If $A=\left[\begin{array}{lll}1 & a & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7\end{array}\right]$ and…
If $A=\left[\begin{array}{lll}1 & a & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7\end{array}\right]$ and $A^{-1}=\left[\begin{array}{ccc}13 & 2 & -7 \\ -3 & b & 2 \\ -2 & 0 & 1\end{array}\right]$, then the values of $a$ and $b$ are respectively
$2,-1$
1,2
2,1
$-1,2$
Solution
$|A|=1 \times(7-20)-a(10-7)+3(4-2)=3 a-7$
Now $-3=-\left\{\frac{-3}{3 a-7}\right\} \Rightarrow a=2$
Also $\mathrm{b}=\frac{7-6}{3 \mathrm{a}-7}=\frac{1}{3 \times 2-7}=-1$
Hence $\mathrm{a}=2$ and $\mathrm{b}=-1$