Question 13 Marks
Find $x$ and $y$ if $\left[\begin{array}{ll}x & 3 x \\ y & 4 y\end{array}\right]\left[\begin{array}{l}2 \\ 1\end{array}\right]=\left[\begin{array}{c}5 \\ 12\end{array}\right]$
Answer$\begin{array}{l}{\left[\begin{array}{ll}x & 3 x \\ y & 4 y\end{array}\right]\left[\begin{array}{l}2 \\ 1\end{array}\right]=\left[\begin{array}{c}5 \\ 12\end{array}\right]}\end{array} $
$ {\left[\begin{array}{l}2 x+3 x \\ 2 y+4 y\end{array}\right]=\left[\begin{array}{c}5 \\ 12\end{array}\right]} $
$ {\left[\begin{array}{l}5 x \\ 6 y\end{array}\right] =\left[\begin{array}{c}5 \\ 12\end{array}\right]}$
Comparing the corresponding elements we get
$5 x=5 \Rightarrow x=1$
$6 y=12\Rightarrow x=1$
$6 y=12\Rightarrow y=2$
View full question & answer→Question 23 Marks
If $A=\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]$ and $B=\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]$ Find $(AB).B$
Answer$A=\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]$ and $B=\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]$
$\begin{array}{l}A B=\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]\end{array} $
$ =\left[\begin{array}{ll}2+2 & 1+4 \\ 4+1 & 2+2\end{array}\right] $
$ =\left[\begin{array}{ll}4 & 5 \\ 5 & 4\end{array}\right]$
$\begin{array}{l}(A B) B=\left[\begin{array}{ll}4 & 5 \\ 5 & 4\end{array}\right]\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right] \end{array} $
$ =\left[\begin{array}{cc}8+5 & 4+10 \\ 10+4 & 5+8\end{array}\right] $
$ =\left[\begin{array}{ll}13 & 14 \\ 14 & 13\end{array}\right]$
View full question & answer→Question 33 Marks
If $A=\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]$ and $B=\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]$ Find $A(B A)$
Answer$A=\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right], B=\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]$
$\begin{array}{l}B A=\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right] \end{array} $
$ =\left[\begin{array}{ll}2+2 & 4+1 \\ 1+4 & 2+2\end{array}\right] $
$ =\left[\begin{array}{ll}4 & 5 \\ 5 & 4\end{array}\right]$
$\begin{array}{l}A(B A)=\left[\begin{array}{ll}1 & 2 \\ 4 & 5\end{array}\right],\left[\begin{array}{ll}4 & 5 \\ 5 & 4\end{array}\right] \end{array} $
$ =\left[\begin{array}{cc}4+10 & 5+8 \\ 8+5 & 10+4\end{array}\right] $
$ =\left[\begin{array}{ll}14 & 13 \\ 13 & 14\end{array}\right]$
View full question & answer→Question 43 Marks
If $\left[\begin{array}{ll}a & 3 \\ 4 & 1\end{array}\right]+\left[\begin{array}{cc}2 & b \\ 1 & -2\end{array}\right]-\left[\begin{array}{cc}1 & 1 \\ -2 & c\end{array}\right]=\left[\begin{array}{ll}5 & 0 \\ 7 & 3\end{array}\right],$ Find the values of $a, b$ and $c$
Answer$\begin{array}{l}{\left[\begin{array}{ll}a & 3 \\ 4 & 1\end{array}\right]+\left[\begin{array}{cc}2 & b \\ 1 & -2\end{array}\right]-\left[\begin{array}{cc}1 & 1 \\ -2 & c\end{array}\right]=\left[\begin{array}{ll}5 & 0 \\ 7 & 3\end{array}\right]} \end{array} $
$ {\left[\begin{array}{cc}a+1 & 2+b \\ 7 & -1-c\end{array}\right]=\left[\begin{array}{ll}5 & 0 \\ 7 & 3\end{array}\right]}$
Comparing the corresponding elements we get
$a + 1 = 5 = > a = 4$
$2 + b = 0 = > b = -2$
$-1 - c = 3 = > c = -4$
View full question & answer→Question 53 Marks
Evaluate $\left[\begin{array}{cc}\cos 45^{\circ} & \sin 30^{\circ} \\ \sqrt{2} \cos 0^{\circ} & \sin 0^{\circ}\end{array}\right]\left[\begin{array}{ll}\sin 45^{\circ} & \cos 90^{\circ} \\ \sin 90^{\circ} & \cot 45^{\circ}\end{array}\right]$
Answer$\left[\begin{array}{cc}\cos 45^{\circ} & \sin 30^{\circ} \\ \sqrt{2} \cos 0^{\circ} & \sin 0^{\circ}\end{array}\right]\left[\begin{array}{ll}\sin 45^{\circ} & \cos 90^{\circ} \\ \sin 90^{\circ} & \cot 45^{\circ}\end{array}\right]$
$\begin{array}{l}=\left[\begin{array}{cc}\frac{1}{\sqrt{2}} & \frac{1}{2} \\ \sqrt{2} & 0\end{array}\right]\left[\begin{array}{cc}\frac{1}{\sqrt{2}} & 0 \\ 1 & 1\end{array}\right] \end{array} $
$ =\left[\begin{array}{cc}\frac{1}{2}+\frac{1}{2} & 0+\frac{1}{2} \\ 1+0 & 0+0\end{array}\right] $
$ =\left[\begin{array}{cc}1 & 0.5 \\ 1 & 0\end{array}\right]$
View full question & answer→Question 63 Marks
if $[x, y]\left[\begin{array}{l}x \\ y\end{array}\right]=[25]$ and $[-x, y]\left[\begin{array}{c}2 x \\ y\end{array}\right]=[-2]$ find $x$ and $y$ if $x, y \ \varepsilon \ W ($whole numbers$)$
Answer$\left[\begin{array}{ll}x & y\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=25$
$x^2+y^2=25$
and
$-2 x^2+y^2=-2$
$x, y\ \& \ W ($whole number$)$
It can be observed that the above two equations are satified when $x = 3$ and $y= 4$
View full question & answer→Question 73 Marks
Find $x$ and $y$ if $[3 \ x \ 8]\left[\begin{array}{ll}1 & 4 \\ 3 & 7\end{array}\right]-3\left[\begin{array}{ll}2 & -7\end{array}\right]=5\left[\begin{array}{ll}3 & 2 y\end{array}\right]$
Answer$[3 \ x, 8]\left[\begin{array}{ll}1 & 4 \\ 3 & 7\end{array}\right]-3\left[\begin{array}{ll}2 & -7\end{array}\right]=5\left[\begin{array}{ll}3 & 2 y\end{array}\right]$
$ {[3 x+24,12 x+56]-[6,-21]=\left[15 10 y\right]}$
$ {[3 x+24-6,12 x+56+21]=\left[15 10 y\right]}$
$ {[3 x+1812 x+77]=\left[15 10 y\right]}$
Comaring the corresspoing elements we get
$3 x+18=15$
$ =3 x=-3$
$ =x=-1$
$ 12 x+77=10 y$
$ =10 y=-12+77=65$
$ =y=6.5$
View full question & answer→Question 83 Marks
Given matrix $A =\left[\begin{array}{c}4 \sin 30^{\circ}, \cos 0^{\circ} \\ \cos 0^{\circ}, 4 \sin 30^{\circ}\end{array}\right]$ and $B=\left[\begin{array}{l}4 \\ 5\end{array}\right]$ If $AX = B$. Write the order of matrix $X$.
