 2B) = \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix} + \begin{bmatrix} 6 & -8 \\ 4 & 14 \end{bmatrix} \)
\( A - 2B + 4A + 2B = \begin{bmatrix} 4+6 & 3+(-8) \\ 1+4 & 1+14 \end{bmatrix} \)
\( 5A = \begin{bmatrix} 10 & 3-8 \\ 5 & 15 \end{bmatrix} \)
\( 5A = \begin{bmatrix} 10 & -5 \\ 5 & 15 \end{bmatrix} \)
Divide by 5 (or multiply by \( \frac{1}{5} \)) to find A:
\( A = \frac{1}{5} \begin{bmatrix} 10 & -5 \\ 5 & 15 \end{bmatrix} \)
\( A = \begin{bmatrix} \frac{10}{5} & \frac{-5}{5} \\ \frac{5}{5} & \frac{15}{5} \end{bmatrix} \)
\( A = \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix} \)
Now substitute the matrix A into equation (i) to find B:
\( 2 \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix} + B = \begin{bmatrix} 3 & -4 \\ 2 & 7 \end{bmatrix} \)
\( \begin{bmatrix} 4 & -2 \\ 2 & 6 \end{bmatrix} + B = \begin{bmatrix} 3 & -4 \\ 2 & 7 \end{bmatrix} \)
Subtract \( \begin{bmatrix} 4 & -2 \\ 2 & 6 \end{bmatrix} \) from both sides to find B:
\( B = \begin{bmatrix} 3 & -4 \\ 2 & 7 \end{bmatrix} - \begin{bmatrix} 4 & -2 \\ 2 & 6 \end{bmatrix} \)
\( B = \begin{bmatrix} 3-4 & -4-(-2) \\ 2-2 & 7-6 \end{bmatrix} \)
\( B = \begin{bmatrix} -1 & -4+2 \\ 0 & 1 \end{bmatrix} \)
\( B = \begin{bmatrix} -1 & -2 \\ 0 & 1 \end{bmatrix} \)
So, \( A = \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix} \) and \( B = \begin{bmatrix} -1 & -2 \\ 0 & 1 \end{bmatrix} \).
In simple words: This is like solving two regular equations for two unknowns. Multiply one equation by a number so that when you add or subtract the equations, one matrix (like B) disappears. Solve for the remaining matrix (A), then plug it back into one of the original equations to find the other matrix (B).

🎯 Exam Tip: When solving simultaneous matrix equations, use the same algebraic methods (substitution or elimination) as you would for scalar equations. Be careful with scalar multiplication and matrix addition/subtraction steps to avoid calculation errors.

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