$\displaystyle \text{Question 1: Given } A = \begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix} , B=\begin{bmatrix} a & b \\ 0 & c \end{bmatrix} \text{ and that }AB = A+B \\ \\ \text{Find the value of } a, b \text{and } c$

$\displaystyle AB = A+B$

$\displaystyle \begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix} . \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix} + \begin{bmatrix} a & b \\ 0 & c \end{bmatrix}$

$\displaystyle \begin{bmatrix} 3a & 3b \\ 0 & 4c \end{bmatrix} = \begin{bmatrix} 3+a & b \\ 0 & 4+c \end{bmatrix}$

$\displaystyle \Rightarrow 3a = 3+a \ or \ a = \frac{3}{2}$

Similarly, $\displaystyle 3b=b \Rightarrow b = 0$

$\displaystyle \text{and } 4c = 4+c \ or \ c = \frac{4}{3}$

$\displaystyle \text{Hence } a =\frac{3}{2}, \ b= 0 \ c= \frac{4}{3}$

$\displaystyle \\$

$\displaystyle \text{Question 2: If } P = \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} , \text{and } Q=\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \text{ then find: }$

$\displaystyle \text{ i) } P^2-Q^2 \hspace{1.0cm} \text{ ii) } (P+Q)(P-Q)$      Also find if $\displaystyle P^2-Q^2 = (P+Q)(P-Q)$

$\displaystyle \text{ i) } P^2-Q^2$

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

$\displaystyle \text{ ii) } (P+Q)(P-Q)$

$\displaystyle = (\begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix})(\begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} -\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} ) = \begin{bmatrix} 2 & 2 \\ 4 & 0 \end{bmatrix} . \begin{bmatrix} 0 & 2 \\ 0 & -2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 8 \end{bmatrix}$

$\displaystyle \text{Therefore} P^2-Q^2 \neq (P+Q)(P-Q)$

$\displaystyle \\$

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

$\displaystyle \text{ i) } ABC \hspace{1.0cm} \text{ ii) } ACB \text{. Find whether} ABC = ACB$

$\displaystyle \text{ i) } ABC = \begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix} . \begin{bmatrix} 3 & 4 \\ -1 & -2 \end{bmatrix} . \begin{bmatrix} -3 & 1 \\ 0 & -2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 10 & 17 \end{bmatrix}. \begin{bmatrix} -3 & 1 \\ 0 & -2 \end{bmatrix} =\begin{bmatrix} -15 & -7 \\ -30 & -14 \end{bmatrix}$

$\displaystyle \text{ ii) } ACB =\begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}. \begin{bmatrix} -3 & 1 \\ 0 & -2 \end{bmatrix}.\begin{bmatrix} 3 & 4 \\ -1 & -2 \end{bmatrix} = \begin{bmatrix} -6 & 0 \\ -12 & 0 \end{bmatrix}.\begin{bmatrix} 3 & 4 \\ -1 & -2 \end{bmatrix} = \begin{bmatrix} -18 & -24 \\ -36 & -48 \end{bmatrix}$

$\displaystyle \text{Therefore} ABC \neq ACB$

$\displaystyle \\$

$\displaystyle \text{Question 4: If } A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} , B=\begin{bmatrix} 6 & 1 \\ 1 & 1 \end{bmatrix} \text{and } C=\begin{bmatrix} -2 & -3 \\ 0 & 1 \end{bmatrix} \\ \\ \text{Find i) } CA+B \hspace{1.0cm} \text{ii) } A+CB \text{ Are these equal.}$

$\displaystyle \text{i) } CA+B = \begin{bmatrix} -2 & -3 \\ 0 & 1 \end{bmatrix} . \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}+\begin{bmatrix} 6 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} -11 & -16 \\ 3 & 4 \end{bmatrix}+\begin{bmatrix} 6 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} -5 & -15 \\ 4 & 5 \end{bmatrix}$

