[4 marks sum]

Question 14 Marks

If $A=\left[\begin{array}{cc}2 & -1 \\ -4 & 5\end{array}\right]$ and $B=\left[\begin{array}{c}-3 \\ 2\end{array}\right]$ find the matrix $C$ such that $A C=B$

Answer

Given
$\begin{aligned}& A=\left[\begin{array}{cc}2 & -1 \\-4 & 5\end{array}\right] \end{aligned}$
$ B=\left[\begin{array}{c}-3 \\2\end{array}\right]$
Let martix $C=\left[\begin{array}{l}x \\ y\end{array}\right]$
$\therefore AC =\left[\begin{array}{cc}2 & -1 \\-4 & 5\end{array}\right]\left[\begin{array}{l}x \\y\end{array}\right]=\left[\begin{array}{c}2 x y \\-4 x+5 y\end{array}\right]$
But $AC = B$
$\therefore\left[\begin{array}{c}2 x-y \\-4 x+5 y\end{array}\right]=\left[\begin{array}{c}-3 \\2\end{array}\right]$
Comparing the corresponding elements
$\begin{array}{r}2 x-y=-3 \\-4 x+5 y=2\end{array}$
Multiplying (i) by 5 and (ii) by 1
$10 x-5 y=-15$
$ -4 x+5 y=2$
$ \text { Adding, we get }$
$ 6 x=-13$
$ \Rightarrow x=\frac{-13}{6}$
Adding, we get
$6 x =-13$
$ \Rightarrow x =\frac{-13}{6}$
Substituting the value of $x$ in $(i)$
$2\left(\frac{-13}{6}\right)-y=-3$
$ \Rightarrow \frac{-13}{3}-y=-3$
$ -y=-3+\frac{13}{3}$
$ =\frac{-9+13}{3}$
$ =\frac{4}{3}$
$\therefore y=-\frac{4}{3}$
$\therefore$ Matrix $C=\left[\begin{array}{c}\frac{-13}{6} \\ -\frac{4}{3}\end{array}\right]$.

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Question 24 Marks

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

Answer

Given that
$
\begin{aligned}
& 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}{cc}
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}{cc}
1+0 & 0+0 \\
2+2 & 0+1
\end{array}\right] \\
& =\left[\begin{array}{cc}
1 & 0 \\
4 & 1
\end{array}\right] \\
& 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}{cc}
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}{cc}
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}
4-3 & 6+0 \\
-2+0 & -3+0
\end{array}\right] \\
& =\left[\begin{array}{cc}
1 & 6 \\
-2 & -3
\end{array}\right] \\
& A^2+A B+B^2=\left[\begin{array}{cc}
1 & 0 \\
4 & 1
\end{array}\right]+\left[\begin{array}{cc}
2 & 3 \\
3 & 6
\end{array}\right]+\left[\begin{array}{cc}
1 & 6 \\
-2 & -3
\end{array}\right] \\
& =\left[\begin{array}{cc}
1+2+1 & 0+3+6 \\
4+3-2 & 1+6+-3
\end{array}\right] \\
& =\left[\begin{array}{cc}
4 & 9 \\
5 & 4
\end{array}\right]
\end{aligned}
$

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Question 34 Marks

Find the values of $a, b, c$ and $d$ if $\left[\begin{array}{ll}a+b & 3 \\ 5+c & a b\end{array}\right]=\left[\begin{array}{cc}6 & d \\ -1 & 8\end{array}\right]$