AnswerGiven $A=\left[\begin{array}{l}4 \sin 30^{\circ}, \cos 0^{\circ} \\ \cos 0^{\circ}, 4 \sin 30^{\circ}\end{array}\right]$ and $B=\left[\begin{array}{l}4 \\ 5\end{array}\right]$
Let the order of matrix $X = m \times n$
Order of matrix $A = 2 \times 2$
Order of matrix $B = 2 \times 1$
Now, $AX = B$
$\therefore m = 2$ and $n = 1$
Thus, order of matrix $X = m \times n = 2 \times 1$ View full question & answer→Question 93 Marks
Evaluate $:\left[\begin{array}{c}4 \sin 30^{\circ}, 2 \cos 60^{\circ} \\ \sin 90^{\circ}, 2 \cos 0^{\circ}\end{array}\right]\left[\begin{array}{ll}4 & 5 \\ 5 & 4\end{array}\right]$
Answer$\left[\begin{array}{c}4 \sin 30^{\circ}, 2 \cos 60^{\circ} \\ \sin 90^{\circ}, 2 \cos 0^{\circ}\end{array}\right]\left[\begin{array}{ll}4 & 5 \\ 5 & 4\end{array}\right]$
$=\left[\begin{array}{cc}4 \times \frac{1}{2} & 2 \times \frac{1}{2} \\ 1 & 2 \times 1\end{array}\right]\left[\begin{array}{ll}4 & 5 \\ 5 & 4\end{array}\right]$
$ =\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]\left[\begin{array}{ll}4 & 5 \\ 5 & 4\end{array}\right]$
$=\left[\begin{array}{cc}8+5 & 10+4 \\ 4+10 & 5+8\end{array}\right] $
$ =\left[\begin{array}{ll}13 & 14 \\ 14 & 13\end{array}\right]$
View full question & answer→Question 103 Marks
Given $A=\left[\begin{array}{cc}3 & -2 \\ -1 & 4\end{array}\right] B=\left[\begin{array}{l}6 \\ 1\end{array}\right], C=\left[\begin{array}{c}-4 \\ 5\end{array}\right]$ and $D=\left[\begin{array}{l}2 \\ 2\end{array}\right]$ Find $AB + 2 C-4 D$
Answer$\begin{array}{l} AB =\left[\begin{array}{cc}3 & -2 \\ -1 & 4\end{array}\right]\left[\begin{array}{l}6 \\ 1\end{array}\right]=\left[\begin{array}{c}18-2 \\ -6+4\end{array}\right]\end{array} =\left[\begin{array}{c}16 \\ -2\end{array}\right] $
$ 2 C =2\left[\begin{array}{c}-4 \\ 5\end{array}\right]=\left[\begin{array}{c}-8 \\ 10\end{array}\right] $
$ 4 D =4\left[\begin{array}{l}2 \\ 2\end{array}\right]$
$=\left[\begin{array}{l}8 \\ 8\end{array}\right] $
$ \text { Now, } AB +2 C -4 D $
$=\left[\begin{array}{c}16 \\ -2\end{array}\right]+\left[\begin{array}{c}-8 \\ 10\end{array}\right]-\left[\begin{array}{l}8 \\ 8\end{array}\right] \\ $
$=\left[\begin{array}{c}16-8-8 \\ -2+10-8\end{array}\right] \\ $
$=\left[\begin{array}{l}0 \\ 0\end{array}\right]$
View full question & answer→Question 113 Marks
Given $A=\left[\begin{array}{ll}p & 0 \\ 0 & 2\end{array}\right], B=\left[\begin{array}{cc}0 & -q \\ 1 & 0\end{array}\right] C=\left[\begin{array}{cc}2 & -2 \\ 2 & 2\end{array}\right]$ and $B A=C^2$ Find the values of $p$ and $q$
Answer$B A=\left[\begin{array}{cc}0 & -q \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}p & 0 \\ 0 & 2\end{array}\right]=\left[\begin{array}{cc}0 & -2 q \\ p & 0\end{array}\right]$
$A=\left[\begin{array}{ll}p & 0 \\ 0 & 2\end{array}\right], B=p\left[\begin{array}{cc}0 & -q \\ 1 & 0\end{array}\right], C=\left[\begin{array}{cc}2 & -2 \\ 2 & 2\end{array}\right]$
$C^2=\left[\begin{array}{cc}2 & -2 \\ 2 & 2\end{array}\right]\left[\begin{array}{cc}2 & -2 \\ 2 & 2\end{array}\right]=\left[\begin{array}{cc}0 & -8 \\ 8 & 0\end{array}\right] $
$ B A=C^2 \Rightarrow\left[\begin{array}{cc}0 & -2 q \\ p & 0\end{array}\right]=\left[\begin{array}{cc}0 & -8 \\ 8 & 0\end{array}\right]$
By comparing,
$-2a = -8 = > q = 4$
And $p = 8$
View full question & answer→Question 123 Marks
Let $A=\left[\begin{array}{ll}4 & -2 \\ 6 & -3\end{array}\right], B=\left[\begin{array}{cc}0 & 2 \\ 1 & -1\end{array}\right]$ and $C=\left[\begin{array}{cc}-2 & 3 \\ 1 & -1\end{array}\right]$ Find $A^2+A+B C$
Answer$A^2=\left[\begin{array}{cc}4 & 2 \\ 6 & -3\end{array}\right]\left[\begin{array}{cc}4 & -2 \\ 6 & -3\end{array}\right]$
$=\left[\begin{array}{ll}16-12 & -8+6 \\ 24-18 & -12+9\end{array}\right]$
$=\left[\begin{array}{ll}4 & -2 \\ 6 & -3\end{array}\right]$
$B C=\left[\begin{array}{cc}0 & 2 \\ 1 & -1\end{array}\right]\left[\begin{array}{cc}-2 & 3 \\ 1 & -1\end{array}\right]$
$=\left[\begin{array}{cc}0+2 & 0-2 \\ -2-1 & 3=1\end{array}\right]$
$=\left[\begin{array}{cc}2 & -2 \\ -3 & 4\end{array}\right]$
