\displaystyle \text{Question 1: Find} x \text{ and } y \text{, if} \begin{bmatrix} 3 & -2 \\ -1 & 4 \end{bmatrix}.\begin{bmatrix} 2x \\ 1 \end{bmatrix} +2 \begin{bmatrix} -4 \\ 5 \end{bmatrix}=4 \begin{bmatrix} 2 \\ y \end{bmatrix} \hspace{1.0cm} [2003]  

Answer: 

\displaystyle \begin{bmatrix} 3 & -2 \\ -1 & 4 \end{bmatrix}.\begin{bmatrix} 2x \\ 1 \end{bmatrix} +2 \begin{bmatrix} -4 \\ 5 \end{bmatrix}=4 \begin{bmatrix} 2 \\ y \end{bmatrix}

\displaystyle \begin{bmatrix} 6x-2 \\ -2x+4 \end{bmatrix} + \begin{bmatrix} -8 \\ 10 \end{bmatrix}= \begin{bmatrix} 8 \\ 4y \end{bmatrix}

Therefore

\displaystyle 6x-10=8 \Rightarrow x = 3

and \displaystyle -2x+14=4y \Rightarrow y = 2

\displaystyle \text{Hence } x = 3 \text{ and } y = 2.

\displaystyle \\

\displaystyle \text{Question 2: Given } \begin{bmatrix} 2 & 1 \\ -3 & 4 \end{bmatrix} . X = \begin{bmatrix} 7 \\ 6 \end{bmatrix}

Find: i) the order of matrix \displaystyle X            ii) the matrix \displaystyle X  \hspace{1.0cm}  [2012]  

Answer: 

\displaystyle \begin{bmatrix} 2 & 1 \\ -3 & 4 \end{bmatrix} . X = \begin{bmatrix} 7 \\ 6 \end{bmatrix}

\displaystyle A_{2 \times 2} . X_{p \times q} = B_{2 \times 1}

Therefore \displaystyle p = 2 \text{ and } q = 1 . Hence the order of Matrix is \displaystyle 2 \times 1

Let \displaystyle X = \begin{bmatrix} x \\ y \end{bmatrix}

Therefore

\displaystyle \begin{bmatrix} 2 & 1 \\ -3 & 4 \end{bmatrix} . \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ 6 \end{bmatrix}

\displaystyle \begin{bmatrix} 2x+y \\ -3x+4y \end{bmatrix} = \begin{bmatrix} 7 \\ 6 \end{bmatrix}

Therefore

\displaystyle 2x+y = 7

and \displaystyle -3x+4y = 6

Solving we get \displaystyle x = 2 \text{ and } y = 3

\displaystyle \text{Hence } X = \begin{bmatrix} 2 \\ 3 \end{bmatrix}

\displaystyle \\

\displaystyle \text{Question 3: If } \begin{bmatrix} a & 3 \\ 4 & 1 \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}

\displaystyle \text{Find the value of } a, b \text{ and } c \hspace{1.0cm} [1981]  

Answer: 

\displaystyle \begin{bmatrix} a & 3 \\ 4 & 1 \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}

\displaystyle \begin{bmatrix} a+2-1 & 3+b-1 \\ 4+1+2 & 1-2-c \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 7 & 3 \end{bmatrix}

\displaystyle \begin{bmatrix} a+1 & b+2 \\ 7 & -c-1 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 7 & 3 \end{bmatrix}

Therefore

\displaystyle a+1 = 5 \Rightarrow 4

\displaystyle b+2 = 0 \Rightarrow b = -2

\displaystyle -c-1=3 \Rightarrow c = -4

\displaystyle \\

\displaystyle \text{Question 4: If } A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}  \text{, and } B = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}

Find: i) \displaystyle A(BA)            ii) \displaystyle (AB) B. \hspace{1.0cm} [1991]  

Answer: 

i) \displaystyle A(BA)

\displaystyle = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} [ \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}. \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}]

\displaystyle =\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}. \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix}

\displaystyle = \begin{bmatrix} 14 & 13 \\ 13 & 14 \end{bmatrix}

ii) \displaystyle (AB) B

\displaystyle = (\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} . \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}) . \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}

\displaystyle = \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix} . \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}

