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If $A$ is a square matrix such that $A(\operatorname{adj} A)=\left[\begin{array}{lll}4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4\end{array}\right]$, then $\operatorname{det}(\operatorname{adj} A)$ is equal to
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$16$
$\begin{aligned} & \text { Given } A(\operatorname{adj} A)=\left[\begin{array}{lll}4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4\end{array}\right] \\ & \text { Now, }|A(\operatorname{adj} A)|=\left|\begin{array}{lll}4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4\end{array}\right| \\ & \Rightarrow|A||\operatorname{adj} A|=4^3 \\ & \because \quad|\operatorname{adj} A|=|A|^{n-1} \\ & \text { Here, } \quad n=3 \\ & \therefore \quad|\operatorname{adj} A|=|A|^2 \\ & \left.\Rightarrow \quad|A| A A\right|^2=64 \\ & \Rightarrow \quad|A|^3=(4)^3 \\ & \Rightarrow \quad|A|=4 \\ & \end{aligned}$
$\begin{aligned} & \text { Now, } \quad|\operatorname{adj} A|=|A|^2=(4)^2 \\ & \therefore \quad|\operatorname{adj} A|=16 \\ & \end{aligned}$
$\begin{aligned} & \text { Now, } \quad|\operatorname{adj} A|=|A|^2=(4)^2 \\ & \therefore \quad|\operatorname{adj} A|=16 \\ & \end{aligned}$
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