## 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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