Question 1: Evaluate: $\displaystyle \text{i) } 3 \begin{bmatrix} 5 \\ -2 \end{bmatrix} = \begin{bmatrix} 15 \\ -6 \end{bmatrix}$ $\displaystyle \text{ii) } 7 \begin{bmatrix} -1 & 2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -7 & 14 \\ 0 & 7 \end{bmatrix}$ $\displaystyle \text{iii) } 2 \begin{bmatrix} -1 & 0 \\ 2 & -3 \end{bmatrix} + \begin{bmatrix} 3 & 3 \\ 5 & 0 \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 4 & -6 \end{bmatrix} + \begin{bmatrix} 3 & 3 \\ 5 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 9 & -6 \end{bmatrix}$ $\displaystyle \text{iv) } 6 \begin{bmatrix} 3 \\ 2 \end{bmatrix} - 2 \begin{bmatrix} -8 \\ 1 \end{bmatrix} = \begin{bmatrix} 18 \\ 12 \end{bmatrix} - \begin{bmatrix} -16 \\ 2 \end{bmatrix} = \begin{bmatrix} 34 \\ -14 \end{bmatrix}$ $\displaystyle \\$

Question 2: Find $\displaystyle x \ and \ y$ if $\displaystyle \text{i) } 3 \begin{bmatrix} 4 & x \end{bmatrix} +2 \begin{bmatrix} y & -3 \end{bmatrix} = \begin{bmatrix} 10 & 0 \end{bmatrix} \hspace{1.0cm} \text{ii) } x \begin{bmatrix} -1 \\ 2 \end{bmatrix} -4 \begin{bmatrix} -2 \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ -8 \end{bmatrix}$ $\displaystyle \text{i) } 3 \begin{bmatrix} 4 & x \end{bmatrix} +2 \begin{bmatrix} y & -3 \end{bmatrix} = \begin{bmatrix} 10 & 0 \end{bmatrix}$ $\displaystyle \Rightarrow \begin{bmatrix} 12 & 3x \end{bmatrix} + \begin{bmatrix} 2y & -6 \end{bmatrix} = \begin{bmatrix} 10 & 0 \end{bmatrix}$

Therefore $\displaystyle 12+2y = \ 10 \ or y = -1$ $\displaystyle 3x-6=0 \ or \ x = 2$ $\displaystyle \text{ii) } x \begin{bmatrix} -1 \\ 2 \end{bmatrix} -4 \begin{bmatrix} -2 \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ -8 \end{bmatrix}$ $\displaystyle \begin{bmatrix} -x \\ 2x \end{bmatrix} - \begin{bmatrix} -8 \\ 4y \end{bmatrix} = \begin{bmatrix} 7 \\ -8 \end{bmatrix}$