\displaystyle = \begin{bmatrix} 14 & 13 \\ 13 & 14 \end{bmatrix}

\displaystyle \\

\displaystyle \text{Question 5: Find } x \text{ and } y \text{, if} \begin{bmatrix} x & 3x \\ y & 4y \end{bmatrix}. \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 12 \end{bmatrix} \hspace{1.0cm} [1992, 2013]  

Answer: 

\displaystyle \begin{bmatrix} x & 3x \\ y & 4y \end{bmatrix}. \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 12 \end{bmatrix}

\displaystyle \begin{bmatrix} 2x + 3x \\ 2y+4y \end{bmatrix}. = \begin{bmatrix} 5 \\ 12 \end{bmatrix}

Therefore

\displaystyle 5x=5 \Rightarrow x = 1

and \displaystyle 6y = 12 \Rightarrow y = 2

\displaystyle \\

\displaystyle \text{Question 6: Given } A = \begin{bmatrix} 2 & -1 \\ 2 & 0 \end{bmatrix} \text{, } B = \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix} \text{ and } C = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}

\displaystyle \text{Find } X  \text{ such that } A+X=2B+C \hspace{1.0cm} [2005]

Answer: 

\displaystyle A+X=2B+C

\displaystyle \begin{bmatrix} 2 & -1 \\ 2 & 0 \end{bmatrix}+X=2\begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix}+ \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}

\displaystyle \begin{bmatrix} 2 & -1 \\ 2 & 0 \end{bmatrix}+X = \begin{bmatrix} -5 & 4 \\ 8 & 2 \end{bmatrix}

\displaystyle X = \begin{bmatrix} -5 & 4 \\ 8 & 2 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ 2 & 0 \end{bmatrix}

\displaystyle X = \begin{bmatrix} -7 & 5 \\ 6 & 2 \end{bmatrix}

\displaystyle \\

\displaystyle \text{Question 7: Find the value of  x  given that } A^2=B \text{, } A = \begin{bmatrix} 2 & 12 \\ 0 & 1 \end{bmatrix} \text{, and } B = \begin{bmatrix} 4 & x \\ 0 & 1 \end{bmatrix} \hspace{1.0cm} [2005]

Answer: 

\displaystyle A^2=B

\displaystyle \begin{bmatrix} 2 & 12 \\ 0 & 1 \end{bmatrix} . \begin{bmatrix} 2 & 12 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 4 & x \\ 0 & 1 \end{bmatrix}

\displaystyle \begin{bmatrix} 4 & 36 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & x \\ 0 & 1 \end{bmatrix}

Therefore \displaystyle x = 36

\displaystyle \\

\displaystyle \text{Question 8: If } A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix} \text{ , and } B = \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix} \text{ and } I matrix of the same order and \displaystyle A^t is the transpose of the matrix, find \displaystyle A^t.B+BI \hspace{1.0cm} [2011]

Answer: 

\displaystyle A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}

\displaystyle A^t = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}

\displaystyle A^t.B+BI

\displaystyle = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix} . \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix}+ \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix} . \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

\displaystyle = \begin{bmatrix} 7 & -1 \\ 17 & -1 \end{bmatrix} + \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix}

\displaystyle = \begin{bmatrix} 11 & -3 \\ 16 & 2 \end{bmatrix}

\displaystyle \\

\displaystyle \text{Question 9: Given } A = \begin{bmatrix} 2 & -6 \\ 2 & 0 \end{bmatrix} \text{, } B = \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix} \text{ and } C = \begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix}

Find the matrix \displaystyle X such that \displaystyle A+2X=2B+C \hspace{1.0cm} [2013]

Answer: 

\displaystyle A+2X=2B+C

\displaystyle \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}

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

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

\displaystyle 2X= \begin{bmatrix} -2 & 4 \\ 8 & 2 \end{bmatrix} -\begin{bmatrix} 2 & -6 \\ 2 & 0 \end{bmatrix}

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

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

\displaystyle \\

\displaystyle \text{Question 10: Let } A = \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix} \text{, } B = \begin{bmatrix} 0 & 2 \\ 1 & -1 \end{bmatrix} \text{ and } C = \begin{bmatrix} -2 & 3 \\ 1 & -1 \end{bmatrix}

Find \displaystyle A^2-A+BC \hspace{1.0cm} [2006]  

Answer: 

