[5 marks sum]

Question 15 Marks

If $A=\left[\begin{array}{ll}0 & -1 \\ 4 & -3\end{array}\right], B =\left[\begin{array}{c}-5 \\ 6\end{array}\right]$ and $3 A \times M =2 B$; Find matrix $M$

Answer

Let the order of matrix $M$ be $a \times b.$
$3A \times m = 2B$
$3\left[\begin{array}{ll}0 & -1 \\ 4 & -3\end{array}\right]_{2 \times 2} \times M_{a \times b}=2\left[\begin{array}{c}-5 \\ 6\end{array}\right]_{2 \times 1}$
Clearly the order of matrix $M$ is $2 \times 1$
Let $M=\left[\begin{array}{l}x \\ y\end{array}\right]$
Then
$3\left[\begin{array}{ll}0 & -1 \\ 4 & -3\end{array}\right] \times\left[\begin{array}{l}x \\ y\end{array}\right]=2\left[\begin{array}{c}-5 \\ 6\end{array}\right] $
$ {\left[\begin{array}{cc}0 & -3 \\ 12 & -9\end{array}\right] \times\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}-10 \\ 2\end{array}\right]} \\ {\left[\begin{array}{c}-3 y \\ 12 x-9 y\end{array}\right]=\left[\begin{array}{c}-10 \\ 12\end{array}\right]}$
Comparing the corressponding elements we get
$-3y = - 10$
$\Rightarrow y=\frac{10}{3}$
$12 x-9 y=12$
$=12 x=42 $
$ \Rightarrow x=\frac{7}{2} $
$ \therefore M=\left[\begin{array}{c}\frac{7}{2} \\ \frac{10}{3}\end{array}\right]$

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Question 25 Marks

Given $\left[\begin{array}{cc}2 & 1 \\ -3 & 4\end{array}\right] x=\left[\begin{array}{l}7 \\ 6\end{array}\right]$ Write the matrix $x$

Answer

$ \begin{array}{l} \text { Let } x=\left[\begin{array}{l} x \\ y \end{array}\right]\end{array} $
$\therefore\left[\begin{array}{cc} 2 & 1 \\ -3 & 4 \end{array}\right] \times\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 7 \\ 6 \end{array}\right] $
$ \Rightarrow\left[\begin{array}{c} 2 x+y \\ -3 x+4 y \end{array}\right]=\left[\begin{array}{l} 7 \\ 6 \end{array}\right] \\ \Rightarrow 2 x+y=7 \ldots(1) \\ -3 x+4 y=6 \ldots(2) $
Multiplying by $4$ in equation $(1)$ and solving with equation $(2)$
$ \begin{array}{l} 8 x+4 y=28 \\ -3 x+4 y=6 \\ ( \pm) \_(z) \_(z) \\ 11 x=22 \\ x=2 \end{array} $
Putting the value of $x$ in equation $(1),$ we get
$ \begin{array}{l} \therefore 2 \times 2+y=7 \\ y=7-4=3 \end{array} $
$\therefore$ The matrix $x =\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}2 \\ 3\end{array}\right]$

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Question 35 Marks

Given matrix $B=\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]$ Find the matrix $X$ if, $X=B^2-4 B$. Hence, solve for $a$ and $b$ given $X\left[\begin{array}{l}a \\ b\end{array}\right]=\left[\begin{array}{c}5 \\ 50\end{array}\right]$

Answer

$B^2=B \times B=\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]$
$=\left[\begin{array}{ll}1 \times 1+1 \times 8 & 1 \times 1+1 \times 3 \\ 8 \times 1+3 \times 8 & 8 \times 1+3 \times 3\end{array}\right]$
$=\left[\begin{array}{cc}9 & 4 \\ 32 & 17\end{array}\right]$
$4 B=4\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]=\left[\begin{array}{cc}4 & 4 \\ 32 & 12\end{array}\right]$
Given : $X=B^2-4 B$
$\Rightarrow X=\left[\begin{array}{cc}9 & 4 \\ 32 & 17\end{array}\right]-\left[\begin{array}{cc}4 & 4 \\ 32 & 12\end{array}\right]$
$=\left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right]$
To find: $a$ and $b \times\left[\begin{array}{l}a \\ b\end{array}\right]=\left[\begin{array}{c}5 \\ 50\end{array}\right] .......$ given
$ \Rightarrow\left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right]\left[\begin{array}{l}a \\ b\end{array}\right]=\left[\begin{array}{c}5 \\ 50\end{array}\right] $
$ \Rightarrow\left[\begin{array}{l}5 a \\ 5 b\end{array}\right]=\left[\begin{array}{c}5 \\ 50\end{array}\right] $
$ \Rightarrow 5\left[\begin{array}{l}a \\ b\end{array}\right]=5\left[\begin{array}{c}1 \\ 10\end{array}\right]$
$\Rightarrow a = 1$ and $b = 10$

