If $A$ is a matrix of order $n\times n$, the value of $|adj\,A|$ is : (a) $|A|^{n-1}$ (b) $|A|^{n}$ (c) $|A|^{1+n}$ (d) $|A|$

If A is a matrix of order n× n, the value of |adjA| is : (a) |A|ⁿ⁻¹ (b) |A|ⁿ (c) |A|¹⁺ⁿ (d) |A|

Question 34 of 44

A

(a)

|A|^{n-1}

(b)

|A|^{n}

(c)

|A|^{1+n}

(d)

|A|

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Using $A(\operatorname{adj}A)=|A|\,I_n$ and taking determinants gives $|\operatorname{adj}A|=|A|^{\,n-1}$ , option (a).

Key identity. For any $n\times n$ matrix,

$A\,(\operatorname{adj}A)=|A|\,I_n.$

Take determinants of both sides.

$|A|\cdot|\operatorname{adj}A|=\big|\,|A|\,I_n\big|=|A|^{\,n}.$

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