07. Matrices Previous Year Questions

A² = [[0, 1], [1, 0]] × [[0, 1], [1, 0]] = [[0(0)+1(1), 0(1)+1(0)], [1(0)+0(1), 1(1)+0(0)]] = [[1, 0], [0, 1]] = I. [1 Mark]

(a) A + C = [[1−5, 2+1], [3+7, 4−4]] = [[−4, 3], [10, 0]]. [1 Mark]
(b) B(A+C) = [[2, 1], [4, 2]] × [[−4, 3], [10, 0]] = [[2(−4)+1(10), 2(3)+0], [4(−4)+2(10), 4(3)+0]] = [[2, 6], [4, 12]]. [1 Mark]
(c) 5B = 5[[2, 1], [4, 2]] = [[10, 5], [20, 10]]. [1 Mark]
(d) B(A+C) − 5B = [[2−10, 6−5], [4−20, 12−10]] = [[−8, 1], [−16, 2]]. [1 Mark]

Correct Option: (c) 8
A² = [[2, 2], [0, 2]] × [[2, 2], [0, 2]] = [[4, (2×2 + 2×2)], [0, 4]] = [[4, 8], [0, 4]].
Comparing: x = 8. [1 Mark]

AB = [[x×4 + 0, 0], [1×4 + 1×y, 1]] = [[4x, 0], [4 + y, 1]]. [2 Marks]
Equating with C = [[4, 0], [x, 1]]:
4x = 4 ⇒ x = 1. [1 Mark]
4 + y = x ⇒ 4 + y = 1 ⇒ y = −3. [1 Mark]

Step 1 Multiplying the matrices on LHS: $\begin{pmatrix} 2x + 0 \\ 0 + 4y \end{pmatrix} = \begin{pmatrix} 2 \\ -8 \end{pmatrix}$.
Step 2 Equating corresponding elements:
$2x = 2 \implies x = 1$
$4y = -8 \implies y = -2$
Therefore, $x = 1, y = -2$. Correct Option: (a) [1 Mark]

Step 1 Add matrices $B$ and $C$:
$B + C = \begin{pmatrix} 1+4 & 2+1 \\ 2+1 & 4+5 \end{pmatrix} = \begin{pmatrix} 5 & 3 \\ 3 & 9 \end{pmatrix}$.1 Mark

Step 2 Matrix multiplication $A(B + C)$:
$A(B + C) = \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix} \begin{pmatrix} 5 & 3 \\ 3 & 9 \end{pmatrix} = \begin{pmatrix} 1(5) + 3(3) & 1(3) + 3(9) \\ 2(5) + 4(3) & 2(3) + 4(9) \end{pmatrix} = \begin{pmatrix} 14 & 30 \\ 22 & 42 \end{pmatrix}$.1 Mark

Step 3 Subtract $14I$ where $14I = \begin{pmatrix} 14 & 0 \\ 0 & 14 \end{pmatrix}$:
$A(B + C) - 14I = \begin{pmatrix} 14 - 14 & 30 - 0 \\ 22 - 0 & 42 - 14 \end{pmatrix} = \mathbf{\begin{pmatrix} 0 & 30 \\ 22 & 28 \end{pmatrix}}$.1 Mark

Step 1 Compute $A^2$:
$$A^2 = \begin{pmatrix} x & 3 \\ y & 3 \end{pmatrix} \begin{pmatrix} x & 3 \\ y & 3 \end{pmatrix} = \begin{pmatrix} x^2 + 3y & 3x + 9 \\ xy + 3y & 3y + 9 \end{pmatrix}$$ [1 Mark]
Step 2 Given $A^2 = 3I = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}$. Equating corresponding elements:
$3x + 9 = 0 \implies 3x = -9 \implies \mathbf{x = -3}$.
$3y + 9 = 3 \implies 3y = -6 \implies \mathbf{y = -2}$. [1 Mark]
Step 3 Verification:
$x^2 + 3y = (-3)^2 + 3(-2) = 9 - 6 = 3$ (True).
$xy + 3y = (-3)(-2) + 3(-2) = 6 - 6 = 0$ (True).
Answer: $x = -3, y = -2$. [1 Mark]

