If $A=\left[\begin{array}{ccc}2 & 0 & -1 \ 5 & 1 & 0 \ 0 & 1 & 3\end{array} ight]$ and $A^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \ \alpha & 6 & -5 \ \beta & -2 & 2\end{array} ight]$, then the values of $\alpha$ and $\beta$ are. respectively

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If $A=\left[\begin{array}{ccc}2 & 0 & -1 \ 5 & 1 & 0 \ 0 & 1 & 3\end{array}ight]$ and $A^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \ \alpha & 6 & -5 \ \beta & -2 & 2\end{array}ight]$, then the values of $\alpha$ and $\beta$ are. respectively

Aicharya · Accepted Answer

We know, $\begin{array}{l} \mathrm{A} \cdot \mathrm{A}^{-1}=\mathrm{I} \ {\left[\begin{array}{ccc} 2 & 0 & -1 \ 5 & 1 & 0 \ 0 & 1 & 3 \end{array}ight]\left[\begin{array}{ccc} 3 & -1 & 1 \ \alpha & 6 & -5 \ \beta & -2 & 2 \end{array}ight]=\left[\begin{array}{lll} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array}ight]} \ {\left[\begin{array}{ccc} 6+0-\beta & -2+0+2 & 2+0-2 \ 15+\alpha+0 & -5+6+0 & 5-5+0 \ 0+\alpha+3 \beta & 0+6-6 & 0-5+6 \end{array}ight]=\left[\begin{array}{lll} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array}ight]} \ {\left[\begin{array}{ccc} 6-\beta & 0 & 0 \ 15+\alpha & 1 & 0 \ \alpha+3 \beta & 0 & 1 \end{array}ight]=\left[\begin{array}{ccc} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array}ight]} \ 6-\beta=1 \Rightarrow \beta=5 	ext { and } 15+\alpha=0 \Rightarrow \alpha=-15 \end{array}$ This problem can also be solved as follows : We have $A=\left[\begin{array}{ccc}2 & 0 & -1 \ 5 & 1 & 0 \ 0 & 1 & 3\end{array}ight] \Rightarrow|A|=2(3)-(5)=1$ $\alpha$ is $a_{21}$ in $A^{-1}$. So we will find cofactor of $a_{12}$ in $A$.

If $A=\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]$ and…

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