$A^2-A=B C=\left[\begin{array}{ll}4 & -2 \\ 6 & -3\end{array}\right]-\left[\begin{array}{cc}4 & -2 \\ 6 & -3\end{array}\right]+\left[\begin{array}{cc}2 & -2 \\ -3 & 4\end{array}\right]$
$=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]+\left[\begin{array}{cc}2 & -2 \\ -3 & 4\end{array}\right]$
$=\left[\begin{array}{cc}2 & -2 \\ -3 & 4\end{array}\right]$
View full question & answer→Question 133 Marks
Find the value of $x$, given that $A^2=B$
$A=\left[\begin{array}{cc}2 & 12 \\ 0 & 1\end{array}\right]$ and $B=\left[\begin{array}{ll}4 & x \\ 0 & 1\end{array}\right]$
Answer$A=\left[\begin{array}{cc}2 & 12 \\ 0 & 1\end{array}\right]$
$A^2=\left[\begin{array}{cc}2 & 12 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}2 & 12 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}4+0 & 24+12 \\ 0+0 & 0+1\end{array}\right]=\left[\begin{array}{cc}4 & 36 \\ 0 & 1\end{array}\right]$
Given $A^2=B$
$\therefore\left[\begin{array}{cc}4 & 36 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}4 & x \\ 0 & 1\end{array}\right]$
Comparing the corresponding elements we get
x = 36
View full question & answer→Question 143 Marks
Given $A=\left[\begin{array}{cc}2 & -1 \\ 2 & 0\end{array}\right] B=\left[\begin{array}{cc}-3 & 2 \\ 4 & 0\end{array}\right]$ and $C=\left[\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right]$ Find the matrix $X$ such that $A+X=2 B=C$
AnswerGiven $A + X = 2B + C$
$\left[\begin{array}{cc}2 & -1 \\ 2 & 0\end{array}\right]+X=2\left[\begin{array}{cc}-3 & 2 \\ 4 & 0\end{array}\right]+\left[\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right]$
$\left[\begin{array}{cc}2 & -1 \\ 2 & 0\end{array}\right]+X=\left[\begin{array}{cc}-6 & 4 \\ 8 & 0\end{array}\right]+\left[\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right]$
$\begin{array}{l}{\left[\begin{array}{cc}2 & -1 \\ 2 & 0\end{array}\right]+X=\left[\begin{array}{cc}-5 & 4 \\ 8 & 2\end{array}\right]} \end{array} $
$ X=\left[\begin{array}{cc}-5 & 4 \\ 8 & 2\end{array}\right]-\left[\begin{array}{cc}2 & -1 \\ 2 & 0\end{array}\right] $
$ X=\left[\begin{array}{cc}-7 & 5 \\ 6 & 2\end{array}\right]$
View full question & answer→Question 153 Marks
If matrix $X=\left[\begin{array}{cc}-3 & 4 \\ 2 & -3\end{array}\right]\left[\begin{array}{c}2 \\ -2\end{array}\right]$ and $2 X-3 Y=\left[\begin{array}{c}10 \\ -8\end{array}\right];$ Find the matrix $X$ and $Y$
Answer$\begin{array}{l}X=\left[\begin{array}{cc}-3 & 4 \\ 2 & -3\end{array}\right]\left[\begin{array}{c}2 \\ -2\end{array}\right]\end{array} $
$ =\left[\begin{array}{c}-6-8 \\ 4+6\end{array}\right] $
$ =\left[\begin{array}{c}-14 \\ 10\end{array}\right]$
$\begin{array}{l}\text { Given } 2 X-3 Y=\left[\begin{array}{c}10 \\ -8\end{array}\right] \end{array} $
$ 2\left[\begin{array}{c}-14 \\ 10\end{array}\right]-3 Y=\left[\begin{array}{c}10 \\ -8\end{array}\right] $
$ 3 Y=2\left[\begin{array}{c}-14 \\ 10\end{array}\right]-\left[\begin{array}{c}10 \\ -8\end{array}\right] $
$ 3 Y=\left[\begin{array}{c}-38 \\ 28\end{array}\right] $
$ Y=\frac{1}{3}\left[\begin{array}{c}-38 \\ 28\end{array}\right]$
View full question & answer→Question 163 Marks
If $A=\left[\begin{array}{ll}a & 0 \\ 0 & 2\end{array}\right], B =\left[\begin{array}{cc}0 & -b \\ 1 & 0\end{array}\right], M=\left[\begin{array}{cc}1 & -1 \\ 1 & 1\end{array}\right]$ and $B A=M^2$ find the values of $a$ and $b$.
Answer$\begin{array}{l}B A=\left[\begin{array}{cc}0 & -b \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}a & 0 \\ 0 & 2\end{array}\right] \end{array} $
$ =\left[\begin{array}{cc}0+0 & 0-2 b \\ a+0 & 0+0\end{array}\right] $
$ =\left[\begin{array}{cc}0 & -2 b \\ a & 0\end{array}\right]$
$\begin{array}{l}M^2=\left[\begin{array}{cc}1 & -1 \\ 1 & 1\end{array}\right]\left[\begin{array}{cc}1 & -1 \\ 1 & 1\end{array}\right]\end{array} $
$ =\left[\begin{array}{cc}1-1 & -1-1 \\ 1+1 & -1+1\end{array}\right] $
$ =\left[\begin{array}{cc}0 & -2 \\ 2 & 0\end{array}\right]$
Given $B A=M^2$
$\left[\begin{array}{cc}0 & -2 b \\ a & 0\end{array}\right]=\left[\begin{array}{cc}0 & -2 \\ 2 & 0\end{array}\right]$
Comparing the corresponding elements we get
$a =2$
$-2 b=-2 $
$\Rightarrow b=1$
View full question & answer→Question 173 Marks
