Prove that |A adj A|=|A|^n. - Vidyadip

Question

Prove that $|A \operatorname{adj} A|=|A|^n$.

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Answer

Let $A=\left[a_{i j}\right]$ be a square matrix of order $n \times n$. If $C_{i j}$ is a cofactor of $a_{i j}$ in $A$, then adj $A=\left[C_{i j}\right]^{\top}=\left[C_{i j}\right]$. Also, it is a matrix of order $n \times n$. Because $A$ and ahj $A$ are matrices of order $n \times n, A \times(\operatorname{adj} A)$ exists and is of order $n \times n$.
$\Rightarrow\left\{\text{A}\times(\text{adj A})\right\}_{\text{ij}}=\sum\limits_{\text{r}-1}^\text{n}\text{A}_{\text{ir }}(\text{adj A})_\text{rj}$
$=\sum\limits_{\text{n}}^{\text{r}-1}\text{a}_{\text{i r}}\text{C}_{\text{r j}}=\begin{cases}|\text{A}| \text{ if i}=\text{j}\\ 0 \text{ otherwise}\end{cases}$
Thus, each diagonal element of A × (adj A) is |A|. Also, the non-diagonal elements are zero.$\Rightarrow\ \text{A}\times(\text{adj A})=\begin{bmatrix}|\text{A}| & 0 & 0 & ......... & 0 \\ 0 & |\text{A}| & 0 & .......... & 0 \\ 0 & 0 & |\text{A}| & 0 ....... & 0 \\ . \\ . \\ . \\ 0 & 0 & 0 & .......... & |\text{A}|\end{bmatrix} $
$\Rightarrow\ |\text{A}\times(\text{adj A})|=\begin{bmatrix}|\text{A}| & 0 & 0 & ......... & 0 \\ 0 & |\text{A}| & 0 & .......... & 0 \\ 0 & 0 & |\text{A}| & 0 ....... & 0 \\ . \\ . \\ . \\ 0 & 0 & 0 & .......... & |\text{A}|\end{bmatrix} $
$=|\text{A}|^\text{n}\begin{bmatrix} 1 & 0 & 0 & ......... & 0 \\ 0 & 1 & 0 & .......... & 0 \\ 0 & 0 & 1 & 0 ....... & 0 \\ . \\ . \\ . \\ 0 & 0 & 0 & .......... & 1\end{bmatrix} $
$=|\text{A}|^\text{n}\text{I}_\text{n}=|\text{A}|^\text{n}$
$\Rightarrow\ |\text{A}\times(\text{adj A})|=|\text{A}|^\text{n}$
Hence proved.

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