Find the adjoint of the matrix A =$\begin{bmatrix} 1 & 2 \\3 & -5 \end{bmatrix}$ and verify the result $A\;(adj\; A)=(adj\;A)\;A=$|$A$|$.I$

Toolbox:
• Let $A = [ a_{ij} ]$ be a square matrix x. Let $A_{ij}$ be the cofactor of $a_{ij}$. Then $[ A_{ij}]$ is the matrix of cofactors and $adj\: A$ ( or adjoint of the matrix A) is given by $adj\: A=[A_{ij}]^T$
• Multiplication of two matrices A and B is possible when the number of columns in the first matrix equals the number of rows in the second matrix B. If A is of type $m$ x $p$ and B is of type $p$ x $x$ then AB is of type $m$ x $n$ .
• $AB = [ c_{ij}]$ where $c_{ij} = \sum^p_{\substack{k=1}}a_{ik}b_{kj}$
Step 1
To find $adj\: A$ where $A = \begin{bmatrix} 1 & 2 \\ 3 & -5 \end{bmatrix}$
$[A_{ij} ] = \begin{bmatrix} -5 & -3 \\ -2 & 1 \end{bmatrix}$
$adj\: A = [ A_{ij}]^T = \begin{bmatrix} -5 & -2 \\ -3 & 1 \end{bmatrix}$
Step 2
To find $A(adj\: A)$ and $(adj\: A)\: A$
$A(adj\: A) = \begin{bmatrix} 1 & 2 \\ 3 & -5 \end{bmatrix} \begin{bmatrix} -5 & -2 \\ -3 & 1 \end{bmatrix}$
$= \begin{bmatrix} (1)(-5)+(2)(-3) & (1)(-2)+2(1) \\ (3)(-5)+(-5)(-3) & (3)(-2)+(-5)(1) \end{bmatrix}$
$= \begin{bmatrix} -5-6 & -2+2 \\ -15+15 & -6-5 \end{bmatrix} = \begin{bmatrix} -11 & 0 \\ 0 & -11 \end{bmatrix}$
$=-11 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = -11\: I$
$(adj\: A )A = \begin{bmatrix} -5 & -2 \\ -3 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & -5 \end{bmatrix}$
$= \begin{bmatrix} (-5)(1)+(-2)(-5) & (-5)(2)+(-2)(-5) \\ (-3)(1)+(1)(3) & (-3)(2)+1(-5) \end{bmatrix}$
$= \begin{bmatrix} -5-6 & -10+10 \\ -3+3 & -6-5 \end{bmatrix} = \begin{bmatrix} -11 & 0 \\ 0 & -11 \end{bmatrix} = -11\: I$
It can be seen that $A(adj\: A) = (adj\: A)A$
Step 3
To find $| A | I$
$| A | = \begin{vmatrix} 1 & 2 \\ 3 & -5 \end{vmatrix} = -5-6 = -11$
$| A | I = -11\: I$
It can be seen from step 2 that $A ( adj\: A) = (adj\: A)A = | A | \: I$
Hence verified