Answer

$
\left[\begin{array}{ll}
a+b & 3 \\
5+c & a b
\end{array}\right]=\left[\begin{array}{cc}
6 & d \\
-1 & 8
\end{array}\right]
$
Comparing the corresponding terms, we get.
$
\begin{aligned}
& 3=d \Rightarrow d=3 \\
& \Rightarrow 5+c=-1 \\
& \Rightarrow c=-1-5 \\
& \Rightarrow c=-6 \\
& a+b=6 \text { and } a b=8 \\
& \therefore(a-b)^2=(a+b)^2-4 a b \\
& =(6)^2-4 \times 8 \\
& =36-32 \\
& =4 \\
& =( \pm 2)^2 \\
& \therefore a-b, b= \pm 2 \\
& \text { (i) If } a-b=2 \\
& a+b=6
\end{aligned}
$
Adding, we get
$
\begin{aligned}
& 2 a=8 \\
& \Rightarrow a=4 \\
& a+b=6 \\
& \Rightarrow 4+b=6 \\
& \Rightarrow b=6-4=2 \\
& \therefore a=4, b=2 \\
& \text { (ii) If } a-b=-2 \\
& a+b=6
\end{aligned}
$
Adding, we get,
$
\begin{aligned}
& 2 a=4 \\
& \Rightarrow a=\frac{4}{2}=2 \\
& a+b=6 \\
& \Rightarrow 2+b=6 \\
& \Rightarrow b=6-2=4 \\
& \therefore a=2, b=4
\end{aligned}
$
Hence $a=4, b=2$, or $a=2, b=4$
$
c=-6 \text { and } d=3
$

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Question 44 Marks

Find the matrix $B$ if $A=\left[\begin{array}{ll}4 & 1 \\ 2 & 3\end{array}\right]$ and $A^2=A+2 B$

Answer

$
\begin{aligned}
& A =\left[\begin{array}{ll}
4 & 1 \\
2 & 3
\end{array}\right] \\
& \text { Let } B =\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right] \\
& A ^2= A \times A =\left[\begin{array}{ll}
4 & 1 \\
2 & 3
\end{array}\right]\left[\begin{array}{ll}
4 & 1 \\
2 & 3
\end{array}\right] \\
& =\left[\begin{array}{cc}
16+2 & 4+3 \\
8+6 & 2+9
\end{array}\right] \\
& =\left[\begin{array}{cc}
18 & 7 \\
14 & 11
\end{array}\right] \\
& A +2 B =\left[\begin{array}{ll}
4 & 1 \\
2 & 3
\end{array}\right]+2\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right] \\
& =\left[\begin{array}{cc}
4 & 1 \\
2 & 3
\end{array}\right]+\left[\begin{array}{cc}
2 a \\
2 c & 3+2 d
\end{array}\right] \\
& =\left[\begin{array}{cc}
4+2 a & 1+2 b \\
2+2 c & 3+2 d
\end{array}\right] \\
& \because A 2= A +2 B \\
& \therefore\left[\begin{array}{cc}
18 & 7 \\
14 & 11
\end{array}\right]=\left[\begin{array}{cc}
4+2 a & 1+2 b \\
2+2 c & 3+2 d
\end{array}\right]
\end{aligned}
$
Comparing the corresponding elements
$
\begin{aligned}
& 4+2 a=18 \\
& \Rightarrow 2 a=18-4=14 \\
& \therefore a=7 \\
& 1+2 b=7 \\
& \Rightarrow 2 b=7-1=6 \\
& \therefore b=3 \\
& 2+2 c=14 \\
& \Rightarrow 2 c=14-2=12 \\
& \therefore c=6 \\
& 3+2 d=11 \\
& \Rightarrow 2 d=11-3=8 \\
& \therefore d=4
\end{aligned}
$
Hence $a=7, b=3, c=6, d=4$
$
\therefore B=\left[\begin{array}{ll}
7 & 3 \\
6 & 4
\end{array}\right] \text {. }
$

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Question 54 Marks

If $B=\left[\begin{array}{cc}-4 & 2 \\ 5 & -1\end{array}\right]$ and $C=\left[\begin{array}{cc}17 & -1 \\ 47 & -13\end{array}\right]$ find the matrix $A$ such that $A B=C$

