Part 1: Finding (AB)^-1

Given: \( A=\begin{bmatrix}1&2&-2\\ -1&3&0\\ 0&-2&1\end{bmatrix} \) and \( B^{-1}=\begin{bmatrix}3&-1&1\\ -15&6&-5\\ 5&-2&2\end{bmatrix} \)

We need to find \( (AB)^{-1} \). We know that \( (AB)^{-1} = B^{-1}A^{-1} \).

First, we need to find \( A^{-1} \).

|A| = 1(3*1 - 0*(-2)) - 2((-1)*1 - 0*0) + (-2)((-1)*(-2) - 3*0) = 1(3) - 2(-1) - 2(2) = 3 + 2 - 4 = 1

Since |A| = 1, A^-1 exists.

The matrix of cofactors of A is:

Cof(A) = \(\begin{bmatrix}3&1&2\\ -2&1&2\\ 6&2&5\end{bmatrix}\)

Adj(A) = (Cof(A))^T = \(\begin{bmatrix}3&-2&6\\ 1&1&2\\ 2&2&5\end{bmatrix}\)

A^-1 = (1/|A|) * Adj(A) = \(\begin{bmatrix}3&-2&6\\ 1&1&2\\ 2&2&5\end{bmatrix}\)

Now, we can find (AB)^-1 = B^-1A^-1

(AB)^-1 = \(\begin{bmatrix}3&-1&1\\ -15&6&-5\\ 5&-2&2\end{bmatrix}\) \(\begin{bmatrix}3&-2&6\\ 1&1&2\\ 2&2&5\end{bmatrix}\)

(AB)^-1 = \(\begin{bmatrix}9-1+2&-6-1+2&18-2+5\\ -45+6-10&30+6-10&-90+12-25\\ 15-2+4&-10-2+4&30-4+10\end{bmatrix}\)

(AB)^-1 = \(\begin{bmatrix}10&-5&21\\ -49&26&-103\\ 17&-8&36\end{bmatrix}\)

Part 2: Solving the system of equations by matrix method

Given system of equations:

\( x+2y+3z=6 \)

\( 2x-y+z=2 \)

\( 3x+2y-2z=3 \)

We can write this system in matrix form as AX = B, where

A = \(\begin{bmatrix}1&2&3\\ 2&-1&1\\ 3&2&-2\end{bmatrix}\), X = \(\begin{bmatrix}x\\ y\\ z\end{bmatrix}\), B = \(\begin{bmatrix}6\\ 2\\ 3\end{bmatrix}\)

First, we find the determinant of A:

|A| = 1((-1)*(-2) - 1*2) - 2(2*(-2) - 1*3) + 3(2*2 - (-1)*3) = 1(2-2) - 2(-4-3) + 3(4+3) = 0 - 2(-7) + 3(7) = 0 + 14 + 21 = 35

Since |A| = 35, A^-1 exists.

The matrix of cofactors of A is:

Cof(A) = \(\begin{bmatrix}0&-7&7\\ 10&-11& -4\\ 5&5&-5\end{bmatrix}\)

Adj(A) = (Cof(A))^T = \(\begin{bmatrix}0&10&5\\ -7&-11&5\\ 7&-4&-5\end{bmatrix}\)

A^-1 = (1/|A|) * Adj(A) = (1/35) \(\begin{bmatrix}0&10&5\\ -7&-11&5\\ 7&-4&-5\end{bmatrix}\)

Now, we can find X = A^-1B

X = (1/35) \(\begin{bmatrix}0&10&5\\ -7&-11&5\\ 7&-4&-5\end{bmatrix}\) \(\begin{bmatrix}6\\ 2\\ 3\end{bmatrix}\)

X = (1/35) \(\begin{bmatrix}0+20+15\\ -42-22+15\\ 42-8-15\end{bmatrix}\)

X = (1/35) \(\begin{bmatrix}35\\ -49\\ 19\end{bmatrix}\)

X = \(\begin{bmatrix}1\\ -7/5\\ 19/35\end{bmatrix}\)

Therefore, x = 1, y = -7/5, z = 19/35