\displaystyle A^2-A+BC

\displaystyle = \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix}.\begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix}-\begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix}+\begin{bmatrix} 0 & 2 \\ 1 & -1 \end{bmatrix}. \begin{bmatrix} -2 & 3 \\ 1 & -1 \end{bmatrix}

\displaystyle = \begin{bmatrix} 4 & -2 \\ 6 & 3 \end{bmatrix} - \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix} + \begin{bmatrix} 2 & -2 \\ -4 & 4 \end{bmatrix}

\displaystyle = \begin{bmatrix} 2 & -2 \\ -1 & 11 \end{bmatrix}

\displaystyle \\

\displaystyle \text{Question 11: Let } A = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \text{, } B = \begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix}

Find \displaystyle A^2+AB+B^2 \hspace{1.0cm} [2007]  

Answer: 

\displaystyle A^2+AB+B^2

\displaystyle =\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}.\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}+\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}.\begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix}+\begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix}.\begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix}

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

\displaystyle =\begin{bmatrix} 4 & 9 \\ 5 & 4 \end{bmatrix}

\displaystyle \\

\displaystyle \text{Question 12: Given } A = \begin{bmatrix} p & 0 \\ 0 & 2 \end{bmatrix} \text{, } B = \begin{bmatrix} 0 & -q \\ 1 & 0 \end{bmatrix} \text{ and } C = \begin{bmatrix} 2 & -2 \\ 2 & 2 \end{bmatrix} \text{ and } BA=C^2

Find the value of \displaystyle p \text{ and } q \hspace{1.0cm} [2008]

Answer: 

\displaystyle BA=C^2

\displaystyle \begin{bmatrix} 0 & -q \\ 1 & 0 \end{bmatrix}.\begin{bmatrix} p & 0 \\ 0 & 2 \end{bmatrix}=\begin{bmatrix} 2 & -2 \\ 2 & 2 \end{bmatrix} .\begin{bmatrix} 2 & -2 \\ 2 & 2 \end{bmatrix}

\displaystyle \begin{bmatrix} 0 & -2q \\ p & 0 \end{bmatrix} = \begin{bmatrix} 0 & -8 \\ 8 & 0 \end{bmatrix}

Therefore

\displaystyle -2q = -8 \Rightarrow q = 4

and \displaystyle p = 8

\displaystyle \\

\displaystyle \text{Question 13: Given } A = \begin{bmatrix} 3 & -2 \\ -1 & 4 \end{bmatrix} \text{, } B = \begin{bmatrix} 6 \\ 1 \end{bmatrix} \text{, } C = \begin{bmatrix} -4 \\ 5 \end{bmatrix} \text{ and } D = \begin{bmatrix} 2 \\ 2 \end{bmatrix}

Find \displaystyle AB+2C-4D \hspace{1.0cm} [2010]

Answer: 

\displaystyle AB+2C-4D

\displaystyle = \begin{bmatrix} 3 & -2 \\ -1 & 4 \end{bmatrix}. \begin{bmatrix} 6 \\ 1 \end{bmatrix} + 2 \begin{bmatrix} -4 \\ 5 \end{bmatrix} - 4 \begin{bmatrix} 2 \\ 2 \end{bmatrix}

\displaystyle = \begin{bmatrix} 16 \\ -2 \end{bmatrix}+\begin{bmatrix} -8 \\ 10 \end{bmatrix}- \begin{bmatrix} 8 \\ 8 \end{bmatrix}

\displaystyle = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

\displaystyle \\

\displaystyle \text{Question 14: Evaluate } \begin{bmatrix} 4\sin{30^o} & 2\cos{60^o} \\ \sin{90^o} & 2\cos{0^o} \end{bmatrix} . \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix} \hspace{1.0cm} [2010]  

Answer: 

\displaystyle \begin{bmatrix} 4\sin{30^o} & 2\cos{60^o} \\ \sin{90^o} & 2\cos{0^o} \end{bmatrix} . \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix}

\displaystyle = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}. \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix}

\displaystyle = \begin{bmatrix} 13 & 14 \\ 14 & 13 \end{bmatrix}

\displaystyle \\

\displaystyle \text{Question 15: If } A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \text{, } I= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

