[5 marks sum]

Question 15 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 25 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 [5 marks sum]

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