Browse Questions

# For $A=\begin{bmatrix} -1 & 2 & -2 \\4 & -3 & 4 \\4 & -4 & 5 \end{bmatrix}$ show that $A=A^{-1}$

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 $A = A^{-1}$ then $AA = AA^{-1} = I$ or $A^2=I$
It is sufficient for us to show that $A^2=I$
Step 2
$A^2 = \begin{bmatrix} -1 & 2 & -2 \\ 4 & -3 & 4 \\ 4 & -4 & 5 \end{bmatrix} \begin{bmatrix} -1 & 2 & -2 \\ 4 & -3 & 4 \\ 4 & -4 & 5 \end{bmatrix}$
$= \begin{bmatrix} (-1)(-1)+(2)(4)+(2)(4) & (-1)(2)+(2)(-3)+(-2)(-4) & (-1)(-2)+(2)(4)+(-2)(5) \\ (4)(-1)+(-3)(4)+(4)(4) & (4)(2)+(-3)(-3)+(4)(-4) & (4)(-2)+(-3)(4)+(4)(5) \\ (4)(-1)+(-4)(4)+(5)(4) & (4)(2)+(-4)(-3)+(5)(-4) & (4)(-2)+(-4)(4)+(5)(5) \end{bmatrix}$
$= \begin{bmatrix} 1+8-8 & -2-6+8 & 2+8-10 \\ -4-12+16 & 8+9-16 & -8-12+20 \\ -4-16+20 & 8+12-20 & -8-16+25 \end{bmatrix}$
$= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$
From step 1 it follows that $A^{-1} = A$
Note : This method is shorter and quicker than actually evaluating $A^{-1}$.