$\displaystyle \text{ ii) } A+CB = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} -2 & -3 \\ 0 & 1 \end{bmatrix}. \begin{bmatrix} 6 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} +\begin{bmatrix} -15 & -5 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} -14 & -3 \\ 4 & 4 \end{bmatrix}$

$\displaystyle \\$

$\displaystyle \text{Question 5: If } A = \begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix} , B=\begin{bmatrix} 3 \\ -11 \end{bmatrix} \text{, find matrix } X \text{ such that } AX=B$

$\displaystyle A_{2 \times 2} \times X_{m \times n} = B_{2 \times 1}$

$\displaystyle \Rightarrow m = 2 \ and \ n = 1$

$\displaystyle \text{Let } X = \begin{bmatrix} a \\ b \end{bmatrix}$

$\displaystyle AX=B$

$\displaystyle \begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix}. \begin{bmatrix} a \\ b \end{bmatrix}=\begin{bmatrix} 3 \\ -11 \end{bmatrix}$

$\displaystyle \begin{bmatrix} 2a+b\\ a+3b \end{bmatrix}= \begin{bmatrix} 3 \\ -11 \end{bmatrix}$

Therefore

$\displaystyle 2a+b = 3$

$\displaystyle a+3b=-11$

Solving the above two equations $\displaystyle a = 4 \ and \ b = -5$

$\displaystyle \\$

$\displaystyle \text{Question 6: If } A = \begin{bmatrix} 4 & 2 \\ 1 & 1 \end{bmatrix} \text{, find } (A-2I)(A-3I)$

$\displaystyle I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\displaystyle (A-2I)(A-3I)$

$\displaystyle = (\begin{bmatrix} 4 & 2 \\ 1 & 1 \end{bmatrix}-2\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix})(\begin{bmatrix} 4 & 2 \\ 1 & 1 \end{bmatrix}-3\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix})$

$\displaystyle = (\begin{bmatrix} 4 & 2 \\ 1 & 1 \end{bmatrix}-\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix})(\begin{bmatrix} 4 & 2 \\ 1 & 1 \end{bmatrix}-\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix})$

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

$\displaystyle = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}$

$\displaystyle \\$

$\displaystyle \text{Question 7: If } A = \begin{bmatrix} 2 & 1 &-1 \\ 0 & 1 & -2 \end{bmatrix} \text{, find } \text{ i) } A^t.A \hspace{1.0cm} \text{ ii) } A.A^t$

$\displaystyle A = \begin{bmatrix} 2 & 1 &-1 \\ 0 & 1 & -2 \end{bmatrix}$

$\displaystyle \text{Therefore} A^t = \begin{bmatrix} 2 & 0 \\ 1 & 1 \\ -1 & -2 \end{bmatrix}$

$\displaystyle \text{ i) } A^t.A = \begin{bmatrix} 2 & 0 \\ 1 & 1 \\ -1 & -2 \end{bmatrix}.\begin{bmatrix} 2 & 1 &-1 \\ 0 & 1 & -2 \end{bmatrix} = \begin{bmatrix} 4& 2 &-2 \\ 2 & 2 & -3 \\ -2 & -3 & 5 \end{bmatrix}$

$\displaystyle \text{ ii) } A.A^t = \begin{bmatrix} 2 & 1 &-1 \\ 0 & 1 & -2 \end{bmatrix}.\begin{bmatrix} 2 & 0 \\ 1 & 1 \\ -1 & -2 \end{bmatrix} = \begin{bmatrix} 6 & 3 \\ 3 & 5 \end{bmatrix}$

$\displaystyle \\$

$\displaystyle \text{Question 8: If } M = \begin{bmatrix} 4 & 1 \\ -1 & 2 \end{bmatrix} \text{, show that } 6M-M^2=9I \text{; where } I \\ \\ \text{ is a } 2 \times 2 \text{ unit matrix. }$