If $M=\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]$ and $I$ is a unit matrix of the same order as that of $M$ Show that $M^2=2 M+3 I$
Answer$\begin{aligned} & M^2=\left(\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right)[(1,2)(2,1)] \end{aligned} $
$ =\left(\begin{array}{ll}1+4 & 2+2 \\ 2+2 & 4+1\end{array}\right) $
$ =\left(\begin{array}{ll}5 & 4 \\ 4 & 5\end{array}\right) $
$ 2 M+3 I=2\left(\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right)+3\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right) $
$ =\left(\begin{array}{ll}2 & 4 \\ 4 & 2\end{array}\right)+\left(\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right) $
$ =\left(\begin{array}{ll}5 & 4 \\ 4 & 5\end{array}\right)$
View full question & answer→Question 183 Marks
Let $A =\left[\begin{array}{cc}2 & 1 \\ 0 & -2\end{array}\right], B=\left[\begin{array}{cc}4 & 1 \\ -3 & -2\end{array}\right]$ and $C=\left[\begin{array}{cc}-3 & 2 \\ -1 & 4\end{array}\right]$ FindA $^2+ AC-5B$
AnswerGiven : $A =\left[\begin{array}{cc}2 & 1 \\ 0 & -2\end{array}\right], B =\left[\begin{array}{cc}4 & 1 \\ -3 & -2\end{array}\right]$ and $C =\left[\begin{array}{ll}-3 & 2 \\ -1 & 4\end{array}\right]$
Now, $A^2=\left[\begin{array}{cc}2 & 1 \\ 0 & -2\end{array}\right]\left[\begin{array}{cc}2 & 1 \\ 0 & -2\end{array}\right]=\left[\begin{array}{ll}4+0 & 2-2 \\ 0+0 & 0+4\end{array}\right]=\left[\begin{array}{ll}4 & 0 \\ 0 & 4\end{array}\right]$
$5 B=\left[\begin{array}{cc}20 & 5 \\ -15 & -10\end{array}\right]$
$A C=\left[\begin{array}{cc}2 & 1 \\ 0 & -2\end{array}\right]\left[\begin{array}{ll}-3 & 2 \\ -1 & 4\end{array}\right]$
$=\left[\begin{array}{cc}-6-1 & 4+4 \\ 0+2 & 0-8\end{array}\right]=\left[\begin{array}{cc}-7 & 8 \\ 2 & -8\end{array}\right]$
$\therefore A^2+A C-5 B=\left[\begin{array}{ll}4 & 0 \\ 0 & 4\end{array}\right]+\left[\begin{array}{cc}-7 & 8 \\ 2 & -8\end{array}\right]-\left[\begin{array}{cc}20 & 5 \\ -15 & -10\end{array}\right] $
$ =\left[\begin{array}{cc}4-7-20 & 0+8-5 \\ 0+2+15 & 4-8+10\end{array}\right] $
$=\left[\begin{array}{cc}-23 & 3 \\ 17 & 6\end{array}\right] .$
View full question & answer→Question 193 Marks
If $A=\left[\begin{array}{ll}3 & x \\ 0 & 1\end{array}\right]$ and $B=\left[\begin{array}{cc}9 & 16 \\ 0 & -y\end{array}\right]$ find $x$ and $y$ when $A^2 = B$
AnswerGiven $A=\left[\begin{array}{ll}3 & x \\ 0 & 1\end{array}\right] B=\left[\begin{array}{cc}9 & 16 \\ 0 & -y\end{array}\right]$ and $A^2=B$
Now, $A^2= A \times A$
$\begin{array}{l}=\left[\begin{array}{ll}3 & x \\ 0 & 1\end{array}\right] \times\left[\begin{array}{ll}3 & x \\ 0 & 1\end{array}\right] \end{array} $
$ =\left[\begin{array}{cc}9 & 3 x+x \\ 0 & 1\end{array}\right] $
$ =\left[\begin{array}{cc}9 & 4 x \\ 0 & 1\end{array}\right]$
We have $A^2 = B$
Two matrices are equal if each and every corresponding element is equal.
$\Rightarrow\left[\begin{array}{cc}9 & 4 x \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}9 & 16 \\ 0 & -y\end{array}\right]$
$\Rightarrow 4 x=16$ and $1=-y$
$\Rightarrow x=4$ and $y=-1$
View full question & answer→Question 203 Marks
Evaluate without using tables :
$\left[\begin{array}{c}2 \cos 60^{\circ}-2 \sin 30^{\circ} \\ -\tan 45^{\circ}, \cos 0^{\circ}\end{array}\right]\left[\begin{array}{c}\cot 45^{\circ}, \cos e c 30^{\circ} \\ \sec 60^{\circ}, \sin 90^{\circ}\end{array}\right]$
Answer$\left[\begin{array}{c}2 \cos 60^{\circ}-2 \sin 30^{\circ} \\ -\tan 45^{\circ}, \cos 0^{\circ}\end{array}\right]\left[\begin{array}{c}\cot 45^{\circ}, \cos e c 30^{\circ} \\ \sec 60^{\circ}, \sin 90^{\circ}\end{array}\right]$
$\begin{aligned} & =\left[\begin{array}{cc}2 \times \frac{1}{2} & -2 \times \frac{1}{2} \\ -1 & 1 \\ 2 & 1\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right] \\ = & {\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right] } \\ = & {\left[\begin{array}{cc}1-2 & 2-1 \\ -1+2 & -2+1\end{array}\right] } \\ = & {\left[\begin{array}{cc}-1 & 1 \\ 1 & -1\end{array}\right] }\end{aligned}$
View full question & answer→Question 213 Marks
if $M =\left[\begin{array}{cc}4 & 1 \\ -1 & 2\end{array}\right]$ show that $6 m-m^2=9 I$ where $I$ is $2 \times 2$ unit matrix.
