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