If A is a square matrix of order n times n,th | vedclass

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Question

(A) For any square matrix $B$,we have $B(\operatorname{adj} B) = |B| I_{n}$.Taking $B = \operatorname{adj} A$,we get $(\operatorname{adj} A)[\operator

Vedclass · Accepted Answer

$|A|^{n-2} A$

Vedclass · Answer

(A) For any square matrix $B$,we have $B(\operatorname{adj} B) = |B| I_{n}$.Taking $B = \operatorname{adj} A$,we get $(\operatorname{adj} A)[\operatorname{adj}(\operatorname{adj} A)] = |\operatorname{adj} A| I_{n}$.Since $|\operatorname{adj} A| = |A|^{n-1}$,we have $(\operatorname{adj} A)[\operatorname{adj}(\operatorname{adj} A)] = |A|^{n-1} I_{n}$.Multiplying both sides by $A$ on the left,we get $A(\operatorname{adj} A)[\operatorname{adj}(\operatorname{adj} A)] = |A|^{n-1} A$.Since $A(\operatorname{adj} A) = |A| I_{n}$,we have $(|A| I_{n})[\operatorname{adj}(\operatorname{adj} A)] = |A|^{n-1} A$.Dividing both sides by $|A|$ (assuming $|A| 
eq 0$),we get $\operatorname{adj}(\operatorname{adj} A) = |A|^{n-2} A$.

If $A$ is a square matrix of order $n \times n$,then $\operatorname{adj}(\operatorname{adj} A)$ is equal to

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