Find \displaystyle A^2-5A+7I \hspace{1.0cm} [2012]

Answer: 

\displaystyle A^2-5A+7I

\displaystyle = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}.\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}-5\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}+7\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

\displaystyle = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}-\begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}+\begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}

\displaystyle = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}

\displaystyle \\

\displaystyle \text{Question 16: If } A = \begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix}, B = \begin{bmatrix} 4 & 1 \\ -3 & -2 \end{bmatrix} \text{ and } C = \begin{bmatrix} -3 & 2 \\ -1 & 4 \end{bmatrix}

\displaystyle \text{Find } A^2 + AC-5B \hspace{1.0cm} [2014]  

Answer: 

\displaystyle A^2 + AC-5B

\displaystyle = \begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix} + \begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix} \begin{bmatrix} -3 & 2 \\ -1 & 4 \end{bmatrix} - 5 \begin{bmatrix} 4 & 1 \\ -3 & -2 \end{bmatrix}

\displaystyle = \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 & 14 \end{bmatrix}

\displaystyle \\

\displaystyle \text{Question 17: Solve for } x \text{ and } y  

\displaystyle \text{i) } \begin{bmatrix} 2 & 5 \\ 5 & 2 \end{bmatrix} . \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -7 \\ 14 \end{bmatrix}

\displaystyle \text{ii) } \begin{bmatrix} x+y & x-4 \end{bmatrix} . \begin{bmatrix} -1 & -2 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} -7 & 14 \end{bmatrix}  

\displaystyle \text{iii) } \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} \hspace{1.0cm} [2014]  

Answer: 

\displaystyle \text{i) } \begin{bmatrix} 2 & 5 \\ 5 & 2 \end{bmatrix} . \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -7 \\ 14 \end{bmatrix}

\displaystyle \Rightarrow \begin{bmatrix} 2x+5y & 5x+2y \end{bmatrix} = \begin{bmatrix} -7 \\ 14 \end{bmatrix}

Therefore

\displaystyle 2x+5y = -7

\displaystyle 5x+2y=14

Solving the above two equations we get

\displaystyle x = 4 \text{ and } y = -3

\displaystyle \text{ii) } \begin{bmatrix} x+y & x-4 \end{bmatrix} . \begin{bmatrix} -1 & -2 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} -7 & 14 \end{bmatrix}

\displaystyle \Rightarrow \begin{bmatrix} x-y-8 & -2y-8 \end{bmatrix} = \begin{bmatrix} -7 & 14 \end{bmatrix}

Therefore

\displaystyle x-y-8=-7 \Rightarrow x-y=1

Also \displaystyle -2y-8=-11 \Rightarrow y = \frac{3}{2}

Substituting we get \displaystyle x = \frac{3}{2}+1 = \frac{5}{2}

\displaystyle \text{iii) } \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}

\displaystyle \Rightarrow \begin{bmatrix} 2 \\ -3+2x \end{bmatrix} + \begin{bmatrix} -6 \\ 3 \end{bmatrix} = \begin{bmatrix} 2y \\ 6 \end{bmatrix}

\displaystyle \begin{bmatrix} -4 \\ 2x \end{bmatrix} = \begin{bmatrix} 2y \\ 6 \end{bmatrix}

Therefore \displaystyle y = -2 and x = \frac{6}{2}=3

\displaystyle \\

\displaystyle \text{Question 18: If } A = \begin{bmatrix} 3 & 5 \\ 4 & -2 \end{bmatrix} \text{ and } B = \begin{bmatrix} 2 \\ 4 \end{bmatrix} , is the product of \displaystyle AB \text{ possible } \hspace{1.0cm} \hspace{1.0cm} [2011]  

Answer: 

The order of matrix \displaystyle A = 2 \times 2 and the order of matrix \displaystyle B \ is \ 2 \times 1

Since the number of columns in \displaystyle A is equal to the number of rows in \displaystyle B , the product \displaystyle AB is possible.

\displaystyle AB = \begin{bmatrix} 3 & 5 \\ 4 & -2 \end{bmatrix} . \begin{bmatrix} 4 \\ 2 \end{bmatrix} = \begin{bmatrix} 6+20 \\ 8-8 \end{bmatrix} = \begin{bmatrix} 26 \\ 0 \end{bmatrix}

\displaystyle \\