ML Aggarwal Class 10 Maths Solutions Chapter 08 Matrices

Exercise 8.1

Question 1. Classify the following matrices:
(i) \( \begin{bmatrix} 2 & -1 \\ 5 & 1 \end{bmatrix} \)
(ii) \( [2 \quad 3 \quad -7] \)
(iii) \( \begin{bmatrix} 3 \\ 0 \\ -1 \end{bmatrix} \)
(iv) \( \begin{bmatrix} 2 & 0 \\ 1 & -4 \\ 0 & 7 \end{bmatrix} \)
(v) \( \begin{bmatrix} 2 & -1 & 7 \\ 2 & 8 & 0 \end{bmatrix} \)
(vi) \( \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \)
Answer:
(i) This is a square matrix with order 2.
(ii) This is a row matrix with order 1 x 3.
(iii) This is a column matrix with order 3 x 1.
(iv) This is a 3 x 2 matrix.
(v) This is a 2 x 3 matrix.
(vi) This is a zero matrix with order 2 x 3.
In simple words: A square matrix has the same number of rows and columns. A row matrix has only one row. A column matrix has only one column. A zero matrix contains all zeros. The order tells you how many rows and columns a matrix has.

Exam Tip: Memorize the standard matrix types - square, row, column, and zero matrices. Always state both the type and order in your answer.

Question 2(i). If a matrix has 4 elements, what are the possible orders it can have?
Answer: The possible orders are 1 x 4, 2 x 2, and 4 x 1. These arise because when you multiply rows by columns, the product must equal 4 - so 1 times 4 equals 4, 2 times 2 equals 4, and 4 times 1 equals 4.
In simple words: If you know how many elements (entries) a matrix has, you can find its possible shapes by thinking about what multiplications give that total.

Exam Tip: Find all factor pairs of the total number of elements - each pair gives one possible order.

Question 2(ii). If a matrix has 8 elements, what are the possible orders it can have?
Answer: The possible orders are 1 x 8, 4 x 2, 2 x 4, and 8 x 1. Since 8 can be factored in multiple ways (1 × 8, 2 × 4, 4 × 2, and 8 × 1), each factorization represents a valid order.
In simple words: Break 8 into two whole number factors - each pair becomes a possible matrix order.

Exam Tip: Always list all factor pairs systematically - do not forget the reversed orders (e.g., 2 x 4 and 4 x 2 are different).

Question 3. Construct a 2 × 2 matrix whose elements aij are given by
(i) aij = 2i - j
(ii) aij = i.j
Answer:
(i) Given aij = 2i - j:

\( a_{11} = 2(1) - 1 = 1, \quad a_{12} = 2(1) - 2 = 0 \)

\( a_{21} = 2(2) - 1 = 3, \quad a_{22} = 2(2) - 2 = 2 \)

Hence, the required matrix = \( \begin{bmatrix} 1 & 0 \\ 3 & 2 \end{bmatrix} \)

(ii) Given aij = i.j:

\( a_{11} = 1 \cdot 1 = 1, \quad a_{12} = 1 \cdot 2 = 2 \)

\( a_{21} = 2 \cdot 1 = 2, \quad a_{22} = 2 \cdot 2 = 4 \)

Hence, the required matrix = \( \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} \)
In simple words: For each position in the matrix, use the formula with that position's row and column numbers to find what number goes there.

Exam Tip: Calculate all four entries separately using the given formula - do not try to memorize a pattern or shortcut.

Question 4. Find the values of x and y if \( \begin{bmatrix} 2x + y \\ 3x - 2y \end{bmatrix} = \begin{bmatrix} 5 \\ 4 \end{bmatrix} \)
Answer: Using the definition of matrix equality, corresponding elements must be equal:

\( 2x + y = 5 \quad [\text{Eq 1}] \)

\( 3x - 2y = 4 \quad [\text{Eq 2}] \)

From Eq 1: \( y = 5 - 2x \)

Substituting into Eq 2:

\( 3x - 2(5 - 2x) = 4 \)

\( 3x - 10 + 4x = 4 \)

\( 7x = 14 \)

\( x = 2 \)

Substituting back: \( y = 5 - 2(2) = 1 \)

Hence, x = 2 and y = 1.
In simple words: Set up equations by matching the elements in the same positions. Then solve the system of equations using substitution or elimination.

Exam Tip: Always use matrix equality correctly - elements in the same position must be equal. Solve systematically and check your answer by substituting back.

Question 5. Find the values of x if \( \begin{bmatrix} 3x + y & -y \\ 2y - x & 3 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ -5 & 3 \end{bmatrix} \)
Answer: Using the definition of matrix equality:

\( -y = 2 \) or \( y = -2 \)

\( 3x + y = 1 \)

Substituting y = -2 into the second equation:

\( 3x - 2 = 1 \)

\( 3x = 3 \)

\( x = 1 \)

Hence, x = 1.
In simple words: Compare elements at matching positions to write equations. Find y first, then use that value to solve for x.

Exam Tip: Look for the simplest equation first - in this case, -y = 2 gives y directly without needing to solve a system.

Question 6. If \( \begin{bmatrix} x + 3 & 4 \\ y - 4 & x + y \end{bmatrix} = \begin{bmatrix} 5 & 4 \\ 3 & 9 \end{bmatrix} \), find the values of x and y.
Answer: Using matrix equality, corresponding elements must match:

\( x + 3 = 5, \quad y - 4 = 3, \quad x + y = 9 \)

\( x = 2, \quad y = 7, \quad x + y = 9 \)

We can verify that x = 2 and y = 7 satisfies all three equations, including the check that x + y = 9.

Hence, x = 2 and y = 7.
In simple words: Equate elements in the same positions to get separate equations. Solve for each variable and then check that your answer works in all the equations.

Exam Tip: When you have three equations with two unknowns, solve the first two and then verify your answer satisfies the third equation as a consistency check.

Question 7. Find the values of x, y and z if \( \begin{bmatrix} x + 2 & 6 \\ 3 & 5z \end{bmatrix} = \begin{bmatrix} -5 & y^2 + y \\ 3 & -20 \end{bmatrix} \)
Answer: Using the definition of matrix equality:

\( x + 2 = -5 \quad \text{or} \quad x = -7 \)

\( y^2 + y = 6 \quad [\text{Eq 1}] \)

\( 5z = -20 \quad \text{or} \quad z = -4 \)

From Eq 1:

\( y^2 + y - 6 = 0 \)

\( y^2 + 3y - 2y - 6 = 0 \)

\( y(y + 3) - 2(y + 3) = 0 \)

\( (y - 2)(y + 3) = 0 \)

\( y = 2 \quad \text{or} \quad y = -3 \)

Hence, the values are:
x = -7, y = 2 and z = -4
OR
x = -7, y = -3 and z = -4
In simple words: Match corresponding matrix entries to create equations. For the quadratic equation in y, factor it to find both solutions.

Exam Tip: When a quadratic arises, do not forget both roots - the answer requires listing both possible solution sets.

Question 8. Find the values of x, y, a and b if \( \begin{bmatrix} x - 2 & y \\ a + 2b & 3a - b \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ 5 & 1 \end{bmatrix} \)
Answer: Using matrix equality:

\( x - 2 = 3 \quad \text{or} \quad x = 5 \)

\( y = 1 \)

\( a + 2b = 5 \quad [\text{Eq 1}] \)

\( 3a - b = 1 \quad [\text{Eq 2}] \)

From Eq 1: \( a = 5 - 2b \)

Substituting into Eq 2:

\( 3(5 - 2b) - b = 1 \)

\( 15 - 6b - b = 1 \)

\( 7b = 14 \)

\( b = 2 \)

Finding a: \( a = 5 - 2(2) = 1 \)

Hence, x = 5, y = 1, a = 1 and b = 2.
In simple words: First find the variables that appear alone (x and y). Then solve the remaining system of two equations in two unknowns.

Exam Tip: Organize your work by handling the simplest equations first, then substitute to eliminate variables from the remaining equations.

Question 9. Find the values of a, b, c and d if \( \begin{bmatrix} a + b & 3 \\ 5 + c & ab \end{bmatrix} = \begin{bmatrix} 6 & d \\ -1 & 8 \end{bmatrix} \)
Answer: Using matrix equality:

\( d = 3 \)

\( 5 + c = -1 \quad \text{or} \quad c = -6 \)

\( a + b = 6 \quad [\text{Eq 1}] \)

\( ab = 8 \quad [\text{Eq 2}] \)

From Eq 1: \( b = 6 - a \)

Substituting into Eq 2:

\( a(6 - a) = 8 \)

\( 6a - a^2 = 8 \)

\( a^2 - 6a + 8 = 0 \)

\( a^2 - 4a - 2a + 8 = 0 \)

\( a(a - 4) - 2(a - 4) = 0 \)

\( (a - 2)(a - 4) = 0 \)

\( a = 2 \quad \text{or} \quad a = 4 \)

Finding corresponding b values:
If a = 2, then b = 6 - 2 = 4
If a = 4, then b = 6 - 4 = 2

Hence, the values are:
a = 2, b = 4, c = -6, d = 3
OR
a = 4, b = 2, c = -6, d = 3
In simple words: Find c and d directly. Then form a quadratic equation from the sum and product relationships, and solve for both variables.

Exam Tip: When you have sum and product of two variables, recognize that they are roots of a quadratic - use \( t^2 - (a+b)t + ab = 0 \).

Exercise 8.2

Question 1. 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:
\( M + 2N = \begin{bmatrix} 2 & 0 \\ 1 & 2 \end{bmatrix} + 2\begin{bmatrix} 2 & 0 \\ -1 & 2 \end{bmatrix} \)

\( = \begin{bmatrix} 2 & 0 \\ 1 & 2 \end{bmatrix} + \begin{bmatrix} 4 & 0 \\ -2 & 4 \end{bmatrix} \)

\( = \begin{bmatrix} 2 + 4 & 0 + 0 \\ 1 - 2 & 2 + 4 \end{bmatrix} \)

\( = \begin{bmatrix} 6 & 0 \\ -1 & 6 \end{bmatrix} \)

Hence, M + 2N = \( \begin{bmatrix} 6 & 0 \\ -1 & 6 \end{bmatrix} \)
In simple words: First multiply N by 2 to get 2N. Then add each entry of M to the matching entry in 2N.

Exam Tip: Perform scalar multiplication before matrix addition. Check each element carefully, especially when dealing with negative numbers.

Question 2. If A = \( \begin{bmatrix} 2 & 0 \\ -3 & 1 \end{bmatrix} \) and B = \( \begin{bmatrix} 0 & 1 \\ -2 & 3 \end{bmatrix} \), find 2A - 3B.
Answer:
\( 2A - 3B = 2\begin{bmatrix} 2 & 0 \\ -3 & 1 \end{bmatrix} - 3\begin{bmatrix} 0 & 1 \\ -2 & 3 \end{bmatrix} \)

\( = \begin{bmatrix} 4 & 0 \\ -6 & 2 \end{bmatrix} - \begin{bmatrix} 0 & 3 \\ -6 & 9 \end{bmatrix} \)

\( = \begin{bmatrix} 4 - 0 & 0 - 3 \\ -6 - (-6) & 2 - 9 \end{bmatrix} \)

\( = \begin{bmatrix} 4 & -3 \\ 0 & -7 \end{bmatrix} \)

Hence, 2A - 3B = \( \begin{bmatrix} 4 & -3 \\ 0 & -7 \end{bmatrix} \)
In simple words: Multiply A by 2 and B by 3 separately. Then subtract the second result from the first by subtracting matching entries.

Exam Tip: Be careful with signs when subtracting - subtracting a negative becomes addition. Double-check by computing 2A and 3B as separate steps first.

Question 3. Simplify: sin A \( \begin{bmatrix} \sin A & -\cos A \\ \cos A & \sin A \end{bmatrix} + \cos A \begin{bmatrix} \cos A & \sin A \\ -\sin A & \cos A \end{bmatrix} \)
Answer:
\( \sin A \begin{bmatrix} \sin A & -\cos A \\ \cos A & \sin A \end{bmatrix} + \cos A \begin{bmatrix} \cos A & \sin A \\ -\sin A & \cos A \end{bmatrix} \)

\( = \begin{bmatrix} \sin^2 A & -\sin A \cos A \\ \sin A \cos A & \sin^2 A \end{bmatrix} + \begin{bmatrix} \cos^2 A & \cos A \sin A \\ -\cos A \sin A & \cos^2 A \end{bmatrix} \)

\( = \begin{bmatrix} \sin^2 A + \cos^2 A & -\sin A \cos A + \cos A \sin A \\ \sin A \cos A - \cos A \sin A & \sin^2 A + \cos^2 A \end{bmatrix} \)

Since \( \sin^2 A + \cos^2 A = 1 \):

\( = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)

Hence, the simplified result = \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
In simple words: Multiply each matrix by its scalar. Then add the matching entries. Use the trigonometric identity to simplify.

Exam Tip: Recognize that the final answer is the identity matrix. This result shows that the two matrices are related through trigonometry - a common pattern in algebra.

Question 4. If A = \( \begin{bmatrix} 1 & 2 \\ -2 & 3 \end{bmatrix} \), B = \( \begin{bmatrix} -2 & -1 \\ 1 & 2 \end{bmatrix} \), and C = \( \begin{bmatrix} 0 & 3 \\ 2 & -1 \end{bmatrix} \), find A + 2B - 3C.
Answer:
\( A + 2B - 3C = \begin{bmatrix} 1 & 2 \\ -2 & 3 \end{bmatrix} + 2\begin{bmatrix} -2 & -1 \\ 1 & 2 \end{bmatrix} - 3\begin{bmatrix} 0 & 3 \\ 2 & -1 \end{bmatrix} \)

\( = \begin{bmatrix} 1 & 2 \\ -2 & 3 \end{bmatrix} + \begin{bmatrix} -4 & -2 \\ 2 & 4 \end{bmatrix} - \begin{bmatrix} 0 & 9 \\ 6 & -3 \end{bmatrix} \)

\( = \begin{bmatrix} 1 + (-4) - 0 & 2 + (-2) - 9 \\ -2 + 2 - 6 & 3 + 4 - (-3) \end{bmatrix} \)

\( = \begin{bmatrix} -3 & -9 \\ -6 & 10 \end{bmatrix} \)

Hence, A + 2B - 3C = \( \begin{bmatrix} -3 & -9 \\ -6 & 10 \end{bmatrix} \)
In simple words: Scale each matrix first. Then combine all three results by adding and subtracting matching entries.