Therefore $\displaystyle -x+8 = 7 \ or \ x = 1$ $\displaystyle 2x-4y = -8 \ or \ y = 10$ $\displaystyle \\$ $\displaystyle \text{Question 3: Given } A = \begin{bmatrix} 2 & 1 \\ 3 & 0 \end{bmatrix} , B = \begin{bmatrix} 1 & 1 \\ 5 & 2 \end{bmatrix} , C = \begin{bmatrix} -3 & -1 \\ 0 & 0 \end{bmatrix} \text{, Find:}$ $\displaystyle \text{i) } 2A-3B+C$ $\displaystyle \text{ii) } A+2C-B$ $\displaystyle \text{i) } 2A-3B+C$ $\displaystyle 2 \begin{bmatrix} 2 & 1 \\ 3 & 0 \end{bmatrix} - 3 \begin{bmatrix} 1 & 1 \\ 5 & 2 \end{bmatrix} + \begin{bmatrix} -3 & -1 \\ 0 & 0 \end{bmatrix}$ $\displaystyle = \begin{bmatrix} 4 & 2 \\ 6 & 0 \end{bmatrix} - \begin{bmatrix} 3 & 3 \\ 15 & 6 \end{bmatrix} + \begin{bmatrix} -3 & -1 \\ 0 & 0 \end{bmatrix}$ $\displaystyle = \begin{bmatrix} -2 & -2 \\ -9 & -6 \end{bmatrix}$ $\displaystyle \text{ii) } A+2C-B$ $\displaystyle \begin{bmatrix} 2 & 1 \\ 3 & 0 \end{bmatrix} + 2 \begin{bmatrix} -3 & -1 \\ 0 & 0 \end{bmatrix} - \begin{bmatrix} 1 & 1 \\ 5 & 2 \end{bmatrix}$ $\displaystyle = \begin{bmatrix} 2 & 1 \\ 3 & 0 \end{bmatrix} + \begin{bmatrix} -6 & -2 \\ 0 & 0 \end{bmatrix} - \begin{bmatrix} 1 & 1 \\ 5 & 2 \end{bmatrix}$ $\displaystyle = \begin{bmatrix} -5 & -2 \\ -2 & -2 \end{bmatrix}$ $\displaystyle \\$ $\displaystyle \text{Question 4: If } \begin{bmatrix} 4 & -2 \\ 4 & 0 \end{bmatrix} +3A = \begin{bmatrix} -2 & -2 \\ 1 & -3 \end{bmatrix} \text{, Find } A$ $\displaystyle \begin{bmatrix} 4 & -2 \\ 4 & 0 \end{bmatrix} +3A = \begin{bmatrix} -2 & -2 \\ 1 & -3 \end{bmatrix}$ $\displaystyle \Rightarrow 3A = \begin{bmatrix} -2 & -2 \\ 1 & -3 \end{bmatrix} - \begin{bmatrix} 4 & -2 \\ 4 & 0 \end{bmatrix}$ $\displaystyle \Rightarrow 3A = \begin{bmatrix} -6 & 0 \\ -3 & -3 \end{bmatrix}$ $\displaystyle \Rightarrow A = \begin{bmatrix} -2 & 0 \\ -1 & -1 \end{bmatrix}$ $\displaystyle \\$ $\displaystyle \text{Question 5: Given } A = \begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix}, B = \begin{bmatrix} -4 & -1 \\ -3 & -2 \end{bmatrix} \text{, Find: }$ $\displaystyle \text{i) } 2A+B \hspace{1.0cm} \text{ ii) Matrix } C \text{ such that } C+B = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$ $\displaystyle \text{i) } 2A+B$ $\displaystyle \Rightarrow 2 \begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix} + \begin{bmatrix} -4 & -1 \\ -3 & -2 \end{bmatrix} = \begin{bmatrix} -2 & 7 \\ 1 & 4 \end{bmatrix}$ $\displaystyle \text{ii) } C+B = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$ $\displaystyle C = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} - \begin{bmatrix} -4 & -1 \\ -3 & -2 \end{bmatrix}$ $\displaystyle C = \begin{bmatrix} 4 & 1 \\ 3 & 2 \end{bmatrix}$ $\displaystyle \\$ $\displaystyle \text{Question 6: If } 2 \begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} + 3 \begin{bmatrix} 1 & 3 \\ y & 2 \end{bmatrix} = \begin{bmatrix} z & -7 \\ 15 & 8 \end{bmatrix} \text{, Find the value of } x, \ y, \text{ and } z$ . $\displaystyle 2 \begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} + 3 \begin{bmatrix} 1 & 3 \\ y & 2 \end{bmatrix} = \begin{bmatrix} z & -7 \\ 15 & 8 \end{bmatrix}$ $\displaystyle \Rightarrow \begin{bmatrix} 6 & 2x \\ 0 & 2 \end{bmatrix} + \begin{bmatrix} 3 & 9 \\ 3y & 6 \end{bmatrix} = \begin{bmatrix} z & -7 \\ 15 & 8 \end{bmatrix}$ $\displaystyle \Rightarrow \begin{bmatrix} 9 & 2x+9 \\ 3y & 8 \end{bmatrix}= \begin{bmatrix} z & -7 \\ 15 & 8 \end{bmatrix}$

