# Verify A (adj A)= (adj A) A = | A | I: $\begin{bmatrix} 2&3 \\ - 4&-6 \\ \end{bmatrix}$

Toolbox:
• For a square matrix of order 2,given by
• $A=\begin{bmatrix}a_{11} & a_{12}\\a_{21} & a_{22}\end{bmatrix}$
• The $adj \;A$ can also be obtained by interchanging $a_{11}$ and $a_{12}$ and by changing signs of $a_{12}$ and $a_{21}$
• (i.e)$adj A=\begin{bmatrix}a_{22} & -a_{12}\\-a_{21} & a_{11}\end{bmatrix}$
$\begin{bmatrix}2 & 3\\-4 & -6\end{bmatrix}$
The value of the determinant ,
$|A|=2\times -6-3\times -4=-12+12=0.$
Hence it is a singular matrix.
We know adj A can be obtained by interchanging $a_{11}$ and $a_{22}$ and by changing signs of $a_{12}$ and $a_{21}$
Therefore $adj\;A=\begin{bmatrix}a_{22} & -a_{12}\\-a_{21} & a_{11}\end{bmatrix}=\begin{bmatrix}-6 & -3\\4 & 2\end{bmatrix}$
Now let us find $A(adj A)$
(i. e)$\begin{bmatrix}2 & 3\\-4 & -6\end{bmatrix}\begin{bmatrix}-6 & -3\\4 & 2\end{bmatrix}$
Matrix multiplication can be obtained by multiplying the rows with the columns.
$\begin{bmatrix}2 & 3\\-4 & -6\end{bmatrix}\begin{bmatrix}-6 & -3\\4 & 2\end{bmatrix}=\begin{bmatrix}-3\times -6+3\times 4& -3\times 2 +3\times 2\\-4\times -6+4\times -6 & -4\times -3+-6\times 2\end{bmatrix}$
$\qquad=\begin{bmatrix}-12+12 & -6+6\\24-24 & 12-12\end{bmatrix}=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}$
Now let us find (adj A) A
(i.e)$\begin{bmatrix}-6 & -3\\4 & 2\end{bmatrix}\begin{bmatrix}2 & 3\\-4 & -6\end{bmatrix}$
Multiplying as we did previously,
$\;\;\;\;=\begin{bmatrix}-6\times 2+-3\times -4 & -6\times 3+-3\times -6\\4\times 2+-4\times 2 &4\times 3+2\times -6\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}-12+12 & -18+18\\8+(-8) & 12-12\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}$
|A|I=0$\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}=0.$
|A|I=0.
Hence verified.
answered Mar 6, 2013