## Linear Algebra and Its Applications, Review Exercise 1.28

Review exercise 1.28. Compute the following matrices:

$\begin{bmatrix} 1&0&0 \\ l&1&0 \\ m&0&1\end{bmatrix}^n \qquad \begin{bmatrix} 1&0&0 \\ l&1&0 \\ m&0&1\end{bmatrix}^{-1} \qquad \begin{bmatrix} 1&0&0 \\ l&1&0 \\ 0&m&1\end{bmatrix}^{-1}$

Answer: For the first matrix we have

$\begin{bmatrix} 1&0&0 \\ l&1&0 \\ m&0&1\end{bmatrix} \begin{bmatrix} 1&0&0 \\ l&1&0 \\ m&0&1\end{bmatrix} = \begin{bmatrix} 1&0&0 \\ 2l&1&0 \\ 2m&0&1\end{bmatrix}$

Based on this, we can guess at the answer for raising the matrix to the power of $n$, and try to prove it by induction. Assume that for some $k \ge 2$ we have

$\begin{bmatrix} 1&0&0 \\ l&1&0 \\ m&0&1\end{bmatrix}^k = \begin{bmatrix} 1&0&0 \\ kl&1&0 \\ km&0&1\end{bmatrix}$

Then for $k+1$ we have

$\begin{bmatrix} 1&0&0 \\ l&1&0 \\ m&0&1\end{bmatrix}^{k+1} = \begin{bmatrix} 1&0&0 \\ l&1&0 \\ m&0&1\end{bmatrix}^k \begin{bmatrix} 1&0&0 \\ l&1&0 \\ m&0&1\end{bmatrix}$

$= \begin{bmatrix} 1&0&0 \\ kl&1&0 \\ km&0&1\end{bmatrix} \begin{bmatrix} 1&0&0 \\ l&1&0 \\ m&0&1\end{bmatrix} = \begin{bmatrix} 1&0&0 \\ (k+1)l&1&0 \\ (k+1)m&0&1\end{bmatrix}$

By induction we then see that for all $n \ge 2$ we have

$\begin{bmatrix} 1&0&0 \\ l&1&0 \\ m&0&1\end{bmatrix}^n = \begin{bmatrix} 1&0&0 \\ nl&1&0 \\ nm&0&1\end{bmatrix}$

Note that this equation also holds true for $n = 1$ and $n = 0$, and we might guess that it holds true for $n = -1$ as well. We can test this as follows:

$\begin{bmatrix} 1&0&0 \\ l&1&0 \\ m&0&1\end{bmatrix} \begin{bmatrix} 1&0&0 \\ -l&1&0 \\ -m&0&1\end{bmatrix} = \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1\end{bmatrix}$

and

$\begin{bmatrix} 1&0&0 \\ -l&1&0 \\ -m&0&1\end{bmatrix} \begin{bmatrix} 1&0&0 \\ l&1&0 \\ m&0&1\end{bmatrix} = \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1\end{bmatrix}$

Therefore we have

$\begin{bmatrix} 1&0&0 \\ l&1&0 \\ m&0&1\end{bmatrix}^{-1} = \begin{bmatrix} 1&0&0 \\ -l&1&0 \\ -m&0&1\end{bmatrix}$

We can find the inverse of the last matrix by Gauss-Jordan elimination. We start by multiplying the first row by $l$ and subtracting it from the second row:

$\begin{bmatrix} 1&0&0&\vline&1&0&0 \\ l&1&0&\vline&0&1&0 \\ 0&m&1&\vline&0&0&1 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&0&0&\vline&1&0&0 \\ 0&1&0&\vline&-l&1&0 \\ 0&m&1&\vline&0&0&1 \end{bmatrix}$

We then multiply the second row by $m$ and subtract it from the third row:

$\begin{bmatrix} 1&0&0&\vline&1&0&0 \\ 0&1&0&\vline&-l&1&0 \\ 0&m&1&\vline&0&0&1 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&0&0&\vline&1&0&0 \\ 0&1&0&\vline&-l&1&0 \\ 0&0&1&\vline&lm&-m&1 \end{bmatrix}$

We thus have

$\begin{bmatrix} 1&0&0 \\ l&1&0 \\ 0&m&1\end{bmatrix}^{-1} = \begin{bmatrix} 1&0&0 \\ -l&1&0 \\ lm&-m&1\end{bmatrix}$

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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