Compute $A^5\mathbf{u}$ Using Linear Combination

Problem 696

Let
\[A=\begin{bmatrix}
-4 & -6 & -12 \\
-2 &-1 &-4 \\
2 & 3 & 6
\end{bmatrix}, \quad \mathbf{u}=\begin{bmatrix}
6 \\
5 \\
-3
\end{bmatrix}, \quad \mathbf{v}=\begin{bmatrix}
-2 \\
0 \\
1
\end{bmatrix}, \quad \text{ and } \mathbf{w}=\begin{bmatrix}
-2 \\
-1 \\
1
\end{bmatrix}.\]

(a) Express the vector $\mathbf{u}$ as a linear combination of $\mathbf{v}$ and $\mathbf{w}$.

(b) Compute $A^5\mathbf{v}$.

(c) Compute $A^5\mathbf{w}$.

(d) Compute $A^5\mathbf{u}$.

 
Add to solve later

Sponsored Links

Solution.

(a) Express the vector $\mathbf{u}$ as a linear combination of $\mathbf{v}$ and $\mathbf{w}$.

Our goal here is to find scalars $c_1, c_2$ such that
\[\mathbf{u}=c_1\mathbf{v}+c_2\mathbf{w}.\] This is the same as the matrix equation
\[\begin{bmatrix}
-2 & -2 \\
0 & -1 \\
1 &1
\end{bmatrix}
\begin{bmatrix}
c_1 \\
c_2
\end{bmatrix}=\begin{bmatrix}
6 \\
5 \\
-3
\end{bmatrix}.\] To solve this, we reduced the augmented matrix as follows:
\begin{align*}
\left[\begin{array}{rr|r}
-2 & -2 & 6 \\
0 &-1 &5 \\
1 & 1 & -3
\end{array}\right] \xrightarrow{R_1 \leftrightarrow R_3}
\left[\begin{array}{rr|r}
1 & 1 & -3 \\
0 &-1 &5 \\
-2 & -2 & 6
\end{array}\right]\\[6pt] \xrightarrow[-R_2]{R_3+2R_1}
\left[\begin{array}{rr|r}
1 & 1 & -3 \\
0 &1 &-5 \\
0 & 0 & 0
\end{array}\right] \xrightarrow{R_1-R_2}
\left[\begin{array}{rr|r}
1 & 0 & 2 \\
0 &1 &-5 \\
0 & 0 & 0
\end{array}\right].
\end{align*}
This yields the solution $c_1=2$ and $c_2=-5$.
Hence, we have the linear combination
\[\mathbf{u}=2\mathbf{v}-5\mathbf{w}.\]

(b) Compute $A^5\mathbf{v}$.

We first compute $A\mathbf{v}$. We have
\[A\mathbf{v}= \begin{bmatrix}
-4 & -6 & -12 \\
-2 &-1 &-4 \\
2 & 3 & 6
\end{bmatrix}
\begin{bmatrix}
-2 \\
0 \\
1
\end{bmatrix}
=\begin{bmatrix}
-4 \\
0 \\
2
\end{bmatrix}=2\begin{bmatrix}
-2 \\
0 \\
1
\end{bmatrix}=2\mathbf{v}.
\] Using this relation $A\mathbf{v}=2\mathbf{v}$, we obtain
\[A^2\mathbf{v}=AA\mathbf{v}=A(2\mathbf{v})=2A\mathbf{v}=2(2\mathbf{v})=2^2\mathbf{v}.\] Next, we have
\[A^3\mathbf{v}=AA^2\mathbf{v}=A(2^2\mathbf{v})=2^2A\mathbf{v}=2^2(a\mathbf{v})=2^3\mathbf{v}.\] Repeating this process, we see that $A^5\mathbf{v}=2^5\mathbf{v}$.
Or, we can find this by computing as follows:
\begin{align*}
A^5\mathbf{v}=A^2A^3\mathbf{v}=A^2(2^3\mathbf{v})=2^3A^2\mathbf{v}=2^3(2^2\mathbf{v})=2^5\mathbf{v}.
\end{align*}
In summary, we have
\[A^5\mathbf{v}=2^5\mathbf{v}=32\begin{bmatrix}
-2 \\
0 \\
1
\end{bmatrix}=\begin{bmatrix}
-64 \\
0 \\
32
\end{bmatrix}.\]

(c) Compute $A^5\mathbf{w}$.

First, we note that
\[A\mathbf{w}= \begin{bmatrix}
-4 & -6 & -12 \\
-2 &-1 &-4 \\
2 & 3 & 6
\end{bmatrix} \begin{bmatrix}
-2 \\
-1 \\
1
\end{bmatrix}=\begin{bmatrix}
2 \\
1 \\
-1
\end{bmatrix}=-\begin{bmatrix}
-2 \\
-1 \\
1
\end{bmatrix}=-\mathbf{w}.\] Using this relation $A\mathbf{w}=-\mathbf{w}$ as in part (a), we obtain
\[A^5\mathbf{w}=(-1)^5\mathbf{w}=-\begin{bmatrix}
-2 \\
-1 \\
1
\end{bmatrix}=\begin{bmatrix}
2 \\
1 \\
-1
\end{bmatrix}.\]

(d) Compute $A^5\mathbf{u}$.

Using the linear combination $\mathbf{u}=2\mathbf{v}-5\mathbf{w}$ obtained part (a), we compute
\begin{align*}
A^5\mathbf{w}&=A^5(2\mathbf{v}-5\mathbf{w})\\
&=2A^5\mathbf{v}-5A^5\mathbf{w}\\[6pt] &=2\begin{bmatrix}
-64 \\
0 \\
32
\end{bmatrix}-5\begin{bmatrix}
2 \\
1 \\
-1
\end{bmatrix} &&\text{by (b), (c)}\\[6pt] &=\begin{bmatrix}
-138 \\
-5 \\
69
\end{bmatrix}.
\end{align*}

Common Mistake

This is a midterm exam problem of Lienar Algebra at the Ohio State University.

One common pitfall is to compute $A^5$, but this is time consuming and it is very likely to make a mistake by hand computaiton.

Click here if solved 19
Add to solve later

Sponsored Links