Exam Tip: Use a row-by-row or entry-by-entry approach to avoid errors. Keep track of signs carefully, especially with subtraction.

Question 5. If A = \( \begin{bmatrix} 0 & -1 \\ 1 & 2 \end{bmatrix} \) and B = \( \begin{bmatrix} 1 & 2 \\ -1 & 1 \end{bmatrix} \), find the matrix X if
(i) 3A + X = B
Answer:
(i) From 3A + X = B:

\( X = B - 3A \)

\( X = \begin{bmatrix} 1 & 2 \\ -1 & 1 \end{bmatrix} - 3\begin{bmatrix} 0 & -1 \\ 1 & 2 \end{bmatrix} \)

\( = \begin{bmatrix} 1 & 2 \\ -1 & 1 \end{bmatrix} - \begin{bmatrix} 0 & -3 \\ 3 & 6 \end{bmatrix} \)

\( = \begin{bmatrix} 1 - 0 & 2 - (-3) \\ -1 - 3 & 1 - 6 \end{bmatrix} \)

\( = \begin{bmatrix} 1 & 5 \\ -4 & -5 \end{bmatrix} \)

Hence, X = \( \begin{bmatrix} 1 & 5 \\ -4 & -5 \end{bmatrix} \)
In simple words: Rearrange the equation to isolate X. Calculate 3A, then subtract it from B.

Exam Tip: Always isolate the unknown matrix first before calculating. Verify by substituting back into the original equation.

Question 5(ii). (Continued) If X - 3B = 2A
Answer:
(ii) From X - 3B = 2A:

\( X = 2A + 3B \)

\( X = 2\begin{bmatrix} 0 & -1 \\ 1 & 2 \end{bmatrix} + 3\begin{bmatrix} 1 & 2 \\ -1 & 1 \end{bmatrix} \)

\( = \begin{bmatrix} 0 & -2 \\ 2 & 4 \end{bmatrix} + \begin{bmatrix} 3 & 6 \\ -3 & 3 \end{bmatrix} \)

\( = \begin{bmatrix} 0 + 3 & -2 + 6 \\ 2 + (-3) & 4 + 3 \end{bmatrix} \)

\( = \begin{bmatrix} 3 & 4 \\ -1 & 7 \end{bmatrix} \)

Hence, X = \( \begin{bmatrix} 3 & 4 \\ -1 & 7 \end{bmatrix} \)
In simple words: Rearrange to get X on one side alone. Then multiply both matrices by their scalars and add.

Exam Tip: Both matrix equations are solved the same way - isolate X, then perform the required scalar multiplication and addition or subtraction.

Question 6. Solve the matrix equation \( \begin{bmatrix} 2 & 1 \\ 5 & 0 \end{bmatrix} - 3X = \begin{bmatrix} -7 & 4 \\ 2 & 6 \end{bmatrix} \)
Answer:
\( \begin{bmatrix} 2 & 1 \\ 5 & 0 \end{bmatrix} - 3X = \begin{bmatrix} -7 & 4 \\ 2 & 6 \end{bmatrix} \)

\( 3X = \begin{bmatrix} 2 & 1 \\ 5 & 0 \end{bmatrix} - \begin{bmatrix} -7 & 4 \\ 2 & 6 \end{bmatrix} \)

\( 3X = \begin{bmatrix} 2 - (-7) & 1 - 4 \\ 5 - 2 & 0 - 6 \end{bmatrix} \)

\( 3X = \begin{bmatrix} 9 & -3 \\ 3 & -6 \end{bmatrix} \)

\( X = \frac{1}{3}\begin{bmatrix} 9 & -3 \\ 3 & -6 \end{bmatrix} \)

\( X = \begin{bmatrix} 3 & -1 \\ 1 & -2 \end{bmatrix} \)

Hence, X = \( \begin{bmatrix} 3 & -1 \\ 1 & -2 \end{bmatrix} \)
In simple words: Rearrange to isolate 3X. Subtract to find 3X, then divide every entry by 3.

Exam Tip: When the unknown has a scalar coefficient, move all other terms to one side first, then divide the entire matrix by the scalar.

Question 7. If \( \begin{bmatrix} 1 & 4 \\ -2 & 3 \end{bmatrix} + 2M = 3\begin{bmatrix} 3 & 2 \\ 0 & -3 \end{bmatrix} \), find the matrix M.
Answer:
\( \begin{bmatrix} 1 & 4 \\ -2 & 3 \end{bmatrix} + 2M = 3\begin{bmatrix} 3 & 2 \\ 0 & -3 \end{bmatrix} \)

\( \begin{bmatrix} 1 & 4 \\ -2 & 3 \end{bmatrix} + 2M = \begin{bmatrix} 9 & 6 \\ 0 & -9 \end{bmatrix} \)

\( 2M = \begin{bmatrix} 9 & 6 \\ 0 & -9 \end{bmatrix} - \begin{bmatrix} 1 & 4 \\ -2 & 3 \end{bmatrix} \)

\( 2M = \begin{bmatrix} 9 - 1 & 6 - 4 \\ 0 - (-2) & -9 - 3 \end{bmatrix} \)

\( 2M = \begin{bmatrix} 8 & 2 \\ 2 & -12 \end{bmatrix} \)

\( M = \frac{1}{2}\begin{bmatrix} 8 & 2 \\ 2 & -12 \end{bmatrix} \)

\( M = \begin{bmatrix} 4 & 1 \\ 1 & -6 \end{bmatrix} \)

Hence, M = \( \begin{bmatrix} 4 & 1 \\ 1 & -6 \end{bmatrix} \)
In simple words: Multiply the right side first. Then move known matrices to one side and unknown to the other. Finally, divide by the coefficient of the unknown.

Exam Tip: Handle scalar multiplication on both sides before isolating the unknown matrix. Divide all entries by the scalar coefficient evenly.

Question 8. Given A = \( \begin{bmatrix} 2 & -6 \\ 2 & 0 \end{bmatrix} \), B = \( \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix} \), and C = \( \begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix} \), find the matrix X such that A + 2X = 2B + C.
Answer:
Putting the values of A, B and C in A + 2X = 2B + C:

\( \begin{bmatrix} 2 & -6 \\ 2 & 0 \end{bmatrix} + 2X = 2\begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix} + \begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix} \)

\( 2X = \begin{bmatrix} -6 & 4 \\ 8 & 0 \end{bmatrix} + \begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix} - \begin{bmatrix} 2 & -6 \\ 2 & 0 \end{bmatrix} \)

\( 2X = \begin{bmatrix} -6 + 4 - 2 & 4 + 0 - (-6) \\ 8 + 0 - 2 & 0 + 2 - 0 \end{bmatrix} \)

\( 2X = \begin{bmatrix} -4 & 10 \\ 6 & 2 \end{bmatrix} \)

\( X = \frac{1}{2}\begin{bmatrix} -4 & 10 \\ 6 & 2 \end{bmatrix} \)

\( X = \begin{bmatrix} -2 & 5 \\ 3 & 1 \end{bmatrix} \)

Hence, X = \( \begin{bmatrix} -2 & 5 \\ 3 & 1 \end{bmatrix} \)
In simple words: Rearrange to get 2X by itself on one side. Compute each scalar multiple, then combine by addition and subtraction. Finally divide by 2.

Exam Tip: Organize your work by handling all scalar multiplications first. Then group all known matrices on one side and the unknown on the other.

Question 9. Find X and Y if X + Y = \( \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} \) and X - Y = \( \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \)
Answer:
Given:

\( X + Y = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} \quad [\text{Eq 1}] \)

\( X - Y = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \quad [\text{Eq 2}] \)

Adding both equations:

\( X + Y + X - Y = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} + \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \)

\( 2X = \begin{bmatrix} 7 + 3 & 0 + 0 \\ 2 + 0 & 5 + 3 \end{bmatrix} \)

\( 2X = \begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix} \)

\( X = \frac{1}{2}\begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix} \)

\( X = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix} \)

From Eq 1:

\( Y = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} - \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix} \)

\( Y = \begin{bmatrix} 7 - 5 & 0 - 0 \\ 2 - 1 & 5 - 4 \end{bmatrix} \)

\( Y = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} \)

Hence, X = \( \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix} \) and Y = \( \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} \)
In simple words: Add the two equations to remove Y and find X. Then substitute X back into the first equation to find Y.

Exam Tip: Use the elimination method - adding the equations removes one unknown, simplifying the problem. Always check your answer by substituting back into both original equations.

Question 10. If 2\( \begin{bmatrix} x & 7 \\ 9 & y-5 \end{bmatrix} + \begin{bmatrix} 6 & -7 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 10 & 7 \\ 22 & 15 \end{bmatrix} \), find the values of x and y.
Answer: Multiply the first matrix by 2: \( \begin{bmatrix} 2x & 14 \\ 18 & 2y-10 \end{bmatrix} \). Add the second matrix: \( \begin{bmatrix} 2x+6 & 14-7 \\ 18+4 & 2y-10+5 \end{bmatrix} = \begin{bmatrix} 2x+6 & 7 \\ 22 & 2y-5 \end{bmatrix} \). This equals \( \begin{bmatrix} 10 & 7 \\ 22 & 15 \end{bmatrix} \). Matching corresponding entries: \( 2x+6 = 10 \) gives \( 2x = 4 \), so \( x = 2 \). Also, \( 2y-5 = 15 \) gives \( 2y = 20 \), so \( y = 10 \).
In simple words: Multiply the first matrix by 2, then add it to the second matrix. Compare the result with the given matrix to find x and y.

Exam Tip: Always perform scalar multiplication before addition. Match entries carefully by position to avoid errors.

Question 11. If 2\( \begin{bmatrix} 3 & 4 \\ 5 & x \end{bmatrix} + \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} z & 0 \\ 10 & 5 \end{bmatrix} \), find the values of x, y and z.
Answer: First, multiply the first matrix by 2: \( \begin{bmatrix} 6 & 8 \\ 10 & 2x \end{bmatrix} \). Add the second matrix: \( \begin{bmatrix} 6+1 & 8+y \\ 10+0 & 2x+1 \end{bmatrix} = \begin{bmatrix} 7 & 8+y \\ 10 & 2x+1 \end{bmatrix} \). This must equal \( \begin{bmatrix} z & 0 \\ 10 & 5 \end{bmatrix} \). Comparing entries: \( 7 = z \), \( 8+y = 0 \), and \( 2x+1 = 5 \). From these: \( z = 7 \), \( y = -8 \), and \( x = 2 \).
In simple words: Double the first matrix, add the second matrix, then compare each position with the answer matrix to get the three unknowns.

Exam Tip: Write out the addition clearly before comparing. One small mistake in positioning can lead to incorrect values.

Question 12. If \( \begin{bmatrix} 5 & 2 \\ -1 & y+1 \end{bmatrix} - 2\begin{bmatrix} 1 & 2x-1 \\ 3 & -2 \end{bmatrix} = \begin{bmatrix} 3 & -8 \\ -7 & 2 \end{bmatrix} \), find the values of x and y.
Answer: Multiply the second matrix by 2: \( 2\begin{bmatrix} 1 & 2x-1 \\ 3 & -2 \end{bmatrix} = \begin{bmatrix} 2 & 4x-2 \\ 6 & -4 \end{bmatrix} \). Now subtract: \( \begin{bmatrix} 5 & 2 \\ -1 & y+1 \end{bmatrix} - \begin{bmatrix} 2 & 4x-2 \\ 6 & -4 \end{bmatrix} = \begin{bmatrix} 5-2 & 2-(4x-2) \\ -1-6 & y+1-(-4) \end{bmatrix} = \begin{bmatrix} 3 & 4-4x \\ -7 & y+5 \end{bmatrix} \). This equals \( \begin{bmatrix} 3 & -8 \\ -7 & 2 \end{bmatrix} \). From the second element of the first row: \( 4-4x = -8 \), so \( 4x = 12 \), giving \( x = 3 \). From the second element of the second row: \( y+5 = 2 \), so \( y = -3 \).
In simple words: Scale the second matrix, subtract it from the first, and match the result with the given answer to find x and y.

Exam Tip: Watch the signs carefully when subtracting, especially with negative numbers in parentheses.

Question 13. 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 values of a, b and c.
Answer: Combine the first two matrices by adding: \( \begin{bmatrix} a+2 & 3+b \\ 5 & 0 \end{bmatrix} \). Subtract the third matrix: \( \begin{bmatrix} a+2-1 & 3+b-1 \\ 5-(-2) & 0-c \end{bmatrix} = \begin{bmatrix} a+1 & b+2 \\ 7 & -c \end{bmatrix} \). This equals \( \begin{bmatrix} 5 & 0 \\ 7 & 3 \end{bmatrix} \). Matching elements: \( a+1 = 5 \) gives \( a = 4 \), \( b+2 = 0 \) gives \( b = -2 \), and \( -c = 3 \) gives \( c = -3 \).
In simple words: Perform addition and subtraction in order from left to right, then match each position with the result matrix.

Exam Tip: Keep track of all operations step by step. A mistake in one step affects all subsequent comparisons.

