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If $A$ is a square matrix of order $n \times n$, then $\operatorname{adj}(\operatorname{adj} A)$ is equal to
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Verified Answer
The correct answer is:
$|A|^{n-2} A$
For any square matrix $B$, we have $B(\operatorname{adj} B)=|B| I_{n}$
On taking $B=\operatorname{adj} A$, we get $(\operatorname{adj} A)[\operatorname{adj}(\operatorname{adj} A)]=|\operatorname{adj} A| I_{n}$
$$
\Rightarrow \quad \operatorname{adj} A[\operatorname{adj}(\operatorname{adj} A)]=|A|^{n-1} I_{n}
$$
$\left(\because|\operatorname{adj} A|=|A|^{n-1}\right)$
$\Rightarrow(A \operatorname{adj} A)[\operatorname{adj}(\operatorname{adj} A)]=|A|^{n-1} A$
$\Rightarrow\left(|A| I_{n}\right)[\operatorname{adj}(\operatorname{adj} A)]=|A|^{n-1} A$
$\Rightarrow$
$(\operatorname{adj} A)=|A|^{n-2} A$
On taking $B=\operatorname{adj} A$, we get $(\operatorname{adj} A)[\operatorname{adj}(\operatorname{adj} A)]=|\operatorname{adj} A| I_{n}$
$$
\Rightarrow \quad \operatorname{adj} A[\operatorname{adj}(\operatorname{adj} A)]=|A|^{n-1} I_{n}
$$
$\left(\because|\operatorname{adj} A|=|A|^{n-1}\right)$
$\Rightarrow(A \operatorname{adj} A)[\operatorname{adj}(\operatorname{adj} A)]=|A|^{n-1} A$
$\Rightarrow\left(|A| I_{n}\right)[\operatorname{adj}(\operatorname{adj} A)]=|A|^{n-1} A$
$\Rightarrow$
$(\operatorname{adj} A)=|A|^{n-2} A$
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