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Let $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]$ and $10 B=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right]$
If $B$ is the inverse of $A$, then the value of $\alpha$ is
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If $B$ is the inverse of $A$, then the value of $\alpha$ is
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The correct answer is:
5
5
$A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]$
$\therefore|A|=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]$
$=1(1+3)+1(2+3)+1(2-1)=4+5+1=10$
$\Rightarrow \operatorname{Adj} A=\left[\begin{array}{ccc}4 & -5 & 1 \\ 2 & 0 & -2 \\ 2 & 5 & 3\end{array}\right]^T=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]$
$\Rightarrow \quad B=A^{-1}=\frac{1}{10}\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]\left[\because A^{-1}=\frac{1}{|A|} \operatorname{Adj} A\right]$
$\Rightarrow 10 B=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]$
Hence, $\alpha=5$
$\therefore|A|=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]$
$=1(1+3)+1(2+3)+1(2-1)=4+5+1=10$
$\Rightarrow \operatorname{Adj} A=\left[\begin{array}{ccc}4 & -5 & 1 \\ 2 & 0 & -2 \\ 2 & 5 & 3\end{array}\right]^T=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]$
$\Rightarrow \quad B=A^{-1}=\frac{1}{10}\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]\left[\because A^{-1}=\frac{1}{|A|} \operatorname{Adj} A\right]$
$\Rightarrow 10 B=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]$
Hence, $\alpha=5$
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