[2 Mark Question Answer]

Question 12 Marks

If $A =\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ and $I =\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ show that $A^2 - (a + d) A = (bc - ad) I.$

Answer

Here $A^2 - (a + d) A$
$\begin{array}{l}=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]-(a+d)\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\end{array}$
$=\left[\begin{array}{ll}a^2+b c & a b+b d \\ a c+d c & c b+d^2\end{array}\right]-\left[\begin{array}{ll}a^2+a d & a b+b d \\ a c+d c & a d+d^2\end{array}\right] $
$ =\left[\begin{array}{cc}b c-a d & 0 \\ 0 & b c-a d\end{array}\right]$
$=(b c-a d)\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$= (bc - ad) I.$
Hence proved

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Question 22 Marks

Find $X$ and $Y,$ if $\left[\begin{array}{cc}2 x & x \\ y & 3 y\end{array}\right]\left[\begin{array}{l}3 \\ 2\end{array}\right]=\left[\begin{array}{c}16 \\ 9\end{array}\right]$

Answer

$\begin{array}{l}{\left[\begin{array}{cc}2 x & x \\ y & 3 y\end{array}\right]\left[\begin{array}{l}3 \\ 2\end{array}\right]=\left[\begin{array}{c}16 \\ 9\end{array}\right]} \end{array}$
$ {\left[\begin{array}{c}6 x+2 x \\ 3 y+6 y\end{array}\right]=\left[\begin{array}{c}16 \\ 9\end{array}\right]} $
$ {\left[\begin{array}{l}8 x \\ 9 y\end{array}\right]=\left[\begin{array}{c}16 \\ 9\end{array}\right]} $
$8 x =16, x =2 9, y =9, y =1 .$

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Question 32 Marks

Given that $A =\left[\begin{array}{ll}3 & 0 \\ 0 & 4\end{array}\right]$ and $B = \left[\begin{array}{ll}a & b \\ 0 & c\end{array}\right]$ and that $AB = A + B,$ find the values of $a, b$ and $c.$

Answer

$AB = A + B$
$\begin{array}{l}{\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}{ll}3 & 0 \\ 0 & 4\end{array}\right]+\left[\begin{array}{ll}a & b \\ 0 & c\end{array}\right]} \end{array} $
$ {\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]}$
$3 a=3+a$
$2 a=3$
$a=\frac{3}{2}$
$3 b=b$
$2 b=0$
$b=0$
$4 c=4+c$
$3 c=4$
$c=\frac{4}{3} .$

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Question 42 Marks

Find $x, y$ if $\left[\begin{array}{cc}-2 & 0 \\ 3 & 1\end{array}\right]\left[\begin{array}{c}-1 \\ 2 x\end{array}\right]+3\left[\begin{array}{c}-2 \\ 1\end{array}\right]=2\left[\begin{array}{l}y \\ 3\end{array}\right]$.

Answer

$\left[\begin{array}{cc}-2 & 0 \\ 3 & 1\end{array}\right]\left[\begin{array}{c}-1 \\ 2 x\end{array}\right]+3\left[\begin{array}{c}-2 \\ 1\end{array}\right]=2\left[\begin{array}{l}y \\ 3\end{array}\right]$
$\left[\begin{array}{c}2+0 \\ -3+2 x\end{array}\right]+\left[\begin{array}{c}-6 \\ 3\end{array}\right]=\left[\begin{array}{c}2 y \\ 6\end{array}\right]$
$\left[\begin{array}{c}-4 \\ 2 x\end{array}\right]=\left[\begin{array}{c}2 y \\ 6\end{array}\right]$
$\Rightarrow 2 y =-4,2 x =6 $
$\Rightarrow y =-2, x =3$
Thus required values is $x=3, y=-2$.

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Question 52 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 $BA = C^2.$ Find the values of $p$ and $q.$

Answer

$ 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] $
$ BA =\left[\begin{array}{cc}0 & -q \\ 1 & 0\end{array}\right]\left[\begin{array}{cc}p & 0 \\ 0 & 2\end{array}\right]=\left[\begin{array}{cc}0 & -2 q \\ p & 0\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]$
$BA = C^2$
$\Rightarrow -2q = -8$
$\Rightarrow q = 4$
$p = 8$
$\Rightarrow p = 8.$

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Question 62 Marks

If $A =\left[\begin{array}{cc}3 & 5 \\ 4 & -2\end{array}\right]$ and $B =\left[\begin{array}{l}2 \\ 4\end{array}\right]$, is the product $AB$ possible ? Given a reason. If yes, find $AB.$

Answer

$A =\left[\begin{array}{cc}3 & 5 \\ 4 & -2\end{array}\right]_{2 \times 2}$ and $B =\left[\begin{array}{l}2 \\ 4\end{array}\right]_{2 \times 1}$
the product $AB$ is possible as the number of columns in $A$ are equal to the number of rows in $B.$
Now $AB =\left[\begin{array}{cc}3 & 5 \\ 4 & -2\end{array}\right]\left[\begin{array}{l}2 \\ 4\end{array}\right]$
$\begin{array}{l}=\left[\begin{array}{c}3 \times 2+5 \times 4 \\ 4 \times 2+(-2) \times 4\end{array}\right] \\\end{array} $
$=\left[\begin{array}{c}26 \\ 0\end{array}\right] .$

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Question 72 Marks

Construct a $2 \times 2$ matrix whose elements $a_{ij}$ are given by
$\frac{(i+2 j)^2}{2}$.