Answer

$
\begin{aligned}
& B =\left[\begin{array}{cc}
-4 & 2 \\
5 & -1
\end{array}\right] \\
& C =\left[\begin{array}{cc}
17 & -1 \\
47 & -13
\end{array}\right] \\
& \text { and } AB = C \\
& \text { Let } A =\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right] \\
& \text { Then } AB =\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right] \times\left[\begin{array}{cc}
-4 & 2 \\
5 & -1
\end{array}\right] \\
& =\left[\begin{array}{ll}
-4 a+5 b & 2 a-b \\
-4 c+5 d & 2 c-d
\end{array}\right] \\
& \because AB = C \\
& \therefore\left[\begin{array}{ll}
-4 a+5 b & 2 a-b \\
-4 c+5 d & 2 c-d
\end{array}\right]=\left[\begin{array}{cc}
17 & -1 \\
47 & 13
\end{array}\right]
\end{aligned}
$
Comparing corresponding elements, we get
$
\begin{aligned}
& \because-4 a+5 b=17....(1) \\
& 2 a-b=-1 .....(2)\\
& -4 c+5 d=47.....(3) \\
& 2 c-d=-13......(4)
\end{aligned}
$
Multiplying (1) by 1 and (2) by 2
$
\begin{aligned}
-4 a+5 b & =17 \\
4 a-2 b & =-2
\end{aligned}
$
Adding
$
\begin{aligned}
& 3 b=15 \\
& \Rightarrow b=\frac{15}{3}=5 \\
& 2 a-b=-1 \\
& \Rightarrow 2 a-5=-1 \\
& \Rightarrow 2 a=-1+5=4 \\
& \Rightarrow a=\frac{4}{2}+2 \\
& \therefore a=2, b=5
\end{aligned}
$
Again multiplying (iii) by 1 and (iv) by 2 ,
$
\begin{aligned}
& -4 c+5 d=47 \\
& 4 c-2 d=-26
\end{aligned}
$
Adding
$
\begin{aligned}
& 3 d=21 \\
& \Rightarrow d=\frac{21}{3}=7
\end{aligned}
$
and
$
\begin{aligned}
& 2 c-d=-13 \\
& \Rightarrow 2 c-7=-13 \\
& \Rightarrow 2 c=-13+7=-6 \\
& \Rightarrow c=\frac{-6}{2}=-3 \\
& \therefore c=-3, d=7
\end{aligned}
$
Now matrix $A=\left[\begin{array}{cc}2 & 5 \\ -3 & 7\end{array}\right]$.

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Question 64 Marks

If $A=\left[\begin{array}{cc}3 & -4 \\ -1 & 2\end{array}\right]$, find matrix $B$ such that $B A=1$, where $I$ is unity matrix of order 2

Answer

$
A=\left[\begin{array}{cc}
3 & -4 \\
-1 & 2
\end{array}\right]
$
$BA = I$, where $I$ is unity matrix of order 2
$
\therefore I=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]
$Let $B =\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$
$\therefore BA =\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \times\left[\begin{array}{cc}3 & -4 \\ -1 & 2\end{array}\right]$
$=\left[\begin{array}{ll}3 a-b & -4 a+2 b \\ 3 c-d & -4 c+2 d\end{array}\right]$
$
\therefore\left[\begin{array}{ll}
3 a-b & -4 a+2 b \\
3 c-d & -4 c+2 d
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]
$
Comparing the corresponding terms, we get
$
\begin{aligned}
& 3 a-b=1, \\
& -4 a+2 b=0 \\
& \Rightarrow 2 b=4 a \\
& \Rightarrow b=2 a \\
& \therefore 3 a-b=1 \\
& \Rightarrow 3 a-2 a=1 \\
& \Rightarrow a=1 \\
& \text { and } \\
& b=2 a \\
& \Rightarrow b=2 \times 1=2 \\
& \therefore a=1, b=2
\end{aligned}
$
and
$3 c-d=0$
$\Rightarrow d=3 c$
$-4 c +2 d =1$
$\Rightarrow-4 c+2 \times 3 c=1$
$\Rightarrow-4 c+6 c=1$
$\Rightarrow 2 x =1$
$\Rightarrow c=\frac{1}{2}$
and
$d =3 c =3 \times \frac{1}{2}=\frac{3}{2}$
Hence $a=1, b=2, c=\frac{1}{2}, d=\frac{3}{2}$
$\therefore$ Matrix $B=\left[\begin{array}{ll}1 & 2 \\ \frac{1}{2} & \frac{3}{2}\end{array}\right]$.

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Mathematics [4 marks sum]

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