Question 14. If \( A = \begin{bmatrix} 2 & a \\ -3 & 5 \end{bmatrix} \), \( B = \begin{bmatrix} -2 & 3 \\ 7 & b \end{bmatrix} \), \( C = \begin{bmatrix} c & 9 \\ -1 & -11 \end{bmatrix} \) and 5A + 2B = C, find the values of a, b and c.
Answer: Calculate 5A: \( 5\begin{bmatrix} 2 & a \\ -3 & 5 \end{bmatrix} = \begin{bmatrix} 10 & 5a \\ -15 & 25 \end{bmatrix} \). Calculate 2B: \( 2\begin{bmatrix} -2 & 3 \\ 7 & b \end{bmatrix} = \begin{bmatrix} -4 & 6 \\ 14 & 2b \end{bmatrix} \). Add them: \( \begin{bmatrix} 10-4 & 5a+6 \\ -15+14 & 25+2b \end{bmatrix} = \begin{bmatrix} 6 & 5a+6 \\ -1 & 25+2b \end{bmatrix} \). This equals C: \( \begin{bmatrix} c & 9 \\ -1 & -11 \end{bmatrix} \). Matching elements: \( 6 = c \), \( 5a+6 = 9 \) gives \( 5a = 3 \) so \( a = \frac{3}{5} \), and \( 25+2b = -11 \) gives \( 2b = -36 \) so \( b = -18 \).
In simple words: Find 5A and 2B separately, add them together, and compare the result with matrix C.

Exam Tip: Work with each matrix operation carefully. Mistakes in scalar multiplication carry through to the final answer.

Exercise 8.3

Question 1. If \( A = \begin{bmatrix} 3 & 5 \\ 4 & -2 \end{bmatrix} \) and \( B = \begin{bmatrix} 2 \\ 4 \end{bmatrix} \), is the product AB possible? Give a reason. If yes find AB.
Answer: Yes, the product is possible. Matrix A is 2 - 3 (2 rows, 2 columns) and matrix B is 2 - 1 (2 rows, 1 column). For multiplication to be possible, the number of columns in the first matrix must equal the number of rows in the second matrix. Here, A has 2 columns and B has 2 rows, so the rule is satisfied. The result will be a 2 - 1 matrix. Computing: \( AB = \begin{bmatrix} 3 \times 2 + 5 \times 4 \\ 4 \times 2 + (-2) \times 4 \end{bmatrix} = \begin{bmatrix} 6+20 \\ 8-8 \end{bmatrix} = \begin{bmatrix} 26 \\ 0 \end{bmatrix} \).
In simple words: Two matrices can be multiplied when the number of columns in the first equals the number of rows in the second. Then multiply and add as usual.

Exam Tip: Always check the dimensions before attempting multiplication. Stating the reason why multiplication is possible earns marks even if the arithmetic has a small error.

Question 2. If \( A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix} \), \( B = \begin{bmatrix} 1 & -1 \\ -3 & 2 \end{bmatrix} \), find AB and BA. Is AB = BA?
Answer: Calculate AB: \( AB = \begin{bmatrix} 2 \times 1 + 5 \times (-3) & 2 \times (-1) + 5 \times 2 \\ 1 \times 1 + 3 \times (-3) & 1 \times (-1) + 3 \times 2 \end{bmatrix} = \begin{bmatrix} 2-15 & -2+10 \\ 1-9 & -1+6 \end{bmatrix} = \begin{bmatrix} -13 & 8 \\ -8 & 5 \end{bmatrix} \). Calculate BA: \( BA = \begin{bmatrix} 1 \times 2 + (-1) \times 1 & 1 \times 5 + (-1) \times 3 \\ (-3) \times 2 + 2 \times 1 & (-3) \times 5 + 2 \times 3 \end{bmatrix} = \begin{bmatrix} 2-1 & 5-3 \\ -6+2 & -15+6 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ -4 & -9 \end{bmatrix} \). Since \( \begin{bmatrix} -13 & 8 \\ -8 & 5 \end{bmatrix} \neq \begin{bmatrix} 1 & 2 \\ -4 & -9 \end{bmatrix} \), we have AB ≠ BA. Matrix multiplication is not commutative in general.
In simple words: Calculate both AB and BA separately, then compare them. Usually they are different.

Exam Tip: Remember that matrix multiplication is not commutative. Always calculate both products when asked to compare them.

Question 3. If \( A = \begin{bmatrix} 3 & 7 \\ 2 & 4 \end{bmatrix} \), \( B = \begin{bmatrix} 0 & 2 \\ 5 & 3 \end{bmatrix} \) and \( C = \begin{bmatrix} 1 & -5 \\ -4 & 6 \end{bmatrix} \), find AB - 5C.
Answer: First, calculate AB: \( AB = \begin{bmatrix} 3 \times 0 + 7 \times 5 & 3 \times 2 + 7 \times 3 \\ 2 \times 0 + 4 \times 5 & 2 \times 2 + 4 \times 3 \end{bmatrix} = \begin{bmatrix} 0+35 & 6+21 \\ 0+20 & 4+12 \end{bmatrix} = \begin{bmatrix} 35 & 27 \\ 20 & 16 \end{bmatrix} \). Next, multiply C by 5: \( 5C = \begin{bmatrix} 5 & -25 \\ -20 & 30 \end{bmatrix} \). Subtract: \( AB - 5C = \begin{bmatrix} 35-5 & 27-(-25) \\ 20-(-20) & 16-30 \end{bmatrix} = \begin{bmatrix} 30 & 52 \\ 40 & -14 \end{bmatrix} \).
In simple words: Multiply A and B, then multiply C by 5, and finally subtract the scaled C from the product AB.

Exam Tip: Do not try to simplify before calculating. Perform operations in order: first AB, then 5C, then the subtraction.

Question 4. If \( A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \), find A(BA).
Answer: First, calculate BA: \( BA = \begin{bmatrix} 2 \times 1 + 1 \times 2 & 2 \times 2 + 1 \times 1 \\ 1 \times 1 + 2 \times 2 & 1 \times 2 + 2 \times 1 \end{bmatrix} = \begin{bmatrix} 2+2 & 4+1 \\ 1+4 & 2+2 \end{bmatrix} = \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix} \). Now, calculate A(BA) = A × BA: \( A(BA) = \begin{bmatrix} 1 \times 4 + 2 \times 5 & 1 \times 5 + 2 \times 4 \\ 2 \times 4 + 1 \times 5 & 2 \times 5 + 1 \times 4 \end{bmatrix} = \begin{bmatrix} 4+10 & 5+8 \\ 8+5 & 10+4 \end{bmatrix} = \begin{bmatrix} 14 & 13 \\ 13 & 14 \end{bmatrix} \).
In simple words: First find BA, then multiply that result by A from the left to get A(BA).

Exam Tip: Be careful with the order. A(BA) means first find BA, then multiply A by that result, not the other way around.

Question 5. Given the matrices \( A = \begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix} \), \( B = \begin{bmatrix} 3 & 4 \\ -1 & -2 \end{bmatrix} \) and \( C = \begin{bmatrix} -3 & 1 \\ 0 & -2 \end{bmatrix} \), find the products of (i) ABC (ii) ACB and state whether they are equal.
Answer:
(i) First, find AB: \( AB = \begin{bmatrix} 2 \times 3 + 1 \times (-1) & 2 \times 4 + 1 \times (-2) \\ 4 \times 3 + 2 \times (-1) & 4 \times 4 + 2 \times (-2) \end{bmatrix} = \begin{bmatrix} 6-1 & 8-2 \\ 12-2 & 16-4 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 10 & 12 \end{bmatrix} \). Then multiply by C: \( ABC = \begin{bmatrix} 5 \times (-3) + 6 \times 0 & 5 \times 1 + 6 \times (-2) \\ 10 \times (-3) + 12 \times 0 & 10 \times 1 + 12 \times (-2) \end{bmatrix} = \begin{bmatrix} -15+0 & 5-12 \\ -30+0 & 10-24 \end{bmatrix} = \begin{bmatrix} -15 & -7 \\ -30 & -14 \end{bmatrix} \).
(ii) First, find AC: \( AC = \begin{bmatrix} 2 \times (-3) + 1 \times 0 & 2 \times 1 + 1 \times (-2) \\ 4 \times (-3) + 2 \times 0 & 4 \times 1 + 2 \times (-2) \end{bmatrix} = \begin{bmatrix} -6+0 & 2-2 \\ -12+0 & 4-4 \end{bmatrix} = \begin{bmatrix} -6 & 0 \\ -12 & 0 \end{bmatrix} \). Then multiply by B: \( ACB = \begin{bmatrix} (-6) \times 3 + 0 \times (-1) & (-6) \times 4 + 0 \times (-2) \\ (-12) \times 3 + 0 \times (-1) & (-12) \times 4 + 0 \times (-2) \end{bmatrix} = \begin{bmatrix} -18+0 & -24+0 \\ -36+0 & -48+0 \end{bmatrix} = \begin{bmatrix} -18 & -24 \\ -36 & -48 \end{bmatrix} \). Since \( \begin{bmatrix} -15 & -7 \\ -30 & -14 \end{bmatrix} \neq \begin{bmatrix} -18 & -24 \\ -36 & -48 \end{bmatrix} \), we have ABC ≠ ACB.
In simple words: Find ABC by multiplying A and B first, then multiply the result by C. Find ACB by multiplying A and C first, then multiply by B. The two results are different.

Exam Tip: The order matters in matrix multiplication. Even though the operation is associative, changing which matrices multiply first gives different results when you regroup with different matrices.

Question 6. Evaluate: \( \begin{bmatrix} 4\sin 30° & 2\cos 60° \\ \sin 90° & 2\cos 0° \end{bmatrix} \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix} \).
Answer: Substitute the trigonometric values: \( \sin 30° = \frac{1}{2} \), \( \cos 60° = \frac{1}{2} \), \( \sin 90° = 1 \), and \( \cos 0° = 1 \). The first matrix becomes \( \begin{bmatrix} 4 \times \frac{1}{2} & 2 \times \frac{1}{2} \\ 1 & 2 \times 1 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \). Now multiply: \( \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix} = \begin{bmatrix} 2 \times 4 + 1 \times 5 & 2 \times 5 + 1 \times 4 \\ 1 \times 4 + 2 \times 5 & 1 \times 5 + 2 \times 4 \end{bmatrix} = \begin{bmatrix} 8+5 & 10+4 \\ 4+10 & 5+8 \end{bmatrix} = \begin{bmatrix} 13 & 14 \\ 14 & 13 \end{bmatrix} \).
In simple words: Replace the trigonometric expressions with their numerical values, then multiply the resulting matrices.

Exam Tip: Always evaluate trigonometric values first before starting matrix multiplication to avoid confusion.

Question 7. If \( A = \begin{bmatrix} -1 & 3 \\ 2 & 4 \end{bmatrix} \) and \( B = \begin{bmatrix} 2 & -3 \\ -4 & -6 \end{bmatrix} \), find the matrix AB + BA.
Answer: Calculate AB: \( AB = \begin{bmatrix} (-1) \times 2 + 3 \times (-4) & (-1) \times (-3) + 3 \times (-6) \\ 2 \times 2 + 4 \times (-4) & 2 \times (-3) + 4 \times (-6) \end{bmatrix} = \begin{bmatrix} -2-12 & 3-18 \\ 4-16 & -6-24 \end{bmatrix} = \begin{bmatrix} -14 & -15 \\ -12 & -30 \end{bmatrix} \). Calculate BA: \( BA = \begin{bmatrix} 2 \times (-1) + (-3) \times 2 & 2 \times 3 + (-3) \times 4 \\ (-4) \times (-1) + (-6) \times 2 & (-4) \times 3 + (-6) \times 4 \end{bmatrix} = \begin{bmatrix} -2-6 & 6-12 \\ 4-12 & -12-24 \end{bmatrix} = \begin{bmatrix} -8 & -6 \\ -8 & -36 \end{bmatrix} \). Add them: \( AB + BA = \begin{bmatrix} -14+(-8) & -15+(-6) \\ -12+(-8) & -30+(-36) \end{bmatrix} = \begin{bmatrix} -22 & -21 \\ -20 & -66 \end{bmatrix} \).
In simple words: Find both AB and BA separately by multiplying in different orders, then add the two results together.

Exam Tip: Even when finding AB + BA, calculate each product fully before adding. This reduces errors.

Question 8. If \( A = \begin{bmatrix} 1 & -2 \\ 2 & -1 \end{bmatrix} \) and \( B = \begin{bmatrix} 3 & 2 \\ -2 & 1 \end{bmatrix} \), find 2B - A².
Answer: Calculate 2B: \( 2B = 2\begin{bmatrix} 3 & 2 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 4 \\ -4 & 2 \end{bmatrix} \). Calculate A²: \( A² = \begin{bmatrix} 1 & -2 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} 1 \times 1 + (-2) \times 2 & 1 \times (-2) + (-2) \times (-1) \\ 2 \times 1 + (-1) \times 2 & 2 \times (-2) + (-1) \times (-1) \end{bmatrix} = \begin{bmatrix} 1-4 & -2+2 \\ 2-2 & -4+1 \end{bmatrix} = \begin{bmatrix} -3 & 0 \\ 0 & -3 \end{bmatrix} \). Subtract: \( 2B - A² = \begin{bmatrix} 6 & 4 \\ -4 & 2 \end{bmatrix} - \begin{bmatrix} -3 & 0 \\ 0 & -3 \end{bmatrix} = \begin{bmatrix} 6-(-3) & 4-0 \\ -4-0 & 2-(-3) \end{bmatrix} = \begin{bmatrix} 9 & 4 \\ -4 & 5 \end{bmatrix} \).
In simple words: Find 2B by doubling all entries. Find A² by multiplying A by itself. Then subtract the second result from the first.

Exam Tip: When squaring a matrix, ensure you multiply the matrix by itself correctly. Watch for sign changes when subtracting A² from 2B.