Step 1 Compute $A^2$ and $B^2$:
$$A^2 = \begin{pmatrix} 3 & 0 \\ 5 & 1 \end{pmatrix} \begin{pmatrix} 3 & 0 \\ 5 & 1 \end{pmatrix} = \begin{pmatrix} 9 & 0 \\ 20 & 1 \end{pmatrix}$$
$$B^2 = \begin{pmatrix} -4 & 2 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} -4 & 2 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 16 + 2 & -8 + 0 \\ -4 + 0 & 2 + 0 \end{pmatrix} = \begin{pmatrix} 18 & -8 \\ -4 & 2 \end{pmatrix}$$ [1 Mark]
Step 2 Compute $2AB$:
$$AB = \begin{pmatrix} 3 & 0 \\ 5 & 1 \end{pmatrix} \begin{pmatrix} -4 & 2 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} -12 & 6 \\ -19 & 10 \end{pmatrix} \implies 2AB = \begin{pmatrix} -24 & 12 \\ -38 & 20 \end{pmatrix}$$ [1 Mark]
Step 3 Combine $A^2 - 2AB + B^2$:
$$= \begin{pmatrix} 9 & 0 \\ 20 & 1 \end{pmatrix} - \begin{pmatrix} -24 & 12 \\ -38 & 20 \end{pmatrix} + \begin{pmatrix} 18 & -8 \\ -4 & 2 \end{pmatrix} = \begin{pmatrix} 9 - (-24) + 18 & 0 - 12 - 8 \\ 20 - (-38) - 4 & 1 - 20 + 2 \end{pmatrix} = \mathbf{\begin{pmatrix} 51 & -20 \\ 54 & -17 \end{pmatrix}}$$ [1 Mark]

Step 1 Multiply scalar elements into each matrix:
$\sin A \begin{bmatrix} \sin A & -\cos A \\ \cos A & \sin A \end{bmatrix} = \begin{bmatrix} \sin^2 A & -\sin A\cos A \\ \sin A\cos A & \sin^2 A \end{bmatrix}$ [1 Mark]
$\cos A \begin{bmatrix} \cos A & \sin A \\ -\sin A & \cos A \end{bmatrix} = \begin{bmatrix} \cos^2 A & \sin A\cos A \\ -\sin A\cos A & \cos^2 A \end{bmatrix}$ [1 Mark]
Step 2 Add corresponding matrix entries:
$= \begin{bmatrix} \sin^2 A + \cos^2 A & -\sin A\cos A + \sin A\cos A \\ \sin A\cos A - \sin A\cos A & \sin^2 A + \cos^2 A \end{bmatrix}$
Using Pythagorean identity $\sin^2 A + \cos^2 A = 1$:
$= \mathbf{\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}}$ (or Identity matrix $I$). [1 Mark]

Step 1 (i) Determine order of $M$:
Let $A = \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix}$, which has order $2 \times 2$.
RHS $= 6I = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}$, which has order $2 \times 2$.
For product $A_{2 \times 2} \times M_{r \times c} = (6I)_{2 \times 2}$ to be defined:
Number of rows of $M$ must equal columns of $A \implies r = 2$.
Result has order $2 \times c = 2 \times 2 \implies c = 2$.
$\mathbf{\therefore \text{Order of matrix } M = 2 \times 2}$. [1 Mark]
Step 2 (ii) Let $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$:
$\begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 4a + 2c & 4b + 2d \\ -a + c & -b + d \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}$. [1 Mark]
Step 3 Solve simultaneous linear systems:
From column 1:
$-a + c = 0 \implies c = a$.
$4a + 2c = 6 \implies 4a + 2a = 6 \implies 6a = 6 \implies a = 1 \implies c = 1$.
From column 2:
$4b + 2d = 0 \implies 2d = -4b \implies d = -2b$.
$-b + d = 6 \implies -b + (-2b) = 6 \implies -3b = 6 \implies b = -2 \implies d = 4$. [1 Mark]
Step 4 Form matrix $M$:
$\mathbf{M = \begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix}}$. [1 Mark]

Solution Compare corresponding matrix entries:
$m(-1) = 3 \implies \mathbf{m = -3}$, and $m(2) = -6 \implies m = -3$.
Correct Answer: (b) -3 [1 Mark]

Step 1 (a) Evaluate $A^2$ and $5A$:
$A^2 = \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} 3(3) + (-5)(-4) & 3(-5) + (-5)(2) \\ -4(3) + 2(-4) & -4(-5) + 2(2) \end{bmatrix} = \mathbf{\begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix}}$. [1 Mark]
$5A = 5 \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} = \mathbf{\begin{bmatrix} 15 & -25 \\ -20 & 10 \end{bmatrix}}$. [0.5 Mark]
Step 2 (b) Evaluate $A^2 - 5A$ and solve for $k$:
$A^2 - 5A = \begin{bmatrix} 29 - 15 & -25 - (-25) \\ -20 - (-20) & 24 - 10 \end{bmatrix} = \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix}$.
$kI = k \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix} \implies \mathbf{k = 14}$. [1.5 Marks]

