If A is an invertible matrix of order n,then | vedclass

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(C) We know that for any square matrix $A$ of order $n$,the property of the adjoint matrix is given by $A(\operatorname{adj} A) = |A|I_n$,where $I_n$

Vedclass · Accepted Answer

$|A|^{n-1}$

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(C) We know that for any square matrix $A$ of order $n$,the property of the adjoint matrix is given by $A(\operatorname{adj} A) = |A|I_n$,where $I_n$ is the identity matrix of order $n$. Taking the determinant on both sides,we get $|A(\operatorname{adj} A)| = ||A|I_n|$. Using the property $|AB| = |A||B|$ and $|kA| = k^n|A|$ for a matrix of order $n$,we have $|A| \cdot |\operatorname{adj} A| = |A|^n \cdot |I_n|$. Since $|I_n| = 1$,we have $|A| \cdot |\operatorname{adj} A| = |A|^n$. Since $A$ is invertible,$|A| 
eq 0$,so we can divide by $|A|$ to get $|\operatorname{adj} A| = |A|^{n-1}$.

If $A$ is an invertible matrix of order $n$,then the determinant of $\operatorname{adj} A$ is equal to :

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