Question 9. If \( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \), \( B = \begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix} \) and \( C = \begin{bmatrix} 5 & 1 \\ 7 & 4 \end{bmatrix} \), compute (i) A(B + C) (ii) (B + C)A.
Answer:
(i) First, find B + C: \( B + C = \begin{bmatrix} 2+5 & 1+1 \\ 4+7 & 2+4 \end{bmatrix} = \begin{bmatrix} 7 & 2 \\ 11 & 6 \end{bmatrix} \). Multiply by A: \( A(B+C) = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 7 & 2 \\ 11 & 6 \end{bmatrix} = \begin{bmatrix} 1 \times 7 + 2 \times 11 & 1 \times 2 + 2 \times 6 \\ 3 \times 7 + 4 \times 11 & 3 \times 2 + 4 \times 6 \end{bmatrix} = \begin{bmatrix} 7+22 & 2+12 \\ 21+44 & 6+24 \end{bmatrix} = \begin{bmatrix} 29 & 14 \\ 65 & 30 \end{bmatrix} \).
(ii) B + C is the same as above: \( \begin{bmatrix} 7 & 2 \\ 11 & 6 \end{bmatrix} \). Multiply by A from the right: \( (B+C)A = \begin{bmatrix} 7 & 2 \\ 11 & 6 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 7 \times 1 + 2 \times 3 & 7 \times 2 + 2 \times 4 \\ 11 \times 1 + 6 \times 3 & 11 \times 2 + 6 \times 4 \end{bmatrix} = \begin{bmatrix} 7+6 & 14+8 \\ 11+18 & 22+24 \end{bmatrix} = \begin{bmatrix} 13 & 22 \\ 29 & 46 \end{bmatrix} \).
In simple words: For part (i), add B and C first, then multiply A from the left. For part (ii), add B and C, then multiply A from the right. The results are different because multiplication is not commutative.

Exam Tip: The distributive property holds for matrices, so A(B + C) = AB + AC and (B + C)A = BA + CA, but A(B + C) is usually not equal to (B + C)A.

Question 10. If A = \( \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} \), B = \( \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} \) and C = \( \begin{bmatrix} 1 & 3 \\ 3 & 1 \end{bmatrix} \), find the matrix C(B - A).
Answer: Start by subtracting matrix A from matrix B. The subtraction gives \( B - A = \begin{bmatrix} 1 & -1 \\ 1 & -1 \end{bmatrix} \). Next, multiply matrix C by this result. Carrying out the multiplication: \( C(B - A) = \begin{bmatrix} 1 & 3 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 4 & -4 \\ 4 & -4 \end{bmatrix} \).
In simple words: Subtract A from B, then multiply the outcome by C. When you follow these steps, the final answer is a matrix with 4 and -4 in the correct positions.

Exam Tip: Always perform matrix subtraction element-by-element first, then use the standard row-by-column rule for multiplication. Double-check your arithmetic at each step to avoid sign errors.

Question 11(i). Let A = \( \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \) and B = \( \begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix} \), find A² + AB + B².
Answer: First, calculate A² by multiplying A by itself: \( A^2 = \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix} \). Next, compute AB: \( AB = \begin{bmatrix} 2 & 3 \\ 3 & 6 \end{bmatrix} \). Then find B² by multiplying B by itself: \( B^2 = \begin{bmatrix} 1 & 6 \\ -2 & -3 \end{bmatrix} \). Finally, add all three matrices together: \( A^2 + AB + B^2 = \begin{bmatrix} 4 & 9 \\ 5 & 4 \end{bmatrix} \).
In simple words: Square matrix A, square matrix B, and multiply A and B together. Then add all three results to get your final answer.

Exam Tip: Work through each operation separately and verify your intermediate results before combining. Matrix multiplication is not commutative, so AB is different from BA.

Question 11(ii). If A = \( \begin{bmatrix} 3 & 0 \\ 5 & 1 \end{bmatrix} \) and B = \( \begin{bmatrix} -4 & 2 \\ 1 & 0 \end{bmatrix} \), find A² - 2AB + B².
Answer: Compute A² by multiplying A by itself: \( A^2 = \begin{bmatrix} 9 & 0 \\ 20 & 1 \end{bmatrix} \). Next, calculate 2AB: first find AB, then multiply by 2, giving \( 2AB = \begin{bmatrix} -24 & 12 \\ -38 & 20 \end{bmatrix} \). Calculate B² by multiplying B by itself: \( B^2 = \begin{bmatrix} 18 & -8 \\ -4 & 2 \end{bmatrix} \). Now substitute into the expression: \( A^2 - 2AB + B^2 = \begin{bmatrix} 9 & 0 \\ 20 & 1 \end{bmatrix} - \begin{bmatrix} -24 & 12 \\ -38 & 20 \end{bmatrix} + \begin{bmatrix} 18 & -8 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} 51 & -20 \\ 54 & -17 \end{bmatrix} \).
In simple words: Find A squared, find B squared, and multiply A times B then double it. Substitute these into the formula to arrive at the final result.

Exam Tip: Keep track of signs when subtracting matrices - a negative sign in front of a matrix reverses all its elements' signs. Verify that matrix dimensions match before each operation.

Question 12. Let A = \( \begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix} \), B = \( \begin{bmatrix} 4 & 1 \\ -3 & -2 \end{bmatrix} \) and C = \( \begin{bmatrix} -3 & 2 \\ -1 & 4 \end{bmatrix} \), find A² + AC - 5B.
Answer: Begin by calculating A²: \( A^2 = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} \). Next, find the product AC: \( AC = \begin{bmatrix} -7 & 8 \\ 2 & -8 \end{bmatrix} \). Then compute 5B by multiplying each element of B by 5: \( 5B = \begin{bmatrix} 20 & 5 \\ -15 & -10 \end{bmatrix} \). Finally, combine these results: \( A^2 + AC - 5B = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} + \begin{bmatrix} -7 & 8 \\ 2 & -8 \end{bmatrix} - \begin{bmatrix} 20 & 5 \\ -15 & -10 \end{bmatrix} = \begin{bmatrix} -23 & 3 \\ 17 & 6 \end{bmatrix} \).
In simple words: Square matrix A, multiply A by C, and scale matrix B by 5. Then add and subtract these results in the order given.

Exam Tip: Scalar multiplication (like 5B) is straightforward - multiply every element. When adding and subtracting matrices, work element-by-element and stay organized to prevent careless mistakes.

Question 13. If A = \( \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix} \), B = \( \begin{bmatrix} 0 & 4 \\ -1 & 7 \end{bmatrix} \) and C = \( \begin{bmatrix} 1 & 0 \\ -1 & 4 \end{bmatrix} \), find AC + B² - 10C.
Answer: Start by multiplying A and C: \( AC = \begin{bmatrix} -1 & 12 \\ -2 & 28 \end{bmatrix} \). Next, calculate B² by multiplying B by itself: \( B^2 = \begin{bmatrix} -4 & 28 \\ -7 & 45 \end{bmatrix} \). Then compute 10C by scaling C: \( 10C = \begin{bmatrix} 10 & 0 \\ -10 & 40 \end{bmatrix} \). Combine all three: \( AC + B^2 - 10C = \begin{bmatrix} -1 & 12 \\ -2 & 28 \end{bmatrix} + \begin{bmatrix} -4 & 28 \\ -7 & 45 \end{bmatrix} - \begin{bmatrix} 10 & 0 \\ -10 & 40 \end{bmatrix} = \begin{bmatrix} -15 & 40 \\ 1 & 33 \end{bmatrix} \).
In simple words: Multiply A by C, square matrix B, and multiply C by 10. Add the first two results and subtract the third to get your answer.

Exam Tip: Organize your working by clearly labeling each intermediate result. This makes it easier to spot errors and helps the examiner follow your logic.

Question 14. If A = \( \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \), find A² and A³. Also state which of these is equal to A.
Answer: Calculate A² by multiplying A by itself: \( A^2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \). Notice that A² is the identity matrix. Now find A³ by multiplying A² by A: \( A^3 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \). Comparing the three matrices, A³ is equal to A. This happens because A is an involutory matrix - when squared it produces the identity, so the next power restores the original matrix.
In simple words: When you multiply this matrix by itself, you get the identity matrix. When you multiply it one more time, you get back to the original matrix.

Exam Tip: Recognize special matrices like involutory matrices (where A² = I) - they appear frequently in exam questions and have predictable patterns for higher powers.

Question 15. If X = \( \begin{bmatrix} 4 & 1 \\ -1 & 2 \end{bmatrix} \), show that 6X - X² = 9I where I is the unit matrix.
Answer: We need to prove that 6X - X² = 9I. First, calculate X²: \( X^2 = \begin{bmatrix} 4 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 4 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 15 & 6 \\ -6 & 3 \end{bmatrix} \). Next, find 6X: \( 6X = \begin{bmatrix} 24 & 6 \\ -6 & 12 \end{bmatrix} \). Now compute 6X - X²: \( 6X - X^2 = \begin{bmatrix} 24 & 6 \\ -6 & 12 \end{bmatrix} - \begin{bmatrix} 15 & 6 \\ -6 & 3 \end{bmatrix} = \begin{bmatrix} 9 & 0 \\ 0 & 9 \end{bmatrix} \). Finally, note that \( 9I = \begin{bmatrix} 9 & 0 \\ 0 & 9 \end{bmatrix} \). Since both sides are equal, the identity is proven.
In simple words: Calculate X squared, multiply X by 6, and subtract. The result matches 9 times the identity matrix, confirming the relationship.

Exam Tip: When proving matrix identities, show all intermediate steps clearly. The examiner needs to see your calculation of X² and your subtraction to award full marks.

Question 16. Show that \( \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \) is a solution of the matrix equation X² - 2X - 3I = 0 where I is the unit matrix of order 2.
Answer: We are given that I = \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \) and X = \( \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \). First, calculate X²: \( X^2 = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 5 & 4 \\ 4 & 5 \end{bmatrix} \). Now compute the left side: \( X^2 - 2X - 3I = \begin{bmatrix} 5 & 4 \\ 4 & 5 \end{bmatrix} - \begin{bmatrix} 2 & 4 \\ 4 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \). The left side equals the zero matrix (right side), so the equation is satisfied.
In simple words: Calculate X squared, double the X matrix, and subtract both from 3 times the identity. If you get all zeros, then X is a solution.

Exam Tip: Always substitute explicitly and show the complete calculation. Do not just state that "it works" - demonstrate the algebra step-by-step to earn full credit.

Question 17. Find the matrix X of order 2 x 2 which satisfies the equation \( \begin{bmatrix} 3 & 7 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} 0 & 2 \\ 5 & 3 \end{bmatrix} + 2X = \begin{bmatrix} 1 & -5 \\ -4 & 6 \end{bmatrix} \).
Answer: Start by multiplying the two matrices on the left: \( \begin{bmatrix} 3 & 7 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} 0 & 2 \\ 5 & 3 \end{bmatrix} = \begin{bmatrix} 35 & 27 \\ 20 & 16 \end{bmatrix} \). Now the equation becomes: \( \begin{bmatrix} 35 & 27 \\ 20 & 16 \end{bmatrix} + 2X = \begin{bmatrix} 1 & -5 \\ -4 & 6 \end{bmatrix} \). Isolate 2X: \( 2X = \begin{bmatrix} 1 & -5 \\ -4 & 6 \end{bmatrix} - \begin{bmatrix} 35 & 27 \\ 20 & 16 \end{bmatrix} = \begin{bmatrix} -34 & -32 \\ -24 & -10 \end{bmatrix} \). Divide by 2 to find X: \( X = \begin{bmatrix} -17 & -16 \\ -12 & -5 \end{bmatrix} \).
In simple words: Multiply the first two matrices, move everything except 2X to the other side, then divide all entries by 2.

Exam Tip: When solving for an unknown matrix, treat it like an algebraic equation - isolate the matrix term first, then perform any remaining operations like scalar division.

Question 18. If A = \( \begin{bmatrix} 1 & 1 \\ x & x \end{bmatrix} \), find the value of x so that A² = O.
Answer: We need to find x such that A² equals the zero matrix. Calculate A² by multiplying A by itself: \( A^2 = \begin{bmatrix} 1 & 1 \\ x & x \end{bmatrix} \begin{bmatrix} 1 & 1 \\ x & x \end{bmatrix} = \begin{bmatrix} 1 + x & 1 + x \\ x + x^2 & x + x^2 \end{bmatrix} \). For A² to equal the zero matrix, all entries must be zero. From the first entry: \( 1 + x = 0 \), which gives x = -1. Verify this works in the second entry: \( x + x^2 = -1 + (-1)^2 = -1 + 1 = 0 \). Both conditions are satisfied, so x = -1.
In simple words: Square the matrix and set all entries equal to zero. Solve the equations you get to find the value of x.

Exam Tip: Always verify your answer by substituting it back into at least one other equation from the matrix. This confirms that your value truly makes A² equal to the zero matrix.

Question 19. Find x and y, if \( \begin{bmatrix} 2x & x \\ y & 3y \end{bmatrix} \begin{bmatrix} 3 \\ 2 \end{bmatrix} = \begin{bmatrix} 16 \\ 9 \end{bmatrix} \).
Answer: Perform the matrix multiplication on the left side: \( \begin{bmatrix} 2x & x \\ y & 3y \end{bmatrix} \begin{bmatrix} 3 \\ 2 \end{bmatrix} = \begin{bmatrix} 6x + 2x \\ 3y + 6y \end{bmatrix} = \begin{bmatrix} 8x \\ 9y \end{bmatrix} \). This result equals the right side: \( \begin{bmatrix} 8x \\ 9y \end{bmatrix} = \begin{bmatrix} 16 \\ 9 \end{bmatrix} \). Comparing the entries, we get 8x = 16 and 9y = 9. Solving these: x = 2 and y = 1.
In simple words: Multiply the matrix by the column vector, and set the result equal to what you want. Then solve for x and y.

Exam Tip: When comparing matrices, match entry-by-entry and solve the resulting equations independently. Double-check your arithmetic on both equations to ensure consistency.

