1. ## Matrix again

1.) If $\displaystyle X = P^{-1}AP$ and $\displaystyle A^3 = I$, prove that $\displaystyle X^3 = I$

Since order of multiplication don't matter when you're multiplying three matrices and more ( I deduced this from the multiplication of matrices (AB)C=A(BC) ),
X = IA
X = A
Multiplying A^2 on both sides,
A^2X = A^3
A^2X = I
This is where I'm stuck..

2.) For A = $\displaystyle \begin{bmatrix} 2 & 1 & -1 \\-1 & 2 & 1 \\0 & 6 & 1 \end{bmatrix}$ and B = $\displaystyle \begin{bmatrix} 4 & 7 & -3 \\-1 & -2 & 1 \\6 & 12 & -5 \end{bmatrix}$
calculate AB and hence solve the system of equations
4a+7b-3c = -8
-a-2b+c = 3
6a+12b-5c = -15

I got the identity matrix for AB.But I do not know how to use it to solve the system of equations, I had applied the identity matric to the system of equations so I got 4a=-8, -2b = 3 and -5c = -15, but this does not seem to be the way..

\displaystyle \displaystyle \begin{align*} X^3 &= \left( P^{-1}AP\right)^3 \\ &= P^{-1}APP^{-1}APP^{-1}AP \\ &= P^{-1}AIAIAP \\ &= P^{-1}AAAP \\ &= P^{-1}A^3P \\ &= P^{-1}IP \\ &= P^{-1}P \\ &= I \end{align*}