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