[3 marks sum]

Question 13 Marks

Find the value of p and q if:
$\left[\begin{array}{cc}2 p+1 & q^2-2 \\ 6 & 0\end{array}\right]=\left[\begin{array}{cc}p+3 & 3 q-4 \\ 5 q-q^2 & 0\end{array}\right]$.

Answer

$2p + 1 = p + 3;$
$2p - p = 3 - 1$
$p = 2 ...(1)$
$q^2 - 2 = 3q - 4$
$q^2 - 3q + 2 = 0$
$q^2 - 2q - q + 2 = 0$
$q(q - 2) - (q - 2) = 0$
$(q - 2) (q - 1) = 0 ...(2)$
$5q - q^2 = 6$
$q^2 - 5q + 6 = 0$
$(q - 3) (q - 2) = 0 ...(3)$
By equation $(2)$ and $(3)$
$q = 2$
$\Rightarrow p = 2, q = 2.$

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Question 23 Marks

Given $A=\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]$ evaluate $A^2- 4A.$

Answer

$\begin{array}{l}A=\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right] \end{array} $
$ A^2=A \times A=\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right] $
$ =\left[\begin{array}{cc}1+8 & 1+3 \\ 8+24 & 8+9\end{array}\right]$
$=\left[\begin{array}{cc}9 & 4 \\ 32 & 17\end{array}\right] $
$ 4 A=4\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right] =\left[\begin{array}{cc}4 & 4 \\ 32 & 12\end{array}\right]$
$\begin{array}{l}A^2-4 A=\left[\begin{array}{cc}9 & 4 \\ 32 & 17\end{array}\right]-\left[\begin{array}{cc}4 & 4 \\ 32 & 12\end{array}\right] \\ \end{array}$
$=\left[\begin{array}{cc}9-4 & 4-4 \\ 32-32 & 17-12\end{array}\right] $
$ A^2-4 A=\left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right] .$

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Question 33 Marks

Evaluate $x,y$ if $\left[\begin{array}{cc}3 & -2 \\ -1 & 4\end{array}\right]\left[\begin{array}{c}2 x \\ 1\end{array}\right]+2\left[\begin{array}{c}-4 \\ 5\end{array}\right]=\left[\begin{array}{c}8 \\ 4 y\end{array}\right]$

Answer

$\begin{array}{l}{\left[\begin{array}{cc}3 & -2 \\ -1 & 4\end{array}\right]\left[\begin{array}{c}2 x \\ 1\end{array}\right]+2\left[\begin{array}{c}-4 \\ 5\end{array}\right]=\left[\begin{array}{c}8 \\ 4 y\end{array}\right]} \end{array}$
${\left[\begin{array}{c}3 \times 2 x+(-2) \times 1 \\ -1 \times 2 x+4 \times 1\end{array}\right]+\left[\begin{array}{c}-8 \\ 10\end{array}\right]=\left[\begin{array}{c}8 \\ 4 y\end{array}\right]} $
${\left[\begin{array}{c}6 x-2 \\ -2 x+4\end{array}\right]+\left[\begin{array}{c}-8 \\ 10\end{array}\right]=\left[\begin{array}{c}8 \\ 4 y\end{array}\right]} $
$ {\left[\begin{array}{c}6 x-2-8 \\ -2 x+4+10\end{array}\right]=\left[\begin{array}{c}8 \\ 4 y\end{array}\right]} $
$ {\left[\begin{array}{c}6 x-10 \\ -2 x+14\end{array}\right]=\left[\begin{array}{c}8 \\ 4 y\end{array}\right]}$
$6x - 10 = 8$
$\Rightarrow 6x = 18, x = 3$
$-2x + 14 = 4y$
$\Rightarrow -2x \ 3 + 14 = 4y$
$y=\frac{14-6}{4}$
$=\frac{8}{4}=2$
$x=3, y=2$

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Question 43 Marks

If $X =\left[\begin{array}{cc}4 & 1 \\ -1 & 2\end{array}\right],$ show that $6X - X^2 = 9I,$ where $I$ is unit matrix.