Step 1 Multiply scalar 2 into the first matrix:
$$\begin{bmatrix} 2x & 14 \\ 18 & 2y - 10 \end{bmatrix} + \begin{bmatrix} 6 & -7 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 10 & 7 \\ 22 & 15 \end{bmatrix}$$ [1 Mark]
Step 2 Add corresponding elements on the LHS:
$$\begin{bmatrix} 2x + 6 & 14 - 7 \\ 18 + 4 & 2y - 10 + 5 \end{bmatrix} = \begin{bmatrix} 2x + 6 & 7 \\ 22 & 2y - 5 \end{bmatrix} = \begin{bmatrix} 10 & 7 \\ 22 & 15 \end{bmatrix}$$
Step 3 Equate corresponding elements:
1) $2x + 6 = 10 \implies 2x = 4 \implies \mathbf{x = 2}$ [1 Mark]
2) $2y - 5 = 15 \implies 2y = 20 \implies \mathbf{y = 10}$ [1 Mark]
Answer: $x = 2$, $y = 10$.

Step 1: Compute AC
$$AC = \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 4 \end{bmatrix} = \begin{bmatrix} 2(1)+3(-1) & 2(0)+3(4) \\ 5(1)+7(-1) & 5(0)+7(4) \end{bmatrix} = \begin{bmatrix} -1 & 12 \\ -2 & 28 \end{bmatrix}$$ [1 Mark]
Step 2: Compute B²
$$B^2 = \begin{bmatrix} 0 & 4 \\ -1 & 7 \end{bmatrix} \begin{bmatrix} 0 & 4 \\ -1 & 7 \end{bmatrix} = \begin{bmatrix} 0-4 & 0+28 \\ 0-7 & -4+49 \end{bmatrix} = \begin{bmatrix} -4 & 28 \\ -7 & 45 \end{bmatrix}$$ [1 Mark]
Step 3: Compute AC + B² - 10C
$$10C = \begin{bmatrix} 10 & 0 \\ -10 & 40 \end{bmatrix}$$ $$AC + B^2 - 10C = \begin{bmatrix} -1 - 4 - 10 & 12 + 28 - 0 \\ -2 - 7 - (-10) & 28 + 45 - 40 \end{bmatrix} = \mathbf{\begin{bmatrix} -15 & 40 \\ 1 & 33 \end{bmatrix}}$$ [1 Mark]
Answer: $\begin{bmatrix} -15 & 40 \\ 1 & 33 \end{bmatrix}$.

Step 1: Compute A²
$$A^2 = \begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 1(1)+3(3) & 1(3)+3(4) \\ 3(1)+4(3) & 3(3)+4(4) \end{bmatrix} = \begin{bmatrix} 10 & 15 \\ 15 & 25 \end{bmatrix}$$ [1 Mark]
Step 2: Compute B²
$$B^2 = \begin{bmatrix} -2 & 1 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} -2 & 1 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} 4-3 & -2+2 \\ 6-6 & -3+4 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$$ [1 Mark]
Step 3: Compute A² - 5B²
$$5B^2 = 5 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}$$ $$5C = A^2 - 5B^2 = \begin{bmatrix} 10 & 15 \\ 15 & 25 \end{bmatrix} - \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} = \begin{bmatrix} 5 & 15 \\ 15 & 20 \end{bmatrix}$$ [1 Mark]
Step 4: Solve for C
$$C = \frac{1}{5} \begin{bmatrix} 5 & 15 \\ 15 & 20 \end{bmatrix} = \mathbf{\begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix}}$$ [1 Mark]
Answer: $C = \begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix}$.

Step 1: Compute B²
$$B^2 = \begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix} = \begin{bmatrix} 1+8 & 1+3 \\ 8+24 & 8+9 \end{bmatrix} = \begin{bmatrix} 9 & 4 \\ 32 & 17 \end{bmatrix}$$ [1 Mark]
Step 2: Compute X = B² - 4B
$$4B = \begin{bmatrix} 4 & 4 \\ 32 & 12 \end{bmatrix}$$ $$X = \begin{bmatrix} 9 & 4 \\ 32 & 17 \end{bmatrix} - \begin{bmatrix} 4 & 4 \\ 32 & 12 \end{bmatrix} = \mathbf{\begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}}$$ [1 Mark]
Step 3: Solve for a and b
$$\begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 5a \\ 5b \end{bmatrix} = \begin{bmatrix} 5 \\ 50 \end{bmatrix}$$ $$5a = 5 \implies \mathbf{a = 1}$$ [1 Mark]
$$5b = 50 \implies \mathbf{b = 10}$$ [1 Mark]
Answers: $X = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}$; $a = 1, b = 10$.