Question 20. Find the values of x and y if \( \begin{bmatrix} x + y & y \\ 2x & x - y \end{bmatrix} \begin{bmatrix} 2 \\ -1 \end{bmatrix} = \begin{bmatrix} 3 \\ 2 \end{bmatrix} \).
Answer: Multiply the matrix by the column vector: \( \begin{bmatrix} x + y & y \\ 2x & x - y \end{bmatrix} \begin{bmatrix} 2 \\ -1 \end{bmatrix} = \begin{bmatrix} 2(x + y) - y \\ 4x - (x - y) \end{bmatrix} = \begin{bmatrix} 2x + y \\ 3x + y \end{bmatrix} \). This equals the right side: \( \begin{bmatrix} 2x + y \\ 3x + y \end{bmatrix} = \begin{bmatrix} 3 \\ 2 \end{bmatrix} \). From the first entry: 2x + y = 3, so y = 3 - 2x. From the second entry: 3x + y = 2. Substitute the expression for y: 3x + (3 - 2x) = 2, which simplifies to x + 3 = 2, giving x = -1. Now find y: y = 3 - 2(-1) = 3 + 2 = 5.
In simple words: Multiply out the left side to get two equations. Solve one for y, substitute into the other, and find both variables.

Exam Tip: Use substitution to reduce two equations to one variable. Always substitute back to verify both values satisfy the original equations.

Question 21. If \( \begin{bmatrix} 1 & 2 \\ 3 & 3 \end{bmatrix} \begin{bmatrix} x & 0 \\ 0 & y \end{bmatrix} = \begin{bmatrix} x & 0 \\ 9 & 0 \end{bmatrix} \), find the values of x and y.
Answer: Multiply the matrices on the left: \( \begin{bmatrix} 1 & 2 \\ 3 & 3 \end{bmatrix} \begin{bmatrix} x & 0 \\ 0 & y \end{bmatrix} = \begin{bmatrix} x + 0 & 0 + 2y \\ 3x + 0 & 0 + 3y \end{bmatrix} = \begin{bmatrix} x & 2y \\ 3x & 3y \end{bmatrix} \). This equals the right side: \( \begin{bmatrix} x & 2y \\ 3x & 3y \end{bmatrix} = \begin{bmatrix} x & 0 \\ 9 & 0 \end{bmatrix} \). From the (1,2) entry: 2y = 0, so y = 0. From the (2,1) entry: 3x = 9, so x = 3. Verify with the (2,2) entry: 3y = 3(0) = 0, which matches the right side.
In simple words: Multiply the two matrices and compare each entry with the result matrix. Solve the equations you get.

Exam Tip: Pay attention to zero entries in the result matrix - they provide direct equations like 2y = 0 that quickly determine variables. Verify your answer in all entries before concluding.

Question 22. 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: When a matrix is multiplied by the identity matrix, the product equals the original matrix. Applying matrix multiplication on the right side:
\( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} a \times 1 + b \times 0 & a \times 0 + b \times 1 \\ c \times 1 + d \times 0 & c \times 0 + d \times 1 \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \)
By comparing both sides, we get a = 3, b = 4, c = 2, and d = 5.
In simple words: When you multiply any matrix by the identity matrix, you get the same matrix back. So a, b, c, d must match the first matrix exactly.

Exam Tip: Remember that multiplying by the identity matrix \( I \) is like multiplying a number by 1 - the result stays unchanged. Use this property to identify corresponding elements quickly.

Question 23. Find the value of x given that A² = B where, \( A = \begin{bmatrix} 2 & 12 \\ 0 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} 4 & x \\ 0 & 1 \end{bmatrix} \).
Answer: Start by calculating A²:
\( A^2 = \begin{bmatrix} 2 & 12 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & 12 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 \times 2 + 12 \times 0 & 2 \times 12 + 12 \times 1 \\ 0 \times 2 + 1 \times 0 & 0 \times 12 + 1 \times 1 \end{bmatrix} = \begin{bmatrix} 4 & 24 + 12 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 36 \\ 0 & 1 \end{bmatrix} \)
Since A² = B, we have \( \begin{bmatrix} 4 & 36 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & x \\ 0 & 1 \end{bmatrix} \). Comparing the elements in the first row and second column, x = 36.
In simple words: Multiply matrix A by itself to get A². Then look at the result and match it with matrix B to find x.

Exam Tip: When two matrices are equal, all matching positions must have the same values. Use this definition of matrix equality to solve for unknowns.

Question 24. 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 A² = B.
Answer: Calculate A² by multiplying A by itself:
\( A^2 = \begin{bmatrix} 2 & x \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & x \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 \times 2 + x \times 0 & 2 \times x + x \times 1 \\ 0 \times 2 + 1 \times 0 & 0 \times x + 1 \times 1 \end{bmatrix} = \begin{bmatrix} 4 & 2x + x \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 3x \\ 0 & 1 \end{bmatrix} \)
Since A² = B, we have \( \begin{bmatrix} 4 & 3x \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 36 \\ 0 & 1 \end{bmatrix} \). By matching elements, 3x = 36, so x = 12.
In simple words: Square the matrix A and then match it against B. The element in the top-right position gives you an equation you can solve for x.

Exam Tip: Always complete the full matrix multiplication step-by-step. Take special care with the (1,2) position where terms add together.

Question 25. If \( A = \begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} 9 & 16 \\ 0 & -y \end{bmatrix} \), find x and y when A² = B.
Answer: Calculate A²:
\( A^2 = \begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 \times 3 + x \times 0 & 3 \times x + x \times 1 \\ 0 \times 3 + 1 \times 0 & 0 \times x + 1 \times 1 \end{bmatrix} = \begin{bmatrix} 9 & 3x + x \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 9 & 4x \\ 0 & 1 \end{bmatrix} \)
Comparing A² with B: \( \begin{bmatrix} 9 & 4x \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 9 & 16 \\ 0 & -y \end{bmatrix} \)
From the matching elements: 4x = 16 and 1 = -y. Therefore, x = 4 and y = -1.
In simple words: Work out A squared, then line it up with B. You get two separate equations - one for x and one for y.

Exam Tip: When a matrix equation contains two unknowns in different positions, solve each equation independently. Always verify your values work in both equations.

Question 26. Find x, y if \( \begin{bmatrix} -2 & 0 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} -1 \\ 2x \end{bmatrix} + 3 \begin{bmatrix} -2 \\ 1 \end{bmatrix} = 2 \begin{bmatrix} y \\ 3 \end{bmatrix} \).
Answer: First, perform the matrix multiplication:
\( \begin{bmatrix} -2 & 0 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} -1 \\ 2x \end{bmatrix} = \begin{bmatrix} (-2)(-1) + 0(2x) \\ 3(-1) + 1(2x) \end{bmatrix} = \begin{bmatrix} 2 \\ -3 + 2x \end{bmatrix} \)
Next, multiply the matrices by their scalars:
\( 3 \begin{bmatrix} -2 \\ 1 \end{bmatrix} = \begin{bmatrix} -6 \\ 3 \end{bmatrix} \) and \( 2 \begin{bmatrix} y \\ 3 \end{bmatrix} = \begin{bmatrix} 2y \\ 6 \end{bmatrix} \)
Now substitute into the equation:
\( \begin{bmatrix} 2 \\ -3 + 2x \end{bmatrix} + \begin{bmatrix} -6 \\ 3 \end{bmatrix} = \begin{bmatrix} 2y \\ 6 \end{bmatrix} \)
\( \begin{bmatrix} 2 - 6 \\ -3 + 2x + 3 \end{bmatrix} = \begin{bmatrix} 2y \\ 6 \end{bmatrix} \)
\( \begin{bmatrix} -4 \\ 2x \end{bmatrix} = \begin{bmatrix} 2y \\ 6 \end{bmatrix} \)
By comparing elements: -4 = 2y, so y = -2, and 2x = 6, so x = 3.
In simple words: Do the multiplication first, then the addition of matrices. Finally, match up the elements on both sides to find x and y.

Exam Tip: Work through scalar multiplication and matrix multiplication in the correct order. Always combine like terms before comparing the final matrices.

Question 27. If \( \begin{bmatrix} a & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 4 & 3 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} b & 11 \\ 4 & c \end{bmatrix} \), find a, b and c.
Answer: Multiply the two matrices on the left side:
\( \begin{bmatrix} a & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 4 & 3 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} a \times 4 + 1 \times (-3) & a \times 3 + 1 \times 2 \\ 1 \times 4 + 0 \times (-3) & 1 \times 3 + 0 \times 2 \end{bmatrix} = \begin{bmatrix} 4a - 3 & 3a + 2 \\ 4 & 3 \end{bmatrix} \)
Comparing with the right side \( \begin{bmatrix} b & 11 \\ 4 & c \end{bmatrix} \), we get three equations:
1. 4a - 3 = b 2. 3a + 2 = 11 3. c = 3
From equation 2: 3a = 9, so a = 3. Substituting into equation 1: b = 4(3) - 3 = 9. From equation 3: c = 3.
In simple words: Multiply the matrices carefully. Then match each position with the result matrix to get three separate equations for a, b, and c.

Exam Tip: When the product contains multiple unknowns, extract all equations from corresponding positions. Solve for variables that appear in simpler equations first.

Question 28. If \( A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \), find x, y so that A² = xA + yI.
Answer: First, calculate A²:
\( A^2 = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 2 \times 2 + 3 \times 1 & 2 \times 3 + 3 \times 2 \\ 1 \times 2 + 2 \times 1 & 1 \times 3 + 2 \times 2 \end{bmatrix} = \begin{bmatrix} 4 + 3 & 6 + 6 \\ 2 + 2 & 3 + 4 \end{bmatrix} = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix} \)
Now compute the right side where \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \):
\( xA + yI = x \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} + y \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2x & 3x \\ x & 2x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 0 & y \end{bmatrix} = \begin{bmatrix} 2x + y & 3x \\ x & 2x + y \end{bmatrix} \)
Comparing with A²: \( \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix} = \begin{bmatrix} 2x + y & 3x \\ x & 2x + y \end{bmatrix} \)
From the (1,2) position: 3x = 12, so x = 4. From the (2,1) position: x = 4 (confirms). From the (1,1) position: 2(4) + y = 7, so y = -1.
In simple words: Square the matrix A and also work out xA + yI using the identity matrix. Match them position by position to build equations for x and y.

Exam Tip: Use the simplest equation first - in this case, the off-diagonal (1,2) element gives x directly without needing to solve a system.

Question 29. If \( P = \begin{bmatrix} 2 & 6 \\ 3 & 9 \end{bmatrix} \) and \( Q = \begin{bmatrix} 3 & x \\ y & 2 \end{bmatrix} \), find x, y such that PQ = O.
Answer: Compute PQ:
\( PQ = \begin{bmatrix} 2 & 6 \\ 3 & 9 \end{bmatrix} \begin{bmatrix} 3 & x \\ y & 2 \end{bmatrix} = \begin{bmatrix} 2 \times 3 + 6 \times y & 2 \times x + 6 \times 2 \\ 3 \times 3 + 9 \times y & 3 \times x + 9 \times 2 \end{bmatrix} = \begin{bmatrix} 6 + 6y & 2x + 12 \\ 9 + 9y & 3x + 18 \end{bmatrix} \)
Since PQ = O (the zero matrix), all elements must equal zero:
- 6 + 6y = 0 \( \implies \) y = -1 - 2x + 12 = 0 \( \implies \) x = -6 - 9 + 9y = 0 \( \implies \) y = -1 (confirms) - 3x + 18 = 0 \( \implies \) x = -6 (confirms)
In simple words: Multiply the two matrices. For the product to be a zero matrix, every single element must equal zero. This gives you equations to solve.

Exam Tip: When a product equals the zero matrix, verify your answer by checking that all four equations are satisfied, not just two.

Question 30. Let \( M \times \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} = [1 \quad 2] \) where M is a matrix.
(i) State the order of the matrix M.
(ii) Find the matrix M.
Answer:
(i) The second matrix has order 2 × 2 and the result has order 1 × 2. Using the multiplication rule: if M has order p × q and is multiplied by a 2 × 2 matrix to give a 1 × 2 result, then M must have order 1 × 2.
In simple words: Look at the sizes of the matrices. The order of M must be 1 × 2 so that when multiplied by a 2 × 2 matrix, you get a 1 × 2 result.
(ii) Let \( M = [x \quad y] \). Then:
\( [x \quad y] \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} = [x \times 1 + y \times 0 \quad x \times 1 + y \times 2] = [x \quad x + 2y] = [1 \quad 2] \)
By comparing elements: x = 1 and x + 2y = 2. Substituting x = 1 into the second equation: 1 + 2y = 2, so y = 1/2. Therefore, \( M = [1 \quad \frac{1}{2}] \).
In simple words: Let M be a 1 × 2 matrix with unknown entries. Multiply it out and match the result with [1 2] to find x and y.

Exam Tip: Always determine matrix order using the multiplication rule (p × q) × (q × r) = p × r. Then use this knowledge to set up the unknown matrix correctly.

Question 31. Given \( \begin{bmatrix} 2 & 1 \\ -3 & 4 \end{bmatrix} X = \begin{bmatrix} 7 \\ 6 \end{bmatrix} \), write:
(i) the order of the matrix X
(ii) the matrix X.
Answer:
(i) The first matrix is 2 × 2 and the result is 2 × 1. Using matrix multiplication rules: (2 × 2) × (p × q) = 2 × 1. This means X must have order 2 × 1.
In simple words: The answer has 2 rows and 1 column. For this to happen when multiplying with a 2 × 2 matrix, X must be a 2 × 1 column matrix.
(ii) Let \( X = \begin{bmatrix} x \\ y \end{bmatrix} \). Then:
\( \begin{bmatrix} 2 & 1 \\ -3 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2x + y \\ -3x + 4y \end{bmatrix} = \begin{bmatrix} 7 \\ 6 \end{bmatrix} \)
This gives two equations:
- 2x + y = 7 \( \implies \) y = 7 - 2x - -3x + 4y = 6
Substitute the first into the second: -3x + 4(7 - 2x) = 6 \( \implies \) -3x + 28 - 8x = 6 \( \implies \) -11x = -22 \( \implies \) x = 2. Then y = 7 - 2(2) = 3. Therefore, \( X = \begin{bmatrix} 2 \\ 3 \end{bmatrix} \).
In simple words: Write X as a column with two unknowns. Multiply it out to get two equations, then solve by substitution.