Answer

Here
$X^2 = X·X$
$\begin{array}{l}=\left[\begin{array}{cc}4 & 1 \\ -1 & 2\end{array}\right] \cdot\left[\begin{array}{cc}4 & 1 \\ -1 & 2\end{array}\right] \\\end{array}$
$ =\left[\begin{array}{cc}16-1 & 4+2 \\ -4-2 & -1+4\end{array}\right]$
$=\left[\begin{array}{cc}15 & 6 \\ -6 & 3\end{array}\right]$
$\text{L.H.S.} = 6X - X^2$
$\begin{array}{l}=6\left[\begin{array}{cc}4 & 1 \\ -1 & 2\end{array}\right]-\left[\begin{array}{cc}15 & 6 \\ -6 & 3\end{array}\right] \\\end{array}$
$ =\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}{cc}24-15 & 6-6 \\ -6+6 & 12-3\end{array}\right] $
$ =\left[\begin{array}{cc}9 & 0 \\ 0 & 9\end{array}\right] $
$=9\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$= 9I = \text{R.H.S.}$
Hence proved.

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Question 53 Marks

If $A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$ and $B=\left[\begin{array}{l}7 \\ 0\end{array}\right]$, find matrix $C$ if $AC = B.$

Answer

Let $C=\left[\begin{array}{l}a \\ b\end{array}\right]$
then $A C=B$
$\begin{array}{l}\Rightarrow\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{l}a \\ b\end{array}\right]=\left[\begin{array}{l}7 \\ 0\end{array}\right] \end{array} $
$ \Rightarrow\left[\begin{array}{c}3 a+b \\ -a+2 b\end{array}\right]=\left[\begin{array}{l}7 \\ 0\end{array}\right]$
$\Rightarrow 3a + b = 7 ...(1)$
$- a + 2b = 0 ...(2)$
From equation $(1),$
$6a + 2b = 14 ...(3)$
From $(3) - (2)$ given
$7a = 14$
$\Rightarrow a = 2$
Put $a = 2$ in $(1),$ we get
$6 + b = 7$
$\Rightarrow b = 7 - 6 = 1$
$\therefore C=\left[\begin{array}{l}2 \\ 1\end{array}\right]$.

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Question 63 Marks

If $A=\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]$ and $I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$, then find k so that $A^2= 8A + kI.$

Answer

We have
$\begin{aligned} & A=\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right] \end{aligned} $
$ A^2=A A=\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right] $
$ =\left[\begin{array}{cc}1 & 0 \\ -8 & 49\end{array}\right]$
And $\begin{aligned} & 8 A+k l=8\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]+k\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \end{aligned} $
$ =\left[\begin{array}{cc}8 & 0 \\ -8 & 56\end{array}\right]+\left[\begin{array}{ll}k & 0 \\ 0 & k\end{array}\right] $
$ =\left[\begin{array}{cc}8+k & 0 \\ -8 & 56+k\end{array}\right]$
Thus $A^2=8 A+k l$
$\Rightarrow\left[\begin{array}{cc}1 & 0 \\ -8 & 49\end{array}\right]=\left[\begin{array}{cc}8+k & 0 \\ -8 & 56+k\end{array}\right]$
$\Rightarrow 1 = 8 + k$
$\Rightarrow k = -7$
Also $56 + k = 49$
$\Rightarrow k = -7.$

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Question 73 Marks

Find $x$ and $y,$ if $\left(\begin{array}{cc}3 & -2 \\ -1 & 4\end{array}\right)\left(\begin{array}{c}2 x \\ 1\end{array}\right)+2\left(\begin{array}{c}-4 \\ 5\end{array}\right)=4\left(\begin{array}{l}2 \\ y\end{array}\right)$

Answer

$\begin{aligned} & \left(\begin{array}{cc}3 & -2 \\ -1 & 4\end{array}\right)\binom{2 x}{1}+2\binom{-4}{5}=4\binom{2}{y} \end{aligned} $
$\begin{aligned} & \binom{6 x-2}{-2 x+4}+\binom{-8}{10}=\binom{8}{4 y} \end{aligned} $
$\begin{aligned} & \binom{6 x-10-8}{-2 x+14+10}=\binom{8}{4 y} \end{aligned} $
Now $6 x-10=8$
$\therefore 6 x=18$
$ \therefore x=\frac{18}{6}=3$
and
$-x+14=4 y$
$ -2 \times 3+14=4 y$
or
$4 y=14-6=8$
$ \therefore y=\frac{8}{4}=2$
$ \therefore x=3, y=2$