Answer

$ \text { We have } a _{ ij }=\frac{(i+2 j)^2}{2}$
$a _{11}=\frac{(1+2 \times 1)^2}{2}=\frac{9}{2}$
$a _{12}=\frac{(1+2 \times 1)^2}{2}=\frac{25}{2}$
$a_{21}=\frac{(2+2 \times 1)^2}{2}=\frac{16}{2}=8$
$a_{22}=\frac{(2+2 \times 2)^2}{2}=18 $
The required matrix
$
\begin{aligned}
A & =\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right] \\
A & =\left[\begin{array}{cc}
\frac{9}{2} & \frac{25}{2} \\
8 & 18
\end{array}\right] .
\end{aligned}
$

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Question 82 Marks

Construct a $2 \times 2$ matrix whose elements $a_{ij}$ are given by $a_{ij}= 2i - j$

Answer

We have $a_{i j}=2 i-j$
Now $a_{11}=2 \times 1-1=1$
$ a_{12}=2 \times 1-2=0$
$a_{21}=2 \times 2-1=3$
$a_{22}=2 \times 2-2=2 $
So the required matrix
$ \begin{array}{l} A =\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] \end{array}$
$A =\left[\begin{array}{ll} 1 & 0 \\ 3 & 2 \end{array}\right] .$

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Question 92 Marks

Find $x$ and $y$ if $:\left(\begin{array}{cc}-3 & 2 \\ 0 & 5\end{array}\right)\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{c}-5 \\ y\end{array}\right)$

Answer

$\begin{array}{l}\left(\begin{array}{cc}-3 & 2 \\ 0 & 5\end{array}\right)\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{c}-5 \\ y\end{array}\right) \end{array} $
$ \Rightarrow\left(\begin{array}{c}-3 x+4 \\ 0-10\end{array}\right)=\left(\begin{array}{c}-5 \\ y\end{array}\right)$
$\Rightarrow-3 x+4=-5 $
$\Rightarrow-3 x=-5-4 $
$ \Rightarrow-3 x=-9 $
$ \Rightarrow x=3$
and
$y = -10$

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Question 102 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

$\begin{array}{l}{\left[\begin{array}{ll}x & 3 x \\ y & 4 y\end{array}\right]\left[\begin{array}{l}2 \\ 1\end{array}\right]=\left[\begin{array}{c}5 \\ 12\end{array}\right]} \end{array}$
$ \Rightarrow\left[\begin{array}{l}2 x+3 x \\ 2 y+4 y\end{array}\right]=\left[\begin{array}{c}5 \\ 12\end{array}\right] $
$ \Rightarrow\left[\begin{array}{l}5 x \\ 6 y\end{array}\right]=\left[\begin{array}{c}5 \\ 12\end{array}\right] $
$\Rightarrow 5 x =5 $
$ \Rightarrow x =1$
and
$\Rightarrow 6y = 12$
$\Rightarrow y = 2.$

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Question 112 Marks

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

Answer

Let $X =\left[\begin{array}{l}a \\ b\end{array}\right]$
$\begin{array}{l}\text { so }\left[\begin{array}{cc}2 & 1 \\ -3 & 4\end{array}\right]\left[\begin{array}{l}a \\ b\end{array}\right]=\left[\begin{array}{l}7 \\ 6\end{array}\right] \end{array} $
$ \Rightarrow\left[\begin{array}{c}2 a+b \\ -3 a+4 b\end{array}\right]=\left[\begin{array}{l}7 \\ 6\end{array}\right] $
$ \Rightarrow 2 a + b =7 \ldots(1) \\ -3 a +4 b =6 \ldots(2) \\ \text { From (1) and (2) } a =2, b =3$
$\Rightarrow X=\left[\begin{array}{l}2 \\ 3\end{array}\right]$.

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Question 122 Marks

Find the value of $x$ given that $A=\left[\begin{array}{cc}2 & 12 \\ 0 & 1\end{array}\right] , B=\left[\begin{array}{ll}4 & x \\ 0 & 1\end{array}\right]$

Answer

$\begin{array}{l} A =\left[\begin{array}{cc}2 & 12 \\ 0 & 1\end{array}\right] \end{array} $
$ \therefore 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 & 36 \\ 0 & 1\end{array}\right] $
$ \because A ^2= B ($given$)$
$ \therefore\left[\begin{array}{cc}4 & 36 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}4 & x \\ 0 & 1\end{array}\right] $
$ \therefore x =36 .$

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Question 132 Marks

$\begin{array}{l}{\left[\begin{array}{cc}2 \sin 30^{\circ} & -2 \cos 60^{\circ} \\ -\cot 45^{\circ} & \sin 90^{\circ}\end{array}\right]} {\left[\begin{array}{cc}\tan 45^{\circ} & \sec 60^{\circ} \\ \operatorname{cosec} 30^{\circ} & \cos 0^{\circ}\end{array}\right]}\end{array}$

Answer

$\begin{array}{l}{\left[\begin{array}{cc}2 \sin 30^{\circ} & -2 \cos 60^{\circ} \\ -\cot 45^{\circ} & \sin 90^{\circ}\end{array}\right]} \\\end{array} $
$ {\left[\begin{array}{rr}\tan 45^{\circ} & \sec 60^{\circ} \\ \operatorname{cosec} 30^{\circ} & \cos 0^{\circ}\end{array}\right]} $
$ =\left[\begin{array}{cc}2 \times \frac{1}{2}-2 \times \frac{1}{2}\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right] $
$=\left[\begin{array}{cc}1 & 1 \\ -1 & 1\end{array}\right]\left[\begin{array}{cc}1 & 2 \\ 2 & 1\end{array}\right]$
$=\left[\begin{array}{cc}-1 & 1 \\ 1 & -1\end{array}\right]$

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Mathematics [2 Mark Question Answer]

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