Answer$M^2=\left[\begin{array}{cc}4 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}4 & 1 \\ -1 & 2\end{array}\right]$
$=\left[\begin{array}{cc}16-1 & 4+2 \\ -4-2 & -1+4\end{array}\right]=\left[\begin{array}{cc}15 & 6 \\ -6 & 3\end{array}\right]$
$\begin{array}{l}6 M-M^2=6\left[\begin{array}{cc}4 & 1 \\ -1 & 2\end{array}\right]-\left[\begin{array}{cc}15 & 6 \\ -6 & 3\end{array}\right] \end{array}$
$=\left[\begin{array}{cc}24 & 6 \\ -6 & 12\end{array}\right]-\left[\begin{array}{cc}15 & 6 \\ -6 & 3\end{array}\right] $
$ =\left[\begin{array}{ll}9 & 0 \\ 0 & 9\end{array}\right] $
$ =9\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$=9 $
hence proved
View full question & answer→Question 223 Marks
If $A=\left[\begin{array}{lll}2 & 1 & -1 \\ 0 & 1 & -2\end{array}\right]$ find $A$. $A^t$ where $A^t$ is the transpose of matrix $A$
Answer$\begin{array}{l}A=\left[\begin{array}{lll}2 & 1 & -1 \\ 0 & 1 & -2\end{array}\right] \end{array} $
$ A^t=\left[\begin{array}{cc}2 & 0 \\ 1 & 1 \\ -1 & -2\end{array}\right] $
$A. A^t=\left[\begin{array}{lll}2 & 1 & -1\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ 1 & 1 \\ -1 & -2\end{array}\right]$
$\begin{array}{l}=\left[\begin{array}{ll}4+1+1 & 0+1+2 \\ 0+1+2 & 0+1+4\end{array}\right] \end{array} $
$=\left[\begin{array}{ll}6 & 3 \\ 3 & 5\end{array}\right]$
View full question & answer→Question 233 Marks
If $A=\left[\begin{array}{ll}4 & 2 \\ 1 & 1\end{array}\right]$ Find $(A-2I)(A - 3I)$
Answer$A-2 I=\left[\begin{array}{ll}4 & 2 \\ 1 & 1\end{array}\right]-2\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}4 & 2 \\ 1 & 1\end{array}\right]-\left[\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right]=\left[\begin{array}{cc}2 & 2 \\ 1 & -1\end{array}\right]$
$A-3 I=\left[\begin{array}{ll}4 & 2 \\ 1 & 1\end{array}\right]-3\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}4 & 2 \\ 1 & 1\end{array}\right]-\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]=\left[\begin{array}{cc}1 & 2 \\ 1 & -2\end{array}\right]$
$\begin{array}{l}(A-2 I)(A-3 I) =\left[\begin{array}{cc}2 & 2 \\ 1 & -1\end{array}\right]\left[\begin{array}{cc}1 & 2 \\ 1 & -2\end{array}\right] \end{array} $
$ =\left[\begin{array}{ll}2+2 & 4-4 \\ 1-1 & 2+2\end{array}\right] $
$ =\left[\begin{array}{ll}4 & 0 \\ 0 & 4\end{array}\right]$
View full question & answer→Question 243 Marks
If $A=\left[\begin{array}{cc}0 & 2 \\ 5 & -2\end{array}\right], B=\left[\begin{array}{cc}1 & -1 \\ 3 & 2\end{array}\right]$ and $I$ is unit matrix of order $2 \times 2$ find $B^2 A$
Answer$\begin{array}{l}B^2=\left[\begin{array}{cc}1 & -1 \\ 3 & 2\end{array}\right]\left[\begin{array}{cc}1 & -1 \\ 3 & 2\end{array}\right] \end{array}$
$ =\left[\begin{array}{cc}1-3 & -1-2 \\ 3+6 & -3+4\end{array}\right] $
$ =\left[\begin{array}{cc}-2 & -3 \\ 9 & 1\end{array}\right] $
$ B^2 A=\left[\begin{array}{cc}-2 & -3 \\ 9 & 1\end{array}\right]\left[\begin{array}{cc}0 & 2 \\ 5 & -2\end{array}\right] $
$ =\left[\begin{array}{cc}0-15 & -4+6 \\ 0+5 & 18-2\end{array}\right] $
$ =\left[\begin{array}{cc}-15 & 2 \\ 5 & 16\end{array}\right]$
View full question & answer→Question 253 Marks
Given the matrices $A=\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right], B=\left[\begin{array}{cc}3 & 4 \\ -1 & -2\end{array}\right]$ and $C=\left[\begin{array}{cc}-3 & 1 \\ 0 & -2\end{array}\right]$ Find $1) \ce{ABC}, 2) \ce{ACB}$ State whether $\ce{ABC = ACB}.$
Answer$A B=\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right]\left[\begin{array}{cc}3 & 4 \\ -1 & -2\end{array}\right]=\left[\begin{array}{cc}6-1 & 8-2 \\ 12-2 & 16-4\end{array}\right]=\left[\begin{array}{cc}5 & 6 \\ 10 & 12\end{array}\right]$
$A B C=\left[\begin{array}{cc}5 & 6 \\ 10 & 12\end{array}\right]\left[\begin{array}{cc}-3 & 1 \\ 0 & -2\end{array}\right]=\left[\begin{array}{cc}-15+0 & 5-12 \\ -30+0 & 10-24\end{array}\right]=\left[\begin{array}{cc}-15 & -7 \\ -30 & -14\end{array}\right]$
$2 AC =\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right]\left[\begin{array}{cc}-3 & 1 \\ 0 & -2\end{array}\right]=\left[\begin{array}{cc}-6+0 & 2-2 \\ -12+0 & 4-4\end{array}\right]=\left[\begin{array}{cc}-6 & 0 \\ -12 & 0\end{array}\right] $
$ A C B=\left[\begin{array}{cc}-6 & 0 \\ -12 & 0\end{array}\right]\left[\begin{array}{cc}3 & 4 \\ -1 & -20\end{array}\right]=\left[\begin{array}{ll}-18-0 & -24-0 \\ -36-0 & -48-0\end{array}\right]=\left[\begin{array}{cc}-18 & -24 \\ -36 & -48\end{array}\right]$
hence $\ce{ABC ? ACB}$
View full question & answer→Question 263 Marks
Given $A=\left[\begin{array}{ll}3 & 0 \\ 0 & 4\end{array}\right], B=\left[\begin{array}{ll}a & b \\ 0 & c\end{array}\right]$ and that $\ce{AB = A + B}.$ Find the values of $a, b$ and $c$
Answer$AB = [(3, 0),(0, 4)][(a, b),(0, c)] $
$= [(3a + 0, 3b + 0),(0 + 0, 0 = 4c)] $
$= [(3a, 3b),(0, 4c)]$
$A+B=\left[\begin{array}{ll}3 & 0 \\ 0 & 4\end{array}\right]+\left[\begin{array}{ll}a & b \\ 0 & c\end{array}\right]=\left[\begin{array}{cc}3+a & b \\ 0 & 4+c\end{array}\right]$
