If $A=\left[\begin{array}{ll}0 & 3 \\ 2 & 0\end{array}\right]$ and $A^{-1}=\lambda(\operatorname{adj} A)$, then $\lambda=$
Given:
$A=\left[\begin{array}{ll}0 & 3 \\ 2 & 0\end{array}\right]$
$A^{-1}=\lambda(\operatorname{adj} A)$
Now,
$A=\left[\begin{array}{ll}0 & 3 \\ 2 & 0\end{array}\right]$
$\Rightarrow|A|=\left|\begin{array}{ll}0 & 3 \\ 2 & 0\end{array}\right|$
$=-6$
As we know that,
$A^{-1}=\frac{\operatorname{adj} A}{|A|}$
$\Rightarrow \lambda(\operatorname{adj} A)=\frac{1}{|A|}(\operatorname{adj} A)$
$\Rightarrow \lambda=\frac{1}{|A|}$
$\Rightarrow \lambda=\frac{1}{-6}$
$\Rightarrow \lambda=-\frac{1}{6}$
Hence, $\lambda=-\frac{1}{6}$.
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