Therefore $\displaystyle z=9$ $\displaystyle 2x+9 = -7 \ or \ x = -8$ $\displaystyle 3y = 15 \ or \ y = 5$ $\displaystyle \\$ $\displaystyle \text{Question 7: Given } A = \begin{bmatrix} -3 & 6 \\ 0 & -9 \end{bmatrix} \text{ and } A^t \text{ is the transpose matrix. Find: }$ $\displaystyle \text{i) } 2A + 3A^t$ $\displaystyle \text{ii) } 2A^t-3A$ $\displaystyle \text{iii) } \frac{1}{2} A - \frac{1}{3}A^t$ $\displaystyle \text{iv) } A^t - \frac{1}{3}A$ $\displaystyle \text{If } A = \begin{bmatrix} -3 & 6 \\ 0 & -9 \end{bmatrix}$ $\displaystyle \text{Then } A^t = \begin{bmatrix} -3 & 0 \\ 6 & -9 \end{bmatrix}$ $\displaystyle \text{i) } 2A + 3A^t$ $\displaystyle 2 \begin{bmatrix} -3 & 6 \\ 0 & -9 \end{bmatrix} +3 \begin{bmatrix} -3 & 0 \\ 6 & -9 \end{bmatrix} = \begin{bmatrix} -15 & 12 \\ 18 & -45 \end{bmatrix}$ $\displaystyle \text{ii) } 2A^t-3A$ $\displaystyle 2 \begin{bmatrix} -3 & 0 \\ 6 & -9 \end{bmatrix} - 3 \begin{bmatrix} -3 & 6 \\ 0 & -9 \end{bmatrix} = \begin{bmatrix} 3 & -18 \\ 12 & 9 \end{bmatrix}$ $\displaystyle \text{iii) } \frac{1}{2} A - \frac{1}{3}A^t$ $\displaystyle \frac{1}{2}\begin{bmatrix} -3 & 6 \\ 0 & -9 \end{bmatrix} - \frac{1}{3}\begin{bmatrix} -3 & 0 \\ 6 & -9 \end{bmatrix} =\begin{bmatrix} -\frac{1}{2} & 3 \\ -2 & -\frac{3}{2} \end{bmatrix}$ $\displaystyle \text{iv) } A^t - \frac{1}{3}A$ $\displaystyle \begin{bmatrix} -3 & 0 \\ 6 & -9 \end{bmatrix}-\frac{1}{3} \begin{bmatrix} -3 & 6 \\ 0 & -9 \end{bmatrix}= \begin{bmatrix} -2 & -2 \\ 6 & -6 \end{bmatrix}$ $\displaystyle \\$ $\displaystyle \text{Question 8: Given } A = \begin{bmatrix} 1 & 1 \\ -2 & 0 \end{bmatrix} \text{ and } B = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} \text{ Solve for: }$ $\displaystyle \text{i) } X+2A=B$ $\displaystyle \text{ii) } 3X+B+2A=0$ $\displaystyle \text{iii) } 3A-2X=X-2B$ $\displaystyle \text{i) } X+2A=B$ $\displaystyle X = B-2A$ $\displaystyle X= \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}-2 \begin{bmatrix} 1 & 1 \\ -2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -3 \\ 5 & 1 \end{bmatrix}$ $\displaystyle \text{ii) } 3X+B+2A=0$ $\displaystyle X = \frac{-1}{3}(B+2A)$ $\displaystyle X = \frac{-1}{3} \Bigg(\begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}+2\begin{bmatrix} 1 & 1 \\ -2 & 0 \end{bmatrix} \Bigg) = \begin{bmatrix} \frac{4}{3} & \frac{1}{3} \\ -1 & \frac{1}{3} \end{bmatrix}$ $\displaystyle \text{iii) } 3A-2X=X-2B$ $\displaystyle X = \frac{1}{3} (3A+2B)$ $\displaystyle X = \frac{1}{3} \Bigg(3\begin{bmatrix} 1 & 1 \\ -2 & 0 \end{bmatrix} +2\begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} \Bigg) = \begin{bmatrix} \frac{-7}{3} & \frac{1}{3} \\ \frac{-4}{3} & \frac{2}{3} \end{bmatrix}$ $\displaystyle \\$ $\displaystyle \text{Question 9: If } M = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \text{ and } N = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \text{, show that } 3M+5N=\begin{bmatrix} 5 \\ 3 \end{bmatrix}$ $\displaystyle 3M+5N=3\begin{bmatrix} 0 \\ 1 \end{bmatrix} +5\begin{bmatrix} 0 \\ 1 \end{bmatrix} =\begin{bmatrix} 5 \\ 3 \end{bmatrix}$

Hence proved.

Question 10: If $\displaystyle I$ is the unit matrix of order $\displaystyle 2 \times 2$ , find the matrix $\displaystyle M$ such that $\displaystyle \text{i) } M-2I= 3\begin{bmatrix} -1 & 0 \\ 4 & 1 \end{bmatrix}$ $\displaystyle \text{ii) } 5M+3I= 4\begin{bmatrix} 2 & -5 \\ 0 & -3 \end{bmatrix}$

Given $\displaystyle I$ is a unit matrix of order $\displaystyle 2 \times 2$ , we have $\displaystyle I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ $\displaystyle \text{i) } M = 2\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + 3\begin{bmatrix} -1 & 0 \\ 4 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 12 & 5 \end{bmatrix}$ $\displaystyle \text{ii) } M= \frac{1}{5} (-3\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + 4\begin{bmatrix} 2 & -5 \\ 0 & -3 \end{bmatrix}) =\begin{bmatrix} 1 & -4 \\ 0 & -3 \end{bmatrix}$ $\displaystyle \\$ $\displaystyle \text{Question 11. If } \begin{bmatrix} 1 & 4 \\ -2 & 3 \end{bmatrix} +2M = 3 \begin{bmatrix} 3 & 2 \\ 0 & -3 \end{bmatrix} \text{ , find the matrix } M$ $\displaystyle 2M =3 \begin{bmatrix} 3 & 2 \\ 0 & -3 \end{bmatrix} - \begin{bmatrix} 1 & 4 \\ -2 & 3 \end{bmatrix}$ $\displaystyle 2M =\begin{bmatrix} 9 & 6 \\ 0 & -9 \end{bmatrix} - \begin{bmatrix} 1 & 4 \\ -2 & 3 \end{bmatrix} = \begin{bmatrix} 8 & 2 \\ 2 & -12 \end{bmatrix}$
Therefore $\displaystyle M =\begin{bmatrix} 4 & 1 \\ 1 & -6 \end{bmatrix}$ $\displaystyle \\$