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Question 45 Marks

Given matrix $A =\left[\begin{array}{l}4 \sin 30^{\circ}, \cos 0^{\circ} \\ \cos 0^{\circ}, 4 \sin 30^{\circ}\end{array}\right]$ and $B=\left[\begin{array}{l}4 \\ 5\end{array}\right]$ If $AX = B$. Find the matrix $'X\ '$

Answer

Let the matrix $X =\left[\begin{array}{l}x \\ y\end{array}\right]$
$AX = B$
$\begin{array}{l}\Rightarrow\left[\begin{array}{c}4 \sin 30^{\circ}, \cos 0^{\circ} \\ \cos 0^{\circ}, 4 \sin 30^{\circ}\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}4 \\ 5\end{array}\right] \end{array}$
$ \Rightarrow\left[\begin{array}{cc}4\left(\frac{1}{2}\right) & 1 \\ 1 & 4\left(\frac{1}{2}\right)\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}4 \\ 5\end{array}\right] $
$\Rightarrow\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}4 \\ 5\end{array}\right] $
$ \Rightarrow\left[\begin{array}{l}2 x+y \\ x+2 y\end{array}\right]=\left[\begin{array}{l}4 \\ 5\end{array}\right]$
$\Rightarrow 2x + y = 4 .............(1)$
$x + 2y = 5 .........(2)$
Multiplying $(1)$ by $2,$ we get
$4x + 2y = 8 ….(3)$
Subtracting $(2)$ from $(3),$ we get
$3x = 3$
$\Rightarrow x = 1$
Substituting the value of $x$ in $(1),$ we get
$2(1) + y = 4$
$\Rightarrow 2 + y = 4$
$\Rightarrow y = 2$
Hence, the matrix $X =\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}1 \\ 2\end{array}\right]$

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Question 55 Marks

Let $A=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right], B=\left[\begin{array}{cc}2 & 3 \\ -1 & 0\end{array}\right]$. Find $A^2+A B+B^2$

Answer

$A=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right], B=\left[\begin{array}{cc}2 & 3 \\ -1 & 0\end{array}\right]$
$A^2=A \times A=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right] \times\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right]$
$=\left[\begin{array}{ll}1 \times 1+0 \times 2 & 1 \times 0+0 \times 1 \\ 2 \times 1+1 \times 2 & 2 \times 0+1 \times 1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 4 & 1\end{array}\right]$
$A B=A \times B=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right] \times\left[\begin{array}{cc}2 & 3 \\ -1 & 0\end{array}\right] $
$=\left[\begin{array}{ll}1 \times 2+0 \times(-1) & 1 \times 3+0 \times 0 \\ 2 \times 2+1 \times(-1) & 2 \times 3+1 \times 0\end{array}\right] =\left[\begin{array}{ll}2 & 3 \\ 3 & 6\end{array}\right]$
$B^2=B \times B=\left[\begin{array}{cc}2 & 3 \\ -1 & 0\end{array}\right] \times\left[\begin{array}{cc}2 & 3 \\ -1 & 0\end{array}\right] $
$ =\left[\begin{array}{cc}2 \times 2+3 \times(-1) & 2 \times 3+3 \times 0 \\ (-1) \times 2+0 \times(-1) & -1 \times 3+0 \times 0\end{array}\right] =\left[\begin{array}{cc}1 & 6 \\ -2 & -3\end{array}\right] $
$\therefore A^2+A B+B^2=\left[\begin{array}{ll}1 & 0 \\ 4 & 1\end{array}\right]+\left[\begin{array}{ll}2 & 3 \\ 3 & 6\end{array}\right]+\left[\begin{array}{cc}1 & 6 \\ -2 & -3\end{array}\right] =\left[\begin{array}{ll}4 & 9 \\ 5 & 4\end{array}\right]$

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Question 65 Marks

if $A=\left[\begin{array}{ll}2 & 1 \\ 1 & 3\end{array}\right]$ and $B=\left[\begin{array}{c}3 \\ -11\end{array}\right]$ Find the matrix $X$ such that $AX = B$

Answer

Let the order of the matrix $X$ be $a \times b$.
$AX = B$
$\left[\begin{array}{ll}2 & 1 \\ 1 & 3\end{array}\right]_{2 \times 2} \times X_{a \times b}=\left[\begin{array}{c}3 \\ -11\end{array}\right]_{2 \times 1}$
Clearly the order of the matrix $X$ is $2 \times 1$
Let $X=[(x),,(y)]$
$\left[\begin{array}{ll}2 & 1 \\ 1 & 3\end{array}\right] \times\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}3 \\ -11\end{array}\right]$
Comparing the two matrices we get
$2x = y = 3 ...(1)$
$x = 3y = -11 ...(2)$
Multiplying $(1)$ with $3$ we get
$6x + 3y = 9 ...(3)$
Substracting $(2)$ from $(3)$ we get
$5x = 20$
$x = 4$
From $(1)$ we have
$y = 3 - 2x = 3 - 8 = -5$
$\therefore X=\left[\begin{array}{c}4 \\ -5\end{array}\right]$

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Mathematics [5 marks sum]

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