# If $A=\Large\frac{1}{3}$$\begin{bmatrix} 2 & 2 & 1 \\-2 & 1 & 2 \\1 & -2 & 2 \end{bmatrix}$ prove that $A^{-1}=A^T$.

Toolbox:
• 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
If $A^{-1} = A^T,$ then $AA^{-1}=AA^T=I$
It is sufficient for us to show $AA^T = I$
Step 2
$A = \frac{1}{3} \begin{bmatrix} 2 & 2 & 1 \\ -2 & 1 & 2 \\ 1 & -2 & 2 \end{bmatrix} A^T = \frac{1}{3}\begin{bmatrix} 2 & -2 & 1 \\ 2 & 1 & -2 \\ 1 & 2 & 2 \end{bmatrix}$
$AA^T = \frac{1}{9} \begin{bmatrix} 2 & 2 & 1 \\ -2 & 1 & 2 \\ 1 & -2 & 2 \end{bmatrix} \begin{bmatrix} 2 & -2 & 1 \\ 2 & 1 & -2 \\ 1 & 2 & 2 \end{bmatrix}$
$= \frac{1}{9} \begin{bmatrix} (2)(2)+(2)(2)+(1)(1) & (2)(-2)+(2)(1)+(1)(2) & (2)(1)+(2)(-2)+(1)(2) \\ (-2)(2)+(1)(2)+(2)(1) & (-2)(-2)+(1)(1)+(2)(2) & (-2)(1)+1(-2)+(2)(2) \\ (1)(2)+(-2)(2)+(2)(1) & (1)(-2)+(-2)(1)+(-1)(2) & (1)(1)+(-2)(-2)+(2)(2) \end{bmatrix}$
$= \frac{1}{9} \begin{bmatrix} 4+4+1 & -4+2+2 & 2-4+2 \\ -4+2+2 & 4+1+4 & -2-2+4 \\ 2-4+2 & -2-2+4 & 1+4+4 \end{bmatrix} = \frac{1}{9} \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix}$
$= \frac{1}{9} . 9 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$
From this, we conclude $A^T = A^{-1}$
Note : This method is shorter and quicker than actually finding $A^{-1}$.