Question

$Adj(AdjA)$ is equal to ___, if $A$ is a square matrix of order $n$.

• A. $|A{|}^{n-1}A$
• B. $|A{|}^{n}A$
• C. $|A{|}^{n-2}A$
• D. None of these

Answer 1

The correct option is C

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

Explanation for the correct option:

Step1. Find the value of $\mathbf{(}\mathit{A}\mathit{d}\mathit{j}\mathbf{}\mathit{A}\mathbf{)}$:

For any square matrix $A$, we know that

${\mathit{A}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\frac{\mathbf{a}\mathbf{d}\mathbf{j}\mathbf{}\mathbf{A}}{\mathbf{\left|}\mathbf{A}\mathbf{\right|}}\mathbf{}\mathbf{.}......\left(i\right)$

Also $A{A}^{-1}=I$, where $I$ is the identity matrix

${\mathit{A}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\mathbf{}\frac{\mathbf{I}}{\mathbf{A}}\mathbf{}\mathbf{.}\mathbf{.}....\left(ii\right)$

From equations (i) and (ii), we get

$\frac{adjA}{\left|A\right|}=\frac{I}{A}$

$\Rightarrow$$AadjA=\left|A\right|......\left(iii\right)$

Step2. Find the value of $Adj(AdjA)$:

Replace $A$ by $adjA$in (iii). we get

$$\begin{gathered} \Rightarrow adjAadj(adjA)=|adjA| \\ \Rightarrow adj(adjA)=\frac{|adjA|}{adjA}........\left(iv\right) \end{gathered}$$

From (iii) $adjA=\frac{\left|A\right|}{A}$

Step 3. Put this value in (iv), and we get

$adj(adjA)=\frac{|adjA|A}{\left|A\right|}$

$\Rightarrow$ $adj(adjA)=\frac{|A{|}^{n-1}A}{\left|A\right|}$ $\left[\mathbf{\because }\mathbf{|}\mathit{a}\mathit{d}\mathit{j}\mathbf{}\mathit{A}\mathbf{|}\mathbf{}\mathbf{=}\mathbf{}\mathbf{|}\mathit{A}{\mathbf{|}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\right]$

$=\mathbf{|}\mathit{A}{\mathbf{|}}^{\mathbf{n}\mathbf{-}\mathbf{2}}\mathbf{}\mathit{A}$

Hence the correct option is C.