## Linear Algebra and Its Applications, Review Exercise 1.10

Review exercise 1.10. Find the inverse of each of the following matrices, or show that the matrix is not invertible. $A = \begin{bmatrix} 1&0&1 \\ 1&1&0 \\ 0&1&1 \end{bmatrix} \quad \rm and \quad A = \begin{bmatrix} 2&1&0 \\ 1&2&1 \\ 0&1&2 \end{bmatrix} \quad \rm and \quad A = \begin{bmatrix} 1&1&-2 \\ 1&-2&1 \\ 2&1&1 \end{bmatrix}$

Answer: We use Gauss-Jordan elimination on the first matrix $A$, starting by multiplying the first row by 1 and subtracting it from the second row: $\begin{bmatrix} 1&0&1&\vline&1&0&0 \\ 1&1&0&\vline&0&1&0 \\ 0&1&1&\vline&0&0&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&0&1&\vline&1&0&0 \\ 0&1&-1&\vline&-1&1&0 \\ 0&1&1&\vline&0&0&1 \end{bmatrix}$

We then multiply the second row times 1 and subtract it from the third row: $\begin{bmatrix} 1&0&1&\vline&1&0&0 \\ 0&1&-1&\vline&-1&1&0 \\ 0&1&1&\vline&0&0&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&0&1&\vline&1&0&0 \\ 0&1&-1&\vline&-1&1&0 \\ 0&0&2&\vline&1&-1&1 \end{bmatrix}$

This completes forward elimination. We start backward elimination by multiplying the third row by $-\frac{1}{2}$ and subtracting it from the second row, and multiplying the third row by $\frac{1}{2}$ and subtracting it from the first row: $\begin{bmatrix} