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 an invertible matrix, $|A| 
eq 0$. Dividing both sides by $|A|$, we 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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