Exam Tip: When solving a 2 × 2 system from a matrix equation, substitution often works better than elimination. Always verify your answer by substituting back.

Question 32. Solve the matrix equation \( \begin{bmatrix} 4 \\ 1 \end{bmatrix} X = \begin{bmatrix} -4 & 8 \\ -1 & 2 \end{bmatrix} \).
Answer: The first matrix is 2 × 1 and the result is 2 × 2. Using the multiplication rule: (2 × 1) × (p × q) = 2 × 2. This means X has order 1 × 2. Let \( X = [x \quad y] \). Then:
\( \begin{bmatrix} 4 \\ 1 \end{bmatrix} [x \quad y] = \begin{bmatrix} 4x & 4y \\ x & y \end{bmatrix} = \begin{bmatrix} -4 & 8 \\ -1 & 2 \end{bmatrix} \)
Comparing elements:
- 4x = -4 \( \implies \) x = -1 - 4y = 8 \( \implies \) y = 2
These values are confirmed by the second row: x = -1 and y = 2. Therefore, \( X = [-1 \quad 2] \).
In simple words: Determine that X is a 1 × 2 row matrix. Multiply it out and match each element on both sides to find x and y.

Exam Tip: Note that when a column vector multiplies a row vector, you get a full matrix. This is different from row × column, which gives a scalar.

Question 33(i). If \( A = \begin{bmatrix} 2 & -1 \\ -4 & 5 \end{bmatrix} \) and \( B = \begin{bmatrix} -3 \\ 2 \end{bmatrix} \), find matrix C such that AC = B.
Answer: The matrix A is 2 × 2 and B is 2 × 1. Using the rule: (2 × 2) × (p × q) = 2 × 1, we find that C has order 2 × 1. Let \( C = \begin{bmatrix} x \\ y \end{bmatrix} \). Then:
\( \begin{bmatrix} 2 & -1 \\ -4 & 5 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2x - y \\ -4x + 5y \end{bmatrix} = \begin{bmatrix} -3 \\ 2 \end{bmatrix} \)
This gives:
- 2x - y = -3 \( \implies \) y = 2x + 3 - -4x + 5y = 2
Substitute: -4x + 5(2x + 3) = 2 \( \implies \) -4x + 10x + 15 = 2 \( \implies \) 6x = -13 \( \implies \) x = -13/6. Then y = 2(-13/6) + 3 = -26/6 + 18/6 = -8/6 = -4/3. Therefore, \( C = \begin{bmatrix} -13/6 \\ -4/3 \end{bmatrix} \).
In simple words: Set up C as a column matrix with unknowns. Multiply it by A and match the result with B to get two equations, then solve.

Exam Tip: Keep fractions in simplified form throughout. Double-check arithmetic when dealing with negative coefficients.

Question 33(ii). If \( A = \begin{bmatrix} 2 & -1 \\ -4 & 5 \end{bmatrix} \) and \( B = [0 \quad -3] \), find matrix C such that CA = B.
Answer: The matrix A is 2 × 2 and B is 1 × 2. Using the rule: (p × q) × (2 × 2) = 1 × 2, we find that C has order 1 × 2. Let \( C = \begin{bmatrix} x \\ y \end{bmatrix} \) (note: C should be written as a row in this context, so treat it as [a b]). Then:
\( \begin{bmatrix} a & b \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -4 & 5 \end{bmatrix} = \begin{bmatrix} 2a - 4b & -a + 5b \end{bmatrix} = [0 \quad -3] \)
This gives:
- 2a - 4b = 0 \( \implies \) a = 2b - -a + 5b = -3
Substitute: -(2b) + 5b = -3 \( \implies \) 3b = -3 \( \implies \) b = -1. Then a = 2(-1) = -2. Therefore, \( C = [-2 \quad -1] \) (or written as a column for clarity in the context of the problem: \( C = \begin{bmatrix} -2 \\ -1 \end{bmatrix} \)).
In simple words: Here C is a row matrix that multiplies A from the left. Set it up, multiply, and match the result with B to get equations for the unknowns.

Exam Tip: Notice the difference between AC and CA - the order of multiplication matters and determines the shape of C. Always verify the final answer by multiplying back.

Question 34. If A = \( \begin{bmatrix} 3 & -4 \\ -1 & 2 \end{bmatrix} \), find the matrix B such that BA = I, where I is the unit matrix of order 2.
Answer: Since BA = I, matrix B has order 2 × 2. Let B = \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \). Multiplying: \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 3 & -4 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)

By matrix equality: 3a - b = 1, -4a + 2b = 0, 3c - d = 0, -4c + 2d = 1.

From the second equation: b = 2a. Substituting into the first: 3a - 2a = 1, so a = 1 and b = 2.

From the third equation: d = 3c. Substituting into the fourth: -4c + 6c = 1, so 2c = 1, giving c = 1/2 and d = 3/2.

Therefore, B = \( \begin{bmatrix} 1 & 2 \\ \frac{1}{2} & \frac{3}{2} \end{bmatrix} \)
In simple words: To find B, we set up equations using matrix multiplication and the fact that the result must equal the identity matrix. We solve for each element by matching the entries.

Exam Tip: When finding inverse matrices, always verify your answer by multiplying BA and checking that you get the identity matrix I.

Question 35. Given \( \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} \) M = 6I, where M is a matrix and I is the unit matrix of order 2 × 2. (i) State the order of matrix M. (ii) Find the matrix M.
Answer:
(i) Since the matrix \( \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} \) is 2 × 2 and 6I is 2 × 2, matrix M must also be of order 2 × 2.

(ii) Let M = \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \). Then \( \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix} \)

Expanding: \( \begin{bmatrix} 4a + 2c & 4b + 2d \\ -a + c & -b + d \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix} \)

This gives us: 4a + 2c = 6, 4b + 2d = 0, -a + c = 0, -b + d = 6.

From -a + c = 0, we get a = c. Substituting into 4a + 2c = 6: 4a + 2a = 6, so 6a = 6 and a = 1. Thus c = 1.

From 4b + 2d = 0, we get d = -2b. Substituting into -b + d = 6: -b - 2b = 6, so -3b = 6 and b = -2. Thus d = 4.

Therefore, M = \( \begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix} \)
In simple words: We multiply the first matrix by M to get 6 times the identity. By setting up equations from the entries, we find that a = 1, b = -2, c = 1, and d = 4.

Exam Tip: Remember that a scalar multiple of the identity matrix has the same value on the diagonal and zeros elsewhere. Use this pattern to solve more efficiently.

Question 36. If B = \( \begin{bmatrix} -4 & 2 \\ 5 & -1 \end{bmatrix} \) and C = \( \begin{bmatrix} 17 & -1 \\ 47 & -13 \end{bmatrix} \), find the matrix A such that AB = C.
Answer: Since AB = C, matrix A has order 2 × 2. Let A = \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \). Then \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} -4 & 2 \\ 5 & -1 \end{bmatrix} = \begin{bmatrix} 17 & -1 \\ 47 & -13 \end{bmatrix} \)

Multiplying out: \( \begin{bmatrix} -4a + 5b & 2a - b \\ -4c + 5d & 2c - d \end{bmatrix} = \begin{bmatrix} 17 & -1 \\ 47 & -13 \end{bmatrix} \)

This gives: -4a + 5b = 17, 2a - b = -1, -4c + 5d = 47, 2c - d = -13.

From 2a - b = -1, we get b = 2a + 1. Substituting: -4a + 5(2a + 1) = 17 gives -4a + 10a + 5 = 17, so 6a = 12 and a = 2. Thus b = 5.

From 2c - d = -13, we get d = 2c + 13. Substituting: -4c + 5(2c + 13) = 47 gives -4c + 10c + 65 = 47, so 6c = -18 and c = -3. Thus d = 7.

Therefore, A = \( \begin{bmatrix} 2 & 5 \\ -3 & 7 \end{bmatrix} \)
In simple words: We write A with unknown entries and multiply by B. By comparing the result with C, we get four equations that we solve to find each element of A.

Exam Tip: Always verify your answer by computing AB and checking that it equals the given matrix C.

Question 37. If A = \( \begin{bmatrix} 4 & -4 \\ -4 & 4 \end{bmatrix} \), find A². If A² = pA, then find the value of p.
Answer: Computing A²: \( \begin{bmatrix} 4 & -4 \\ -4 & 4 \end{bmatrix} \begin{bmatrix} 4 & -4 \\ -4 & 4 \end{bmatrix} \)

\( = \begin{bmatrix} 16 + 16 & -16 - 16 \\ -16 - 16 & 16 + 16 \end{bmatrix} = \begin{bmatrix} 32 & -32 \\ -32 & 32 \end{bmatrix} \)

Now, if A² = pA, then \( \begin{bmatrix} 32 & -32 \\ -32 & 32 \end{bmatrix} = p \begin{bmatrix} 4 & -4 \\ -4 & 4 \end{bmatrix} \)

This means: 32 = 4p, so p = 8.
In simple words: We multiply the matrix by itself to get A². Then we compare this result with pA by dividing corresponding entries to find p.

Exam Tip: When A² = pA, you can find p by dividing any non-zero entry of A² by the corresponding entry of A.

Question 1. If \( \begin{bmatrix} x + 3 & 4 \\ y - 4 & x + y \end{bmatrix} = \begin{bmatrix} 5 & 4 \\ 3 & 9 \end{bmatrix} \), then the values of x and y are
(a) x = 2, y = 7
(b) x = 7, y = 2
(c) x = 3, y = 6
(d) x = -2, y = 7
Answer: (a) x = 2, y = 7
In simple words: When two matrices are equal, matching entries must be the same. From x + 3 = 5, we get x = 2. From y - 4 = 3, we get y = 7. We can verify: x + y = 2 + 7 = 9, which matches the bottom-right entry.

Exam Tip: Always verify your answer by checking that all four entries match the given matrix, not just the ones you used to solve.

Question 2. If \( \begin{bmatrix} x + 2y & -y \\ 3x & 4 \end{bmatrix} = \begin{bmatrix} -4 & 3 \\ 6 & 4 \end{bmatrix} \), then the values of x and y are
(a) x = 2, y = 3
(b) x = 2, y = -3
(c) x = -2, y = 3
(d) x = 3, y = 2
Answer: (b) x = 2, y = -3
In simple words: From -y = 3, we get y = -3. From 3x = 6, we get x = 2. These values satisfy x + 2y = 2 - 6 = -4, confirming our answer.

Exam Tip: Look for equations with single variables first - they are the quickest to solve and reduce your work.

Question 3. If \( \begin{bmatrix} x - 2y & 5 \\ 3 & y \end{bmatrix} = \begin{bmatrix} 6 & 5 \\ 3 & -2 \end{bmatrix} \), then the value of x is
(a) -2
(b) 0
(c) 1
(d) 2
Answer: (d) 2
In simple words: From y = -2, we can substitute into x - 2y = 6 to get x - 2(-2) = 6, which means x + 4 = 6, so x = 2.

Exam Tip: Solve for variables that appear alone in an entry first, then substitute into other equations.

Question 4. If \( \begin{bmatrix} x + 2y & 3y \\ 4x & 2 \end{bmatrix} = \begin{bmatrix} 0 & -3 \\ 8 & 2 \end{bmatrix} \), then the value of x - y is
(a) -3
(b) 1
(c) 3
(d) 5
Answer: (c) 3
In simple words: From 3y = -3, we get y = -1. From 4x = 8, we get x = 2. Therefore x - y = 2 - (-1) = 3.

Exam Tip: When finding a combination like x - y, solve for both variables separately first, then compute the required expression.

Question 5. If \( x \begin{bmatrix} 2 \\ 3 \end{bmatrix} + y \begin{bmatrix} -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 10 \\ 6 \end{bmatrix} \), then the values of x and y are
(a) x = 2, y = 6
(b) x = 2, y = -6
(c) x = 3, y = -4
(d) x = 3, y = -6
Answer: (b) x = 2, y = -6
In simple words: Expanding the left side: \( \begin{bmatrix} 2x \\ 3x \end{bmatrix} + \begin{bmatrix} -y \\ 0 \end{bmatrix} = \begin{bmatrix} 2x - y \\ 3x \end{bmatrix} = \begin{bmatrix} 10 \\ 6 \end{bmatrix} \). From 3x = 6, we get x = 2. From 2x - y = 10, we get 4 - y = 10, so y = -6.

Exam Tip: For equations involving scalar multiples of column matrices, expand the left side first and match entries systematically.

Question 6. If A = \( \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \), then A² =
(a) \( \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} \)
(b) \( \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix} \)
(c) \( \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \)
(d) \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
Answer: (d) \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
In simple words: Multiplying: \( \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 + 1 & 0 + 0 \\ 0 + 0 & 1 + 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \), which is the identity matrix.

Exam Tip: Matrices that square to the identity are called involutions. Their rows and columns are orthogonal, which shows in the clean result.

Question 7. If A = \( \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \), then A² =
(a) A
(b) O
(c) I
(d) 2A
Answer: (b) O
In simple words: Computing A²: \( \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \). The product is the zero matrix O.

Exam Tip: When a matrix squares to the zero matrix, it is called nilpotent. This occurs when the matrix has a specific structure where products of entries sum to zero.

Question 8. If A = [ 0 1 0 0 ], then A² =
(a) [ 2 0 1 1 ]
(b) [ 1 0 1 2 ]
(c) [ 1 0 2 1 ]
(d) none of these
Answer: (c) [ 1 0 2 1 ]
In simple words: When you multiply matrix A by itself, you get the result shown in option (c). This is found by following the rule for matrix multiplication where you multiply rows by columns.

