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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
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The correct answer is:
$2,-1$
$$
|A|=1 \times(7-20)-a(10-7)+3(4-2)=3 a-7
$$
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 $\mathrm{b}=-1$
|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 $\mathrm{b}=-1$
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