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Question 83 Marks

If $A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$ and $I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ find $A^2 - 5A + 7 I.$

Answer

$A^2=A \cdot A $
$ =\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right] $
$ =\left[\begin{array}{cc}9-1 & 3+2 \\ -3-2 & -1+4\end{array}\right] $
$ =\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right] $
$ \Rightarrow A^2-5 A+71 $
$ =\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]-5\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]+7\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] $
$ =\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]-\left[\begin{array}{cc}15 & 5 \\ -5 & 10\end{array}\right]+\left[\begin{array}{ll}7 & 0 \\ 0 & 7\end{array}\right] $
$ =\left[\begin{array}{cc}-7 & 0 \\ 0 & -7\end{array}\right]+\left[\begin{array}{ll}7 & 0 \\ 0 & 7\end{array}\right] $
$=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right] $
$ =0 $

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Question 93 Marks

Find matrices $X$ and $Y, $ if $X+Y=\left[\begin{array}{ll}5 & 2 \\ 0 & 9\end{array}\right]$ and $X Y=\left[\begin{array}{cc}3 & 6 \\ 0 & -1\end{array}\right]$

Answer

We have
$ X+Y=\left[\begin{array}{ll} 5 & 2 \\ 0 & 9 \end{array}\right] $
and $ X-Y=\left[\begin{array}{cc} 3 & 6 \\ 0 & -1 \end{array}\right] $
Now $(X+Y)+(X-Y)=\left[\begin{array}{ll}5 & 2 \\ 0 & 9\end{array}\right]\left[\begin{array}{cc}3 & 6 \\ 0 & -1\end{array}\right]$
$ 2 X=\left[\begin{array}{ll} 8 & 8 \\ 0 & 8 \end{array}\right] $
$X=\frac{1}{2}\left[\begin{array}{ll} 8 & 8 \\ 0 & 8 \end{array}\right]=\left[\begin{array}{ll} 4 & 4 \\ 0 & 4
\end{array}\right]$
Also $(X+Y)-(X-Y)=\left[\begin{array}{ll}5 & 2 \\ 0 & 9\end{array}\right]+\left[\begin{array}{cc}-3 & -6 \\ 0 & 1\end{array}\right]$
$\Rightarrow 2 Y=\left[\begin{array}{cc} 2 & -4 \\ 0 & 10 \end{array}\right]$

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Question 103 Marks

Given $A=\left[\begin{array}{cc}2 & -6 \\ 2 & 0\end{array}\right] B=\left[\begin{array}{cc}-3 & 2 \\ 4 & 0\end{array}\right], C=\left[\begin{array}{ll}4 & 0 \\ 0 & 2\end{array}\right]$ Find the martix $X$ such that $A + 2X = 2B + C.$

Answer

$\begin{array}{l} A =\left[\begin{array}{cc}2 & -6 \\ 2 & 0\end{array}\right] B =\left[\begin{array}{cc}-3 & 2 \\ 4 & 0\end{array}\right], C =\left[\begin{array}{ll}4 & 0 \\ 0 & 2\end{array}\right] \end{array} $
$ A +2 x =2 B + C $
$ {\left[\begin{array}{cc}2 & -6 \\ 2 & 0\end{array}\right]+2 X =2\left[\begin{array}{cc}-3 & 2 \\ 4 & 0\end{array}\right]+\left[\begin{array}{ll}4 & 0 \\ 0 & 2\end{array}\right]} $
$ 2 X =\left[\begin{array}{cc}-6 & 4 \\ 8 & 0\end{array}\right]+\left[\begin{array}{cc}4 & 0 \\ 0 & 2\end{array}\right]-\left[\begin{array}{cc}2 & -6 \\ 2 & 0\end{array}\right] $
$ 2 X =\left[\begin{array}{cc}-6+4-24+0+6 \\ 8+0-20+2-0\end{array}\right]=\left[\begin{array}{cc}-4 & 10 \\ 6 & 2\end{array}\right] $
$ 2 X =2\left[\begin{array}{cc}-2 & 5 \\ 3 & 1\end{array}\right] X =\left[\begin{array}{cc}-2 & 5 \\ 3 & 1\end{array}\right]$

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Mathematics [3 marks sum]

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