Given $AB = A + B$
$\therefore\left[\begin{array}{cc}3 a & 3 b \\ 0 & 4 c\end{array}\right]=\left[\begin{array}{cc}3+a & b \\ 0 & 4+c\end{array}\right]$
Comparing the correspomding elememts we get
$3a = 3 + a$
$2a = 3$
$\Rightarrow a=\frac{3}{2}$
$3 b=b $
$\Rightarrow b=0$
$4 c=4+c $
$\Rightarrow 3 c=4 $
$\Rightarrow c=\frac{4}{3}$
View full question & answer→Question 273 Marks
If $A=\left[\begin{array}{ll}2 & x \\ 0 & 1\end{array}\right]$ and $B=\left[\begin{array}{cc}4 & 36 \\ 0 & 1\end{array}\right]$. Find the vlaue of $x$ given that $A^2=B$
Answer$A^2=\left[\begin{array}{ll}2 & x \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}2 & x \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}4+0 & 2 x+x \\ 0+0 & 0+1\end{array}\right]=\left[\begin{array}{cc}4 & 3 x \\ 0 & 1\end{array}\right]$
Given $A^2=B$
$\left[\begin{array}{cc}4 & 3 x \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}4 & 36 \\ 0 & 1\end{array}\right]$
Comparing the two matrices we get
$3 x=36 \Rightarrow x=12$
View full question & answer→Question 283 Marks
Solve for $x$ and $y \left[\begin{array}{cc}-2 & 0 \\ 3 & 1\end{array}\right]\left[\begin{array}{c}-1 \\ 2 x\end{array}\right]+3\left[\begin{array}{c}-2 \\ 1\end{array}\right]=2\left[\begin{array}{l}y \\ 3\end{array}\right]$
Answer$\begin{array}{l}{\left[\begin{array}{cc}-2 & 0 \\ 3 & 1\end{array}\right]\left[\begin{array}{c}-1 \\ 2 x\end{array}\right]+3\left[\begin{array}{c}-2 \\ 1\end{array}\right]=2\left[\begin{array}{l}y \\ 3\end{array}\right]} \end{array} $
$ \Rightarrow\left[\begin{array}{c}2+0 \\ -3+2 x\end{array}\right]+\left[\begin{array}{c}-6 \\ 3\end{array}\right]=\left[\begin{array}{c}2 y \\ 6\end{array}\right] $
$ \Rightarrow\left[\begin{array}{c}2 \\ -3+2 x\end{array}\right]+\left[\begin{array}{c}-6 \\ 3\end{array}\right]=\left[\begin{array}{c}2 y \\ 6\end{array}\right] $
$ \Rightarrow\left[\begin{array}{c}2-6 \\ -3+2 x+3\end{array}\right]=\left[\begin{array}{c}2 y \\ 6\end{array}\right] $
$ \Rightarrow\left[\begin{array}{c}-4 \\ 2 x\end{array}\right]=\left[\begin{array}{c}2 y \\ 6\end{array}\right]$
$\Rightarrow 2 y=-4$ and $2 x=6$
$\Rightarrow y=-2$ and $x=3$
Thus, the values of $x$ andty are: $3,-2$
View full question & answer→Question 293 Marks
Solve for $x$ and $y :\left[\begin{array}{ll}x+y & x-4\end{array}\right]\left[\begin{array}{cc}-1 & -2 \\ 2 & 2\end{array}\right]=\left[\begin{array}{ll}-7 & -11\end{array}\right]$
Answer$\left[\begin{array}{ll}x+y & x-4\end{array}\right]\left[\begin{array}{cc}-1 & -2 \\ 2 & 2\end{array}\right]=\left[\begin{array}{ll}-7 & -11\end{array}\right]$
$[-x-y+2 x-8 -2 x -2 y+2 x-8]=[-7 -11]$
$[-y+x-8 -2 y-8]=[-7 -11]$
Comparing the corresponding elements, we get,
$-2 y-8=-11 $
$\Longrightarrow-2 y=-3 $
$\Longrightarrow y=\frac{3}{2}$
$ -y+x-8=-7$
$ \Rightarrow-\frac{3}{2}+x-8=-7$
$ \Rightarrow x=1+\frac{3}{2}=\frac{5}{2}$
View full question & answer→Question 303 Marks
If $\left.A=\left[\begin{array}{ll}1 & 4 \\ 2 & 1\end{array}\right], B=\left[\begin{array}{cc}-3 & 2 \\ 4 & 0\end{array}\right]\right]$ and $C=\left[\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right]$ Simplify $A^2+B C$
Answer$A^2=\left[\begin{array}{ll}1 & 4 \\ 2 & 1\end{array}\right]\left[\begin{array}{ll}1 & 4 \\ 2 & 1\end{array}\right]=\left[\begin{array}{ll}1+8 & 4+4 \\ 2+2 & 8+1\end{array}\right]=\left[\begin{array}{ll}9 & 8 \\ 4 & 9\end{array}\right]$
$B C=\left[\begin{array}{cc}-3 & 2 \\ 4 & 0\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right]=\left[\begin{array}{cc}-3+0 & 0+4 \\ 4=0 & 0+0\end{array}\right]=\left[\begin{array}{cc}-3 & 4 \\ 4 & 0\end{array}\right]$
$A^2+B C=\left[\begin{array}{ll}9 & 8 \\ 4 & 9\end{array}\right]+\left[\begin{array}{cc}-3 & 4 \\ 4 & 0\end{array}\right]=\left[\begin{array}{cc}6 & 12 \\ 8 & 9\end{array}\right]$
View full question & answer→Question 313 Marks
If $A=\left[\begin{array}{ll}2 & 1 \\ 0 & 0\end{array}\right], B=\left[\begin{array}{ll}2 & 3 \\ 4 & 1\end{array}\right]$ and $C=\left[\begin{array}{ll}1 & 4 \\ 0 & 2\end{array}\right]$ then show that $(B-A) C$ $=B C-A C)$
Answer$B-A=\left[\begin{array}{ll}2 & 3 \\ 4 & 1\end{array}\right]-\left[\begin{array}{ll}2 & 1 \\ 0 & 0\end{array}\right]=\left[\begin{array}{ll}0 & 2 \\ 4 & 1\end{array}\right]$
$(B-A) C=\left[\begin{array}{ll}0 & 2 \\ 4 & 1\end{array}\right]\left[\begin{array}{ll}1 & 4 \\ 0 & 2\end{array}\right]$
$=\left[\begin{array}{cc}0 & 0+4 \\ 4+0 & 16+2\end{array}\right]=\left[\begin{array}{cc}0 & 4 \\ 4 & 18\end{array}\right]$
$\begin{array}{l}B C=\left[\begin{array}{ll}2 & 3 \\ 4 & 1\end{array}\right]\left[\begin{array}{ll}1 & 4 \\ 0 & 2\end{array}\right]=\left[\begin{array}{cc}2+0 & 8+6 \\ 4+0 & 16+2\end{array}\right]=\left[\begin{array}{ll}2 & 14 \\ 4 & 18\end{array}\right] \end{array}$
