Let $A$ be a $3\times 3$ matrix and let
\[\mathbf{v}=\begin{bmatrix}
1 \\
2 \\
-1
\end{bmatrix} \text{ and } \mathbf{w}=\begin{bmatrix}
2 \\
-1 \\
3
\end{bmatrix}.\] Suppose that $A\mathbf{v}=-\mathbf{v}$ and $A\mathbf{w}=2\mathbf{w}$.
Then find the vector
\[A^5\begin{bmatrix}
-1 \\
8 \\
-9
\end{bmatrix}.\]

Solution.

Note that the given information is only about the vectors $\mathbf{v}$ and $\mathbf{w}$.
Thus, we first need to relate the vector $\begin{bmatrix}
-1 \\
8 \\
-9
\end{bmatrix}$ with $\mathbf{v}$ and $\mathbf{w}$.
So let us find a linear combination:
\[\begin{bmatrix}
-1 \\
8 \\
-9
\end{bmatrix}=a\mathbf{v}+b\mathbf{w}.\] The augmented matrix for this is
\begin{align*}
\left[\begin{array}{rr|r}
1 & 2 & -1 \\
2 &-1 &8 \\
-1 & 3 & -9
\end{array}\right] \xrightarrow[R_3+R_1]{R_2-2R_1}
\left[\begin{array}{rr|r}
1 & 2 & -1 \\
0 &-5 &10 \\
0 & 5 & -10
\end{array}\right] \xrightarrow{-\frac{1}{5}R_2}\\[6pt] \left[\begin{array}{rr|r}
1 & 2 & -1 \\
0 &1 &-2 \\
0 & 5 & -10
\end{array}\right] \xrightarrow[R_3-R_2]{R_1-2R_2}
\left[\begin{array}{rr|r}
1 & 0 & 3 \\
0 &1 &-2 \\
0 & 0 & 0
\end{array}\right].
\end{align*}
It yields that the solution is $a=3$ and $b=-2$.
Thus, we have
\[\begin{bmatrix}
-1 \\
8 \\
9
\end{bmatrix}=3\mathbf{v}-2\mathbf{w}.\]

Then, we compute
\begin{align*}
A^5\begin{bmatrix}
-1 \\
8 \\
9
\end{bmatrix}&=A^5(3\mathbf{v}-2\mathbf{w})=3A^5\mathbf{v}-2A^5\mathbf{w} \tag{*}
\end{align*}

Now, by assumption we have $A\mathbf{v}=-\mathbf{v}$.
Using this repeatedly we see that $A^5\mathbf{v}=(-1)^5\mathbf{v}=-\mathbf{v}$.
Similarly, we have $A^5\mathbf{w}=2^5\mathbf{w}=32\mathbf{w}$ using $A\mathbf{w}=2\mathbf{w}$.
Combining these with (*) yields,
\begin{align*}
A^5\begin{bmatrix}
-1 \\
8 \\
9
\end{bmatrix}&=3(-\mathbf{v})-2(32\mathbf{w})\\[6pt] &=-3\begin{bmatrix}
1 \\
2 \\
-1
\end{bmatrix}-64\begin{bmatrix}
2 \\
-1 \\
3
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
-3-128 \\
-6+64 \\
3-192
\end{bmatrix}
=\begin{bmatrix}
-131 \\
58 \\
-189
\end{bmatrix}.
\end{align*}

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