$\displaystyle 6M-M^2=9I$

$\displaystyle \text{LHS} = 6 \begin{bmatrix} 4 & 1 \\ -1 & 2 \end{bmatrix}- \begin{bmatrix} 4 & 1 \\ -1 & 2 \end{bmatrix}.\begin{bmatrix} 4 & 1 \\ -1 & 2 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 24 & 4 \\ 6- & 12 \end{bmatrix}-\begin{bmatrix} 15 & 6 \\ -6 & 3 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 9 & 0 \\ 0 & 9 \end{bmatrix}$

$\displaystyle = 9 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\displaystyle = 9I$

LHS = RHS. Hence proven.

$\displaystyle \\$

$\displaystyle \text{Question 9: If } P = \begin{bmatrix} 2 & 6 \\ 3 & 9 \end{bmatrix} , Q=\begin{bmatrix} 3 & x \\ y & 2 \end{bmatrix} \text{, find } x \ and \ y \text{ such that } PQ= \text{null matrix}$

$\displaystyle PQ=\text{null matrix}$

Therefore

$\displaystyle \begin{bmatrix} 2 & 6 \\ 3 & 9 \end{bmatrix} . \begin{bmatrix} 3 & x \\ y & 2 \end{bmatrix}= \begin{bmatrix} 0 & 0 \\ 0& 0 \end{bmatrix}$

$\displaystyle \begin{bmatrix} 6+6y & 2x+12 \\ 9+9y & 3x+18 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0& 0 \end{bmatrix}$

Therefore

$\displaystyle 6+6y=0 \Rightarrow y = -1$

$\displaystyle 2x+12=0 \Rightarrow x = -6$

$\displaystyle \text{Hence } x = -6 \ and \ y = -1$

$\displaystyle \\$

$\displaystyle \text{Question 10: Evaluate } \begin{bmatrix} 2 \cos {60^{\circ}} & -2 \sin{30^{\circ}} \\ -\tan{45^{\circ}} & \cos{0^{\circ}} \end{bmatrix}$ . $\displaystyle \begin{bmatrix} \cot{45^{\circ}} & \mathrm{cosec{\ 30^0}} \\ \sec{60^{\circ}} & \sin{90^{\circ}} \end{bmatrix}$

$\displaystyle \begin{bmatrix} 2 \cos {60^{\circ}} & -2 \sin{30^{\circ}} \\ -\tan{45^{\circ}} & \cos{0^{\circ}} \end{bmatrix}$ . $\displaystyle \begin{bmatrix} \cot{45^{\circ}} & \mathrm{cosec{\ 30^{\circ}}} \\ \sec{60^{\circ}} & \sin{90^o} \end{bmatrix}$

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

$\displaystyle = \begin{bmatrix} -1 & 1\\ 1 & -1 \end{bmatrix}$

$\displaystyle \\$

Question 11: State True or False with reason

$\displaystyle \text{ i) } A+B=B+A$ : True > addition of matrices is commutative

$\displaystyle \text{ ii) } A-B=B-A$ : False > subtraction of matrices is not commutative

$\displaystyle \text{ iii) } (B.C).A = B.(C.A)$ : True > Multiplication of matrices is associative

$\displaystyle \text{ iv) } (A+B).C = A.C+B.C$ : True > Multiplication of matrices is distributive over addition

$\displaystyle \text{ v) } A.(B-C)=A.B-A.C$ : True > Multiplication of matrices is distributive over subtraction

$\displaystyle \text{ vi) } (A-B).C=A.C-B.C$ : True > Multiplication of matrices is distributive over subtraction

$\displaystyle \text{ vii) } A^2-B^2=(A+B)(A-B)$ : False > Laws of algebra for factorization is not applicable to matrices

$\displaystyle \text{viii) } (A-B)^2=A^2-2A.B+B^2$ : False > Laws of algebra for factorization is not applicable to matrices