Question

Picture attached below.

Answer 1

Note that both AB and BA will be 2 x 2 square matrices.

Also note that AB doesn't necessarily equal BA as matrix multiplication isn't commutative.

We find each product by multiplying the first row of A by the first column of B then first row of A by second column B. We then repeat the same procedure with the second row of A with the two columns of B. Doing so yields:\[\bf AB=\left[\begin{matrix}5 & 3 \\ 3 & 2\end{matrix}\right]\left[\begin{matrix}2 & -3 \\ -3 & 5\end{matrix}\right]=\left[\begin{matrix}5(2)+3(-3) & 5(-3)+3(5) \\ 3(2)+2(-3) & 3(-3)+2(5)\end{matrix}\right]=\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]=I_2\]As you can see, the result is a 2 x 2 Identity matrix. Now we do the same thing again except in the opposite order, i.e. we find \(\bf BA\):\[\bf BA=\left[\begin{matrix}2 & -3 \\ -3 & 5\end{matrix}\right]\left[\begin{matrix}5 & 3 \\ 3 & 2\end{matrix}\right]=\left[\begin{matrix}2(5)+(-3)3 & 2(3)+2(-3) \\ -3(5)+5(3) & -3(3)+5(2)\end{matrix}\right]=\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right] =I_2\]As you can see, the product either way remains the same. Because multiplying the matrices gives us the same result, which is a 2 x 2 identity matrix, then this must mean that \(\bf B=A^{-1}\).

Answer 2

@sprezzy