$ A C=\left[\begin{array}{ll}2 & 1 \\ 0 & 0\end{array}\right]\left[\begin{array}{ll}1 & 4 \\ 0 & 2\end{array}\right]=\left[\begin{array}{cc}2+0 & 8+2 \\ 0 & 0\end{array}\right]=\left[\begin{array}{cc}2 & 10 \\ 0 & 0\end{array}\right] $
$ B C-A C=\left[\begin{array}{ll}2 & 14 \\ 4 & 18\end{array}\right]-\left[\begin{array}{cc}2 & 10 \\ 0 & 0\end{array}\right]=\left[\begin{array}{cc}0 & 4 \\ 4 & 18\end{array}\right]$
Hence $(B - A)C = BC - AC$
View full question & answer→Question 323 Marks
If $A=\left[\begin{array}{ll}2 & 1 \\ 0 & 0\end{array}\right], B=\left[\begin{array}{ll}2 & 3 \\ 4 & 1\end{array}\right]$ and $C=\left[\begin{array}{ll}1 & 4 \\ 0 & 2\end{array}\right]$ then show that $A(B+ C)=A B+A C$
Answer$B+C=\left[\begin{array}{ll}2 & 3 \\ 4 & 1\end{array}\right]+\left[\begin{array}{ll}1 & 4 \\ 0 & 2\end{array}\right]=\left[\begin{array}{ll}3 & 7 \\ 4 & 3\end{array}\right]$
$A(B+C)=\left[\begin{array}{ll}2 & 1 \\ 0 & 0\end{array}\right]\left[\begin{array}{ll}3 & 7 \\ 4 & 3\end{array}\right]$
$=\left[\begin{array}{cc}6+4 & 14+3 \\ 0 & 0\end{array}\right]=\left[\begin{array}{cc}10 & 1() 7 \\ 0 & 0\end{array}\right]$
$A B=\left[\begin{array}{ll}2 & 1 \\ 0 & 0\end{array}\right]\left[\begin{array}{ll}2 & 3 \\ 4 & 1\end{array}\right]=\left[\begin{array}{cc}4+4 & 6+1 \\ 0 & 0\end{array}\right]=\left[\begin{array}{ll}8 & 7 \\ 0 & 0\end{array}\right]$
$\begin{array}{l}A C=\left[\begin{array}{ll}2 & 1 \\ 0 & 0\end{array}\right]\left[\begin{array}{ll}1 & 4 \\ 0 & 2\end{array}\right]=\left[\begin{array}{cc}2+0 & 8+2 \\ 0 & 0\end{array}\right]=\left[\begin{array}{cc}2 & 10 \\ 0 & 0\end{array}\right] \end{array} $
$ A B+A C=\left[\begin{array}{ll}8 & 7 \\ 0 & 0\end{array}\right]+\left[\begin{array}{cc}2 & 10 \\ 0 & 0\end{array}\right]=\left[\begin{array}{cc}10 & 17 \\ 0 & 0\end{array}\right]$
Hence $A(B + C) = AB + AC$
View full question & answer→Question 333 Marks
If $A=\left[\begin{array}{cc}-1 & 1 \\ a & b\end{array}\right]$ and $A^2=I$; Find $a$ and $b$
Answer$A=\left[\begin{array}{cc}-1 & 1 \\ a & b\end{array}\right]$
$A^2=\left[\begin{array}{cc}-1 & 1 \\ a & b\end{array}\right]\left[\begin{array}{cc}-1 & 1 \\ a & b\end{array}\right]$
$=\left[\begin{array}{cc}1+a & -1+b \\ -a+a b & a+b^2\end{array}\right]$
It is given that $A^2=I$
$\therefore\left[\begin{array}{cc}1+a & -1+b \\ -a+a b & a+b^2\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
Comparing the corresponding elements we get
1 + a = 1
Therefore a = 0
-1 + b = 0
Therefore b = 1
View full question & answer→Question 343 Marks
Find the matrix $A,$ If $B =\left[\begin{array}{ll}2 & 1 \\ 0 & 1\end{array}\right]$ and $B^2=B+\frac{1}{2} A$
Answer$B^2=B+\frac{1}{2} A$
$A=2\left(B^2-B\right)$
$\begin{array}{l}B^2=\left[\begin{array}{ll}2 & 1 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}2 & 1 \\ 0 & 1\end{array}\right] \end{array} $
$ =\left[\begin{array}{llll}4 & +0 & 2 & +1 \\ 0 & +0 & 0 & +1\end{array}\right] $
$=\left[\begin{array}{ll}4 & 3 \\ 0 & 1\end{array}\right] $
$\begin{array}{l}=B^2-B=\left[\begin{array}{ll}4 & 3 \\ 0 & 1\end{array}\right]-\left[\begin{array}{ll}2 & 1 \\ 0 & 1\end{array}\right] \end{array} $
$ =\left[\begin{array}{ll}2 & 2 \\ 0 & 0\end{array}\right]$
$A=2\left(B^2-B\right)$
$=2\left[\begin{array}{ll}2 & 2 \\ 0 & 0\end{array}\right]$
$=\left[\begin{array}{ll}4 & 4 \\ 0 & 0\end{array}\right]$
View full question & answer→Question 353 Marks
Given $A=\left[\begin{array}{cc}1 & 1 \\ -2 & 0\end{array}\right]$ and $B=\left[\begin{array}{cc}2 & -1 \\ 1 & 1\end{array}\right]$ Solve for matrix $X: 3A - 2X = X - 2B$
Answer$3A - 2X = X - 2B$
$3A + 2B = X + 2X$
$3X = 3A + 2B$
$\begin{array}{l}3 X=3\left[\begin{array}{cc}1 & 1 \\ -2 & 0\end{array}\right]+2\left[\begin{array}{cc}2 & -1 \\ 1 & 1\end{array}\right]\end{array} $
$ 3 X=\left[\begin{array}{cc}7 & 1 \\ -4 & 2\end{array}\right] $
$ X=\left[\begin{array}{cc}\frac{7}{3} & \frac{1}{3} \\ -\frac{4}{3} & \frac{2}{3}\end{array}\right]$
View full question & answer→Question 363 Marks
Given $A=\left[\begin{array}{cc}1 & 1 \\ -2 & 0\end{array}\right]$ and $B=\left[\begin{array}{cc}2 & -1 \\ 1 & 1\end{array}\right] 3X + B + 2A = O$
Answer$3X + B + 2A = O$
$3X = -2A - B$
$\begin{array}{l}3 X=-2\left[\begin{array}{cc}1 & 1 \\ -2 & 0\end{array}\right]-\left[\begin{array}{cc}2 & -1 \\ 1 & 1\end{array}\right] \end{array} $
$ 3 X=\left[\begin{array}{cc}-2 & -2 \\ 4 & 0\end{array}\right]-\left[\begin{array}{cc}2 & -1 \\ 1 & 1\end{array}\right] $
$ 3 X=\left[\begin{array}{cc}-4 & -1 \\ 3 & -1\end{array}\right] $
$ X=\left[\begin{array}{cc}\frac{-4}{3} & \frac{-1}{3} \\ 1 & \frac{-1}{3}\end{array}\right]$
View full question & answer→Question 373 Marks
Given $A=\left[\begin{array}{cc}-3 & 6 \\ 0 & -9\end{array}\right]$ and $A^t$ is its transpose matrix. Find $\frac{1}{2} A-\frac{1}{3} A^t$
Answer$\begin{array}{l}A=\left[\begin{array}{cc}-3 & 6 \\ 0 & -9\end{array}\right] \end{array} $
$A^t=\left[\begin{array}{cc}-3 & 0 \\ 6 & -9\end{array}\right] $
