If $A=\left[\begin{array}{lll}1 & 2 & 3 \\ 1 & 1 & a \\ 2 & 4 & 7\end{array}\right]$ and…
If $A=\left[\begin{array}{lll}1 & 2 & 3 \\ 1 & 1 & a \\ 2 & 4 & 7\end{array}\right]$ and $B=\left[\begin{array}{ccc}13 & 2 & b \\ -3 & -1 & 2 \\ -2 & 0 & 1\end{array}\right]$ where matrix $B$ is inverse of matrix $a$, then the value of $a$ and $b$ are
$\mathrm{a}=-5, \mathrm{~b}=7$
$a=7, b=-5$
$a=-7, b=5$
$a=5, b=-7$
Solution
$A=\left|\begin{array}{lll}
1 & 2 & 3 \\
1 & 1 & a \\
2 & 4 & 7
\end{array}\right|=(7-4 a)-2(7-2 a)+3(2)=-1$
matrix B is inverse of matrix A. ' $b$ ' is element $(1 \times 3)$ in matrix B. Here element $(3 \times 1)$ in matrix $A$ is 2 .
Cofactor of $2==(-1)^{3+1}\left|\begin{array}{ll}2 & 3 \\ 1 & \mathrm{a}\end{array}\right|=2 \mathrm{a}-3$
Now $\frac{2 \mathrm{a}-3}{-1}=\mathrm{b} \Rightarrow 2 \mathrm{a}+\mathrm{b}=3$
For element $(1 \times 2)$ in matrix A i.e. 2
Now cofactor of 2 is $(-1)^{1+2}\left|\begin{array}{ll}1 & a \\ 2 & 7\end{array}\right|=-(7-2 a)$ and
$\frac{-(7-2 \mathrm{a})}{-1}=7-2 \mathrm{a}$
Here element $(2 \times 1)$ in matrix B is -3
$\therefore 7-2 \mathrm{a}=-3$
Solving eq. (1) and (2), we get $\mathrm{a}=5, \mathrm{~b}=-7$