Exam Tip: Always multiply matrices step-by-step, element by element. Check that the number of columns in the first matrix matches the number of rows in the second matrix before starting.

Question 9. If A = [ 3 -1 1 2 ], then A² =
(a) [ 8 5 -5 3 ]
(b) [ 8 -5 5 3 ]
(c) [ 8 -5 -5 -3 ]
(d) [ 8 -5 -5 3 ]
Answer: (d) [ 8 -5 -5 3 ]
In simple words: Square the matrix by multiplying it by itself. Work through each position carefully, adding up the products of rows and columns to get the final answer.

Exam Tip: Double-check your arithmetic when multiplying and adding the terms. A single calculation error will make your entire matrix incorrect.

Question 10. If matrix A = [ 2 2 0 2 ] and A² = [ 4 x 0 4 ], then the value of x is:
(a) 2
(b) 4
(c) 8
(d) 10
Answer: (c) 8
In simple words: Find A² by multiplying matrix A by itself. Then match the result with the given matrix to find the missing value x.

Exam Tip: When a matrix element is unknown, compute the product at that position using the standard matrix multiplication rule and set it equal to the given value to solve for the variable.

Question 11. If A = [ 2 -2 -2 2 ], then A² = pA. The value of p is:
(a) 2
(b) 4
(c) -2
(d) -4
Answer: (b) 4
In simple words: Calculate A² by multiplying A by itself. Then compare it with pA (which means p times matrix A). Find the value of p that makes them equal.

Exam Tip: Use the matrix equality definition - two matrices are equal only when every matching element is the same. This helps you set up an equation to find p.

Question 12. The product AB of two matrices A and B is possible if:
(a) A and B have same number of rows.
(b) A and B have same number of columns.
(c) The number of columns of A is equal to the number of rows of B.
(d) The number of rows of A is equal to the number of columns of B.
Answer: (c) The number of columns of A is equal to the number of rows of B.
In simple words: For two matrices to multiply together, the number of columns in the first matrix must equal the number of rows in the second matrix. This is the key rule of matrix multiplication.

Exam Tip: Remember: if A is m × n and B is n × p, then AB gives a result that is m × p. The middle numbers (the two n's) must match.

Question 13. If A = [ -1 2 ] and B = [ 1 0 -2 3 ], which of the following operations is possible?
(a) A - B
(b) A + B
(c) AB
(d) BA
Answer: (c) AB
In simple words: A has order 1 × 2, and B has order 2 × 2. Since A has 2 columns and B has 2 rows, multiplication AB is possible. The other operations need matrices of the same size, which these are not.

Exam Tip: Always check the order (dimensions) of each matrix first. Addition and subtraction need matching dimensions. Multiplication needs the column count of the first to match the row count of the second.

Assertion-Reason Type Questions

Question. Assertion (A): BA = C
Reason (R): Matrix multiplication is not always commutative.
Given: A, B and C are square matrices of order 2 such that AB = C.

(a) Assertion (A) is true, but Reason (R) is false.
(b) Assertion (A) is false, but Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (b) Assertion (A) is false, but Reason (R) is true.
In simple words: The reason is correct - matrix multiplication is not commutative, meaning AB does not always equal BA. However, the assertion is false because just because AB = C does not mean BA will also equal C. The reason explains why the assertion is false.

Exam Tip: In assertion-reason questions, both statements can be true but the reason might not explain the assertion. Here, BA is not necessarily equal to C because matrix multiplication does not follow the commutative property.

Question. Assertion (A): Product BA need not be equal to C.
Reason (R): Matrix multiplication is not associative.
Given: A, B and C are square matrices of order 2 such that AB = C.

(a) Assertion (A) is true, but Reason (R) is false.
(b) Assertion (A) is false, but Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (a) Assertion (A) is true, but Reason (R) is false.
In simple words: The assertion is correct - BA does not have to equal C since AB ≠ BA in general for matrices. However, the reason given is incorrect because matrix multiplication actually is associative. The assertion is true for the wrong reason - it should be because multiplication is not commutative, not because it is not associative.

Exam Tip: Know the key properties: multiplication is associative (A(BC) = (AB)C) but not commutative (AB ≠ BA). Use the correct property to explain why BA need not equal C.

Question. Assertion (A): Product AB of the two matrices A and B is possible.
Reason (R): Number of columns of matrix A is equal to number of rows in matrix B.
Given: A = [ 3 -2 ] and B = [ -1 4 2 0 ]

(a) Assertion (A) is true, but Reason (R) is false.
(b) Assertion (A) is false, but Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (c) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
In simple words: Matrix A has 1 row and 2 columns. Matrix B has 2 rows and 2 columns. Since A has 2 columns and B has 2 rows, they match. This is exactly why AB is possible. The reason correctly explains the assertion.

Exam Tip: Always verify dimensions before multiplying. The assertion is true because the reason condition is satisfied - when columns of the first matrix equal rows of the second matrix, multiplication is defined.

Chapter Test

Question 1. Find the values of a and b if [ a + 3 b² + 2 0 2 ] = [ 2a + 1 3b 0 b² - 5b ]
Answer: By applying the definition of matrix equality, corresponding elements must be equal.

From the first element:
\( a + 3 = 2a + 1 \)
\( a = 2 \)

From the second element:
\( b^2 + 2 = 3b \)
\( b^2 - 3b + 2 = 0 \)

Factoring this equation:
\( b^2 - 2b - b + 2 = 0 \)
\( b(b - 2) - 1(b - 2) = 0 \)
\( (b - 1)(b - 2) = 0 \)
\( b = 1 \text{ or } b = 2 \)

Now checking which value satisfies the fourth element equation \( b^2 - 5b = -6 \):

For \( b = 1 \): \( (1)^2 - 5(1) = 1 - 5 = -4 \neq -6 \) (not valid)

For \( b = 2 \): \( (2)^2 - 5(2) = 4 - 10 = -6 \) (valid)

Therefore, \( a = 2 \) and \( b = 2 \).
In simple words: Set up equations by matching elements in equal matrices. Solve the quadratic to find possible values, then check each value in the other equation to confirm which one works.

Exam Tip: When finding unknowns in matrix equations, substitute back into all equations to verify your solution. A value may satisfy one equation but not another.

Question 2. Find a, b, c and d if 3 [ a b c d ] = [ 4 a + b c + d 3 ] + [ a 6 -1 2d ]
Answer: Start by rewriting the equation:\
\( 3 \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 4 + a & a + b + 6 \\ c + d - 1 & 3 + 2d \end{bmatrix} \)

\( \begin{bmatrix} 3a & 3b \\ 3c & 3d \end{bmatrix} = \begin{bmatrix} 4 + a & a + b + 6 \\ c + d - 1 & 3 + 2d \end{bmatrix} \)

By matrix equality, we get four equations:
(1) \( 3a = 4 + a \) ⟹ \( 2a = 4 \) ⟹ \( a = 2 \)
(2) \( 3b = a + b + 6 \)
(3) \( 3c = c + d - 1 \)
(4) \( 3d = 3 + 2d \) ⟹ \( d = 3 \)

Substituting \( a = 2 \) into equation (2):
\( 3b = 2 + b + 6 \)
\( 2b = 8 \)
\( b = 4 \)

Substituting \( d = 3 \) into equation (3):
\( 3c = c + 3 - 1 \)
\( 2c = 2 \)
\( c = 1 \)

Therefore, \( a = 2, b = 4, c = 1, d = 3 \).
In simple words: Expand the left side by multiplying by 3. Then match each position with the right side to create four separate equations. Solve them one by one, using each solution to help with the others.

Exam Tip: Organize your work by listing each equation separately. Solve the simplest equations first (those with only one unknown), then substitute those values into the remaining equations.

Question 3. Determine the matrices A and B when A + 2B = [ 1 2 6 -3 ] and 2A - B = [ 2 -1 2 -1 ]
Answer: We have two matrix equations:
\( A + 2B = \begin{bmatrix} 1 & 2 \\ 6 & -3 \end{bmatrix} \) ... (Equation 1)
\( 2A - B = \begin{bmatrix} 2 & -1 \\ 2 & -1 \end{bmatrix} \) ... (Equation 2)

Multiply Equation 1 by 2:
\( 2A + 4B = \begin{bmatrix} 2 & 4 \\ 12 & -6 \end{bmatrix} \)

Subtract Equation 2 from this result:
\( (2A + 4B) - (2A - B) = \begin{bmatrix} 2 & 4 \\ 12 & -6 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ 2 & -1 \end{bmatrix} \)
\( 5B = \begin{bmatrix} 0 & 5 \\ 10 & -5 \end{bmatrix} \)
\( B = \begin{bmatrix} 0 & 1 \\ 2 & -1 \end{bmatrix} \)

Substitute B back into Equation 1:
\( A + 2\begin{bmatrix} 0 & 1 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 6 & -3 \end{bmatrix} \)
\( A + \begin{bmatrix} 0 & 2 \\ 4 & -2 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 6 & -3 \end{bmatrix} \)
\( A = \begin{bmatrix} 1 & 0 \\ 2 & -1 \end{bmatrix} \)

Therefore, \( A = \begin{bmatrix} 1 & 0 \\ 2 & -1 \end{bmatrix} \) and \( B = \begin{bmatrix} 0 & 1 \\ 2 & -1 \end{bmatrix} \).
In simple words: You have two equations with two unknowns. Multiply one equation to line up terms, then add or subtract to remove one variable. Solve for the remaining variable, then find the other by substitution.

Exam Tip: This is like solving a system of linear equations, but with matrices. Write out the steps clearly and verify your answer by substituting back into both original equations.

Question 4(i). Find the matrix B if A = [ 4 1 2 3 ] and A² = A + 2B.
Answer: First, calculate A²:
\( A^2 = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix} \)
\( = \begin{bmatrix} 4 \times 4 + 1 \times 2 & 4 \times 1 + 1 \times 3 \\ 2 \times 4 + 3 \times 2 & 2 \times 1 + 3 \times 3 \end{bmatrix} \)
\( = \begin{bmatrix} 16 + 2 & 4 + 3 \\ 8 + 6 & 2 + 9 \end{bmatrix} \)
\( = \begin{bmatrix} 18 & 7 \\ 14 & 11 \end{bmatrix} \)

Using the given equation \( A^2 = A + 2B \):
\( 2B = A^2 - A \)
\( 2B = \begin{bmatrix} 18 & 7 \\ 14 & 11 \end{bmatrix} - \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix} \)
\( 2B = \begin{bmatrix} 18 - 4 & 7 - 1 \\ 14 - 2 & 11 - 3 \end{bmatrix} \)
\( 2B = \begin{bmatrix} 14 & 6 \\ 12 & 8 \end{bmatrix} \)
\( B = \begin{bmatrix} 7 & 3 \\ 6 & 4 \end{bmatrix} \)
In simple words: Square matrix A by multiplying it by itself. Then rearrange the equation to isolate B. Calculate A² - A, divide by 2 to get B.

Exam Tip: Always compute A² carefully, checking each element through the row-by-column multiplication rule. Be cautious with signs and arithmetic.

Question 4(ii). If A = [ 1 2 -3 4 ], B = [ 0 1 -2 5 ] and C = [ -2 0 -1 1 ], find A(4B - 3C).
Answer: First, calculate 4B:
\( 4B = 4\begin{bmatrix} 0 & 1 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix} 0 & 4 \\ -8 & 20 \end{bmatrix} \)

Next, calculate 3C:
\( 3C = 3\begin{bmatrix} -2 & 0 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} -6 & 0 \\ -3 & 3 \end{bmatrix} \)

Now find 4B - 3C:
\( 4B - 3C = \begin{bmatrix} 0 & 4 \\ -8 & 20 \end{bmatrix} - \begin{bmatrix} -6 & 0 \\ -3 & 3 \end{bmatrix} \)
\( = \begin{bmatrix} 0 - (-6) & 4 - 0 \\ -8 - (-3) & 20 - 3 \end{bmatrix} \)
\( = \begin{bmatrix} 6 & 4 \\ -5 & 17 \end{bmatrix} \)

Finally, multiply A by (4B - 3C):
\( A(4B - 3C) = \begin{bmatrix} 1 & 2 \\ -3 & 4 \end{bmatrix} \begin{bmatrix} 6 & 4 \\ -5 & 17 \end{bmatrix} \)
\( = \begin{bmatrix} 1 \times 6 + 2 \times (-5) & 1 \times 4 + 2 \times 17 \\ -3 \times 6 + 4 \times (-5) & -3 \times 4 + 4 \times 17 \end{bmatrix} \)
\( = \begin{bmatrix} 6 - 10 & 4 + 34 \\ -18 - 20 & -12 + 68 \end{bmatrix} \)
\( = \begin{bmatrix} -4 & 38 \\ -38 & 56 \end{bmatrix} \)
In simple words: Follow the order of operations. First, scale matrices B and C by their coefficients. Then subtract them. Finally, multiply the result by A using matrix multiplication rules.

Exam Tip: Work step-by-step and don't try to combine all operations at once. Compute (4B - 3C) completely before multiplying by A. Double-check the final multiplication carefully.

Question 4. If \( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \), \( B = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} \), \( C = \begin{bmatrix} 4 & 1 \\ 1 & 5 \end{bmatrix} \) and \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \), find the value of \( A(4B - 3C) \).
Answer: First, calculate \( 4B \):
\( 4B = 4\begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 4 & 8 \\ 8 & 16 \end{bmatrix} \)

Next, find \( 3C \):
\( 3C = 3\begin{bmatrix} 4 & 1 \\ 1 & 5 \end{bmatrix} = \begin{bmatrix} 12 & 3 \\ 3 & 15 \end{bmatrix} \)

Now compute \( 4B - 3C \):
\( 4B - 3C = \begin{bmatrix} 4 & 8 \\ 8 & 16 \end{bmatrix} - \begin{bmatrix} 12 & 3 \\ 3 & 15 \end{bmatrix} = \begin{bmatrix} -8 & 5 \\ 5 & 1 \end{bmatrix} \)

Finally, multiply by A:
\( A(4B - 3C) = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} -8 & 5 \\ 5 & 1 \end{bmatrix} = \begin{bmatrix} 1(-8) + 2(5) & 1(5) + 2(1) \\ 3(-8) + 4(5) & 3(5) + 4(1) \end{bmatrix} = \begin{bmatrix} 2 & 7 \\ -4 & 19 \end{bmatrix} \)

Therefore, \( A(4B - 3C) = \begin{bmatrix} 2 & 7 \\ -4 & 19 \end{bmatrix} \)
In simple words: Multiply matrix B by 4 and matrix C by 3, then subtract. After that, multiply the result by matrix A to get your answer.

