Question

If 3A = $\left[\begin{array}{ccc}1&2&2\\2&1&-2\\-2&2&-1\end{array}\right]$, then show that A⁻¹ = A'

Answer 1

Please see the attachment

Answer 2

Answer:

In this question,

We have to show to that if 3A = [tex]\mathbf{\begin{bmatrix} 1 &2 &2 \\  2 &1 &-2 \\  -2 &2 &-1  \end{bmatrix}}[/tex] then A⁻¹ = A(transpose)

Proof -

3A = [tex]\mathbf{\begin{bmatrix} 1 &2 &2 \\  2 &1 &-2 \\  -2 &2 &-1  \end{bmatrix}}[/tex]

$A\ =\ \frac{1}{3}\begin{bmatrix}1 &2 &2 \\ 2 &1 &-2 \\ -2 &2 &-1 \end{bmatrix}$

$A^{T} = \frac{1}{3}\begin{bmatrix}1 &2 &-2 \\ 2 &1 &2 \\ 2 &-2 &-1 \end{bmatrix}$

Then $AA^{T}$

$AA^{T}\ =\  \frac{1}{3}\begin{bmatrix}1 &2 &2 \\ 2 &1 &-2 \\ -2 &2 &-1 \end{bmatrix}\times \frac{1}{3}\begin{bmatrix}1 &2 &-2 \\ 2 &1 &2 \\ 2 &-2 &-1 \end{bmatrix}$

$AA^{T}\ =\ \frac{1}{9}\begin{bmatrix}9 &0 &0 \\ 0 &9 &0 \\ 0 &0 &9 \end{bmatrix}$

$AA^{T}\ =\ \begin{bmatrix}1 &0 &0 \\ 0 &1 &0 \\ 0 &0 &1 \end{bmatrix}$

$AA^{T}\ =\ I$

We know that A is invertible

Therefore $\mathbf{A^{T}\ =\ A^{-1}}$

Hence, proved