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

\(
A=\left|\begin{array}{lll}
1 & 2 & 3 \\
1 & 1 & a \\
2 & 4 & 7
\end{array}\right|=(7-4 a)-2(7-2 a)+3(2)=-1
\)
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 \(\mathrm{a}=5, \mathrm{~b}=-7\)