Exam Tip: Work carefully through each multiplication step - errors in scalar multiplication or subtraction early on will affect your final answer. Always verify your matrix dimensions are compatible before multiplying.

Question 5. If \( A = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \), \( B = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} \), \( C = \begin{bmatrix} 4 & 1 \\ 1 & 5 \end{bmatrix} \) and \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \), find \( A(B + C) - 14I \).
Answer: Start by finding \( B + C \):
\( B + C = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} + \begin{bmatrix} 4 & 1 \\ 1 & 5 \end{bmatrix} = \begin{bmatrix} 5 & 3 \\ 3 & 9 \end{bmatrix} \)

Then multiply by A:
\( A(B + C) = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} 5 & 3 \\ 3 & 9 \end{bmatrix} = \begin{bmatrix} 1(5) + 3(3) & 1(3) + 3(9) \\ 2(5) + 4(3) & 2(3) + 4(9) \end{bmatrix} = \begin{bmatrix} 14 & 30 \\ 22 & 42 \end{bmatrix} \)

Calculate \( 14I \):
\( 14I = 14\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix} \)

Finally, subtract:
\( A(B + C) - 14I = \begin{bmatrix} 14 & 30 \\ 22 & 42 \end{bmatrix} - \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix} = \begin{bmatrix} 0 & 30 \\ 22 & 28 \end{bmatrix} \)
In simple words: Add the two matrices B and C first. Next, multiply this sum by A. Then multiply the identity matrix by 14. Subtract this from your product to get the final result.

Exam Tip: Break the problem into smaller steps - addition, then multiplication, then subtraction. This reduces errors and makes checking your work easier.

Question 6. If \( A = \begin{bmatrix} 3 & 2 \\ 0 & 5 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix} \), find each of the following and state if they are equal:
(i) \( (A + B)(A - B) \)
(ii) \( A^2 - B^2 \)
Answer:
(i) First, compute \( A + B \):
\( A + B = \begin{bmatrix} 3 & 2 \\ 0 & 5 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 4 & 2 \\ 1 & 7 \end{bmatrix} \)

Next, find \( A - B \):
\( A - B = \begin{bmatrix} 3 & 2 \\ 0 & 5 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ -1 & 3 \end{bmatrix} \)

Now multiply:
\( (A + B)(A - B) = \begin{bmatrix} 4 & 2 \\ 1 & 7 \end{bmatrix} \begin{bmatrix} 2 & 2 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} 4(2) + 2(-1) & 4(2) + 2(3) \\ 1(2) + 7(-1) & 1(2) + 7(3) \end{bmatrix} = \begin{bmatrix} 6 & 14 \\ -5 & 23 \end{bmatrix} \)

(ii) Calculate \( A^2 \):
\( A^2 = \begin{bmatrix} 3 & 2 \\ 0 & 5 \end{bmatrix} \begin{bmatrix} 3 & 2 \\ 0 & 5 \end{bmatrix} = \begin{bmatrix} 3(3) + 2(0) & 3(2) + 2(5) \\ 0(3) + 5(0) & 0(2) + 5(5) \end{bmatrix} = \begin{bmatrix} 9 & 16 \\ 0 & 25 \end{bmatrix} \)

Compute \( B^2 \):
\( B^2 = \begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 1(1) + 0(1) & 1(0) + 0(2) \\ 1(1) + 2(1) & 1(0) + 2(2) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 3 & 4 \end{bmatrix} \)

Therefore:
\( A^2 - B^2 = \begin{bmatrix} 9 & 16 \\ 0 & 25 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 8 & 16 \\ -3 & 21 \end{bmatrix} \)

Since \( (A + B)(A - B) = \begin{bmatrix} 6 & 14 \\ -5 & 23 \end{bmatrix} \) and \( A^2 - B^2 = \begin{bmatrix} 8 & 16 \\ -3 & 21 \end{bmatrix} \), the two results are not equal. This shows that the distributive property \( (A + B)(A - B) = A^2 - B^2 \) does not hold for matrices because matrix multiplication is not commutative.
In simple words: When you work out both expressions, you get different answers. This happens because with matrices, the order of multiplication matters. The same rule that works for regular numbers doesn't always work for matrices.

Exam Tip: Always compute both sides fully before concluding they are equal or unequal. Note in your answer that matrix multiplication is not commutative, which is why these two expressions give different results.

Question 7. If \( A = \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} \), find \( A^2 - 5A - 14I \), where I is the unit matrix of order 2 × 2.
Answer: First, compute \( A^2 \):
\( A^2 = \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} 3(3) + (-5)(-4) & 3(-5) + (-5)(2) \\ (-4)(3) + 2(-4) & (-4)(-5) + 2(2) \end{bmatrix} = \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix} \)

Calculate \( 5A \):
\( 5A = 5\begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} 15 & -25 \\ -20 & 10 \end{bmatrix} \)

Find \( 14I \):
\( 14I = 14\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix} \)

Now compute the final expression:
\( A^2 - 5A - 14I = \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix} - \begin{bmatrix} 15 & -25 \\ -20 & 10 \end{bmatrix} - \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
In simple words: Square the matrix A, then multiply it by 5, then multiply the identity matrix by 14. Subtract both of these from the squared matrix to reach the zero matrix.

Exam Tip: When the answer is a zero matrix, it often indicates an important algebraic property. Double-check your arithmetic carefully, as getting exactly zeros in all positions requires precise calculation.

Question 8. If \( A = \begin{bmatrix} 3 & 3 \\ p & q \end{bmatrix} \) and \( A^2 = O \), find p and q.
Answer: Since \( A^2 = O \):
\( \begin{bmatrix} 3 & 3 \\ p & q \end{bmatrix} \begin{bmatrix} 3 & 3 \\ p & q \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)

Multiply the matrices:
\( \begin{bmatrix} 3(3) + 3(p) & 3(3) + 3(q) \\ p(3) + q(p) & p(3) + q(q) \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)

\( \begin{bmatrix} 9 + 3p & 9 + 3q \\ 3p + qp & 3p + q^2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)

Using matrix equality, each element on the left equals the corresponding element on the right:

\( 9 + 3p = 0 \Rightarrow 3p = -9 \Rightarrow p = -3 \)

\( 9 + 3q = 0 \Rightarrow 3q = -9 \Rightarrow q = -3 \)

We also have two additional equations: \( 3p + qp = 0 \) and \( 3p + q^2 = 0 \). Let's verify that \( p = -3 \) and \( q = -3 \) satisfy these:

For the first equation: \( 3(-3) + (-3)(-3) = -9 + 9 = 0 \) ✓

For the second equation: \( 3(-3) + (-3)^2 = -9 + 9 = 0 \) ✓

Therefore, \( p = -3 \) and \( q = -3 \).
In simple words: When a matrix times itself gives zero, we can match up elements to find the unknown values. Both p and q turn out to be -3.

Exam Tip: Use the definition of matrix equality - corresponding elements must be identical. Always verify your answers by substituting back into all the equations you derive.

Question 9. If \( \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \), find a, b, c and d.
Answer: Multiply the matrices on the left side:
\( \begin{bmatrix} (-1)(a) + 0(c) & (-1)(b) + 0(d) \\ 0(a) + 1(c) & 0(b) + 1(d) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \)

\( \begin{bmatrix} -a & -b \\ c & d \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \)

By matrix equality, compare corresponding elements:

\( -a = 1 \Rightarrow a = -1 \)

\( -b = 0 \Rightarrow b = 0 \)

\( c = 0 \)

\( d = -1 \)

Therefore, \( a = -1 \), \( b = 0 \), \( c = 0 \), and \( d = -1 \).
In simple words: Perform the multiplication and then match each element in the resulting matrix with the elements in the target matrix to find the unknowns.

Exam Tip: Write out the matrix multiplication carefully element by element. Setting up the problem clearly reduces careless errors.

Question 10. Find a and b if \( \begin{bmatrix} a - b & b - 4 \\ b + 4 & a - 2 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} -2 & -2 \\ 14 & 0 \end{bmatrix} \).
Answer: First, multiply the matrices on the left side:
\( \begin{bmatrix} (a - b)(2) + (b - 4)(0) & (a - b)(0) + (b - 4)(2) \\ (b + 4)(2) + (a - 2)(0) & (b + 4)(0) + (a - 2)(2) \end{bmatrix} = \begin{bmatrix} -2 & -2 \\ 14 & 0 \end{bmatrix} \)

\( \begin{bmatrix} 2a - 2b & 2b - 8 \\ 2b + 8 & 2a - 4 \end{bmatrix} = \begin{bmatrix} -2 & -2 \\ 14 & 0 \end{bmatrix} \)

Comparing corresponding elements:

From the first element: \( 2a - 2b = -2 \Rightarrow a - b = -1 \Rightarrow 2a - 2b = -2 \)

From the second element: \( 2b - 8 = -2 \Rightarrow 2b = 6 \Rightarrow b = 3 \)

From the fourth element: \( 2a - 4 = 0 \Rightarrow 2a = 4 \Rightarrow a = 2 \)

Let's verify with the third element: \( 2b + 8 = 2(3) + 8 = 14 \) ✓

Therefore, \( a = 2 \) and \( b = 3 \).
In simple words: Multiply the first two matrices, then compare each element of the result with the target matrix. Solving the equations gives you the values of a and b.

Exam Tip: Use multiple equations to solve for the unknowns and verify your answer by checking all elements match, not just one or two.

Question 11. If \( A = \begin{bmatrix} \sec 60° & \cos 90° \\ -3 \tan 45° & \sin 90° \end{bmatrix} \) and \( B = \begin{bmatrix} 0 & \cot 45° \\ -2 & 3 \sin 90° \end{bmatrix} \), find:
(i) \( 2A - 3B \)
(ii) \( A^2 \)
(iii) \( BA \)
Answer: First, replace trigonometric values: \( \sec 60° = 2 \), \( \cos 90° = 0 \), \( \tan 45° = 1 \), \( \sin 90° = 1 \), \( \cot 45° = 1 \)

\( A = \begin{bmatrix} 2 & 0 \\ -3 & 1 \end{bmatrix} \), \( B = \begin{bmatrix} 0 & 1 \\ -2 & 3 \end{bmatrix} \)

(i) Calculate \( 2A - 3B \):
\( 2A = \begin{bmatrix} 4 & 0 \\ -6 & 2 \end{bmatrix} \), \( 3B = \begin{bmatrix} 0 & 3 \\ -6 & 9 \end{bmatrix} \)

\( 2A - 3B = \begin{bmatrix} 4 & 0 \\ -6 & 2 \end{bmatrix} - \begin{bmatrix} 0 & 3 \\ -6 & 9 \end{bmatrix} = \begin{bmatrix} 4 & -3 \\ 0 & -7 \end{bmatrix} \)

(ii) Calculate \( A^2 \):
\( A^2 = \begin{bmatrix} 2 & 0 \\ -3 & 1 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} 2(2) + 0(-3) & 2(0) + 0(1) \\ (-3)(2) + 1(-3) & (-3)(0) + 1(1) \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ -9 & 1 \end{bmatrix} \)

(iii) Calculate \( BA \):
\( BA = \begin{bmatrix} 0 & 1 \\ -2 & 3 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} 0(2) + 1(-3) & 0(0) + 1(1) \\ (-2)(2) + 3(-3) & (-2)(0) + 3(1) \end{bmatrix} = \begin{bmatrix} -3 & 1 \\ -13 & 3 \end{bmatrix} \)
In simple words: First substitute the trigonometric values to get simple numbers. Then perform matrix operations normally - addition, subtraction, or multiplication as required.

Exam Tip: Always compute trigonometric values before performing matrix operations. This step eliminates a major source of errors and makes calculations cleaner.

Question 12. Given matrix \( X = \begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix} \) and \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \), prove that \( X^2 = 4X + 5I \).
Answer: To prove this, we'll calculate the left-hand side (L.H.S.) and right-hand side (R.H.S.) separately and show they are equal.

L.H.S. - Calculate \( X^2 \):
\( X^2 = \begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix} = \begin{bmatrix} 1(1) + 1(8) & 1(1) + 1(3) \\ 8(1) + 3(8) & 8(1) + 3(3) \end{bmatrix} = \begin{bmatrix} 1 + 8 & 1 + 3 \\ 8 + 24 & 8 + 9 \end{bmatrix} = \begin{bmatrix} 9 & 4 \\ 32 & 17 \end{bmatrix} \)

R.H.S. - Calculate \( 4X + 5I \):
\( 4X = 4\begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 4 \\ 32 & 12 \end{bmatrix} \)

\( 5I = 5\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} \)

\( 4X + 5I = \begin{bmatrix} 4 & 4 \\ 32 & 12 \end{bmatrix} + \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} = \begin{bmatrix} 9 & 4 \\ 32 & 17 \end{bmatrix} \)

Since L.H.S. = R.H.S., we have proved that \( X^2 = 4X + 5I \).
In simple words: Square the matrix X and separately compute 4 times X plus 5 times the identity matrix. Both calculations give the same result, which proves the equation.

Exam Tip: For proof questions, always calculate both sides independently and show they are identical. Present your work clearly with the L.H.S. and R.H.S. labels so the examiner can follow your logic easily.