$ \frac{1}{2} A-\frac{1}{3} A^t$
$\begin{array}{l}=\frac{1}{2}\left[\begin{array}{cc}-3 & 6 \\ 0 & -9\end{array}\right]-\frac{1}{3}\left[\begin{array}{cc}-3 & 0 \\ 6 & -9\end{array}\right] \end{array} $
$ =\left[\begin{array}{cc}\frac{-3}{2} & 3 \\ 0 & \frac{-9}{2}\end{array}\right]-\left[\begin{array}{cc}-1 & 0 \\ 2 & -3\end{array}\right] $
$ =\left[\begin{array}{cc}\frac{-1}{2} & 3 \\ -2 & \frac{-3}{2}\end{array}\right]$
View full question & answer→Question 383 Marks
Given $A=\left[\begin{array}{cc}-3 & 6 \\ 0 & -9\end{array}\right]$ and $A^t$ is its transpose matrix Find $2 A^t-3 A$
Answer$\begin{array}{l}A=\left[\begin{array}{cc}-3 & 6 \\ 0 & -9\end{array}\right] \end{array} $
$ A^t=\left[\begin{array}{cc}-3 & 0 \\ 6 & -9\end{array}\right]$
$2 A^t-3 A $
$ =2\left[\begin{array}{cc}-3 & 0 \\ 6 & -9\end{array}\right]-3\left[\begin{array}{cc}-3 & 6 \\ 0 & -9\end{array}\right] $
$ =\left[\begin{array}{cc}-6 & 0 \\ 12 & -18\end{array}\right]-\left[\begin{array}{cc}-9 & 18 \\ 0 & -27\end{array}\right] $
$ =\left[\begin{array}{cc}3 & -18 \\ 12 & 9\end{array}\right]$
View full question & answer→Question 393 Marks
Given$A=\left[\begin{array}{cc}-3 & 6 \\ 0 & -9\end{array}\right]$ and $A^t$ its transpose matrix. Find $2 A+3 A^t$
Answer$\begin{array}{l}A=\left[\begin{array}{cc}-3 & 6 \\ 0 & -9\end{array}\right] \end{array} $
$ A^t=\left[\begin{array}{cc}-3 & 0 \\ 6 & -9\end{array}\right] $
$2 A+3 A^t $
$ =2\left[\begin{array}{cc}-3 & 6 \\ 0 & -9\end{array}\right]+3\left[\begin{array}{cc}-3 & 0 \\ 6 & -9\end{array}\right] $
$ =\left[\begin{array}{cc}-6 & 12 \\ 0 & -18\end{array}\right]+\left[\begin{array}{cc}-9 & 0 \\ 18 & -27\end{array}\right] $
$ =\left[\begin{array}{cc}-15 & 12 \\ 18 & -45\end{array}\right]$
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If $2\left[\begin{array}{ll}3 & x \\ 0 & 1\end{array}\right]+3\left[\begin{array}{ll}1 & 3 \\ y & 2\end{array}\right]=\left[\begin{array}{cc}z & -7 \\ 15 & 8\end{array}\right]$ Find the values of $x, y$ and $z.$
Answer$\begin{array}{l}2\left[\begin{array}{ll}3 & x \\ 0 & 1\end{array}\right]+3\left[\begin{array}{ll}1 & 3 \\ y & 2\end{array}\right]=\left[\begin{array}{cc}z & -7 \\ 15 & 8\end{array}\right] \end{array}$
$ {\left[\begin{array}{cc}6 & 2 x \\ 0 & 2\end{array}\right]+\left[\begin{array}{cc}3 & 9 \\ 3 y & 6\end{array}\right]=\left[\begin{array}{cc}z & -7 \\ 15 & 8\end{array}\right]} $
${\left[\begin{array}{cc}9 & 2 x+9 \\ 3 y & 8\end{array}\right]=\left[\begin{array}{cc}z & -7 \\ 15 & 8\end{array}\right]}$
comparing the correspinding elements we get
$2 x+9=-7 $
$\Rightarrow 2 x=-16 $
$\Rightarrow x=-8$
$3 y=15$
$\Rightarrow y=5$
$z=9$
View full question & answer→Question 413 Marks
If $\left[\begin{array}{cc}4 & -2 \\ 4 & 0\end{array}\right]+3 A=\left[\begin{array}{cc}-2 & -2 \\ 1 & -3\end{array}\right]$ Find $A$
Answer$\begin{array}{l}{\left[\begin{array}{cc}4 & -2 \\ 4 & 0\end{array}\right]+3 A=\left[\begin{array}{cc}-2 & -2 \\ 1 & -3\end{array}\right]} \end{array} $
$ 3 A=\left[\begin{array}{cc}-2 & -2 \\ 1 & -3\end{array}\right]-\left[\begin{array}{cc}4 & -2 \\ 4 & 0\end{array}\right] $
$ 3 A=\left[\begin{array}{cc}-2-4-2+2 \\ 1-4 & -3-0\end{array}\right] $
$ 3 A=\left[\begin{array}{cc}-6 & 0 \\ -3-3\end{array}\right] $
$ A=\frac{1}{3}\left[\begin{array}{cc}-6 & 0 \\ -3 & -3\end{array}\right]=\left[\begin{array}{cc}-2 & 0 \\ -1-1\end{array}\right]$
View full question & answer→Question 423 Marks
If $I$ is the unit matrix of order $2 \times 2.$ Find the matrix $M$ such that $5 M+3 I=4\left[\begin{array}{ll}2 & -5 \\ 0 & -3\end{array}\right]$
Answer$\begin{array}{l}5 M+3 I=4\left[\begin{array}{ll}2 & -5 \\ 0 & -3\end{array}\right] \end{array} $
$ 5 M=4\left[\begin{array}{ll}2 & -5 \\ 0 & -3\end{array}\right]-3\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] $
$ 5 M=\left[\begin{array}{ll}8 & -20 \\ 0 & -12\end{array}\right]-\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right] $
$ 5 M=\left[\begin{array}{ll}5 & -20 \\ 0 & -15\end{array}\right] $
$ M=\frac{1}{5}\left[\begin{array}{ll}5 & -20 \\ 0 & -15\end{array}\right]=\left[\begin{array}{ll}1 & -4 \\ 0 & -3\end{array}\right]$
View full question & answer→Question 433 Marks
If $I$ is the unit matrix of order $2 \times 2$ Find the matrix $M$ such that $M-2 I=3\left[\begin{array}{cc}-1 & 0 \\ 4 & 1\end{array}\right]$
Answer$\begin{array}{l}M-2 I=3\left[\begin{array}{cc}-1 & 0 \\ 4 & 1\end{array}\right]\end{array} $
$ M=3\left[\begin{array}{cc}-1 & 0 \\ 4 & 1\end{array}\right]+2 I $
$ M=3\left[\begin{array}{cc}-1 & 0 \\ 4 & 1\end{array}\right]+2\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] $
$ M=\left[\begin{array}{cc}-3 & 0 \\ 12 & 3\end{array}\right]+\left[\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right] $
$ M=\left[\begin{array}{cc}-1 & 0 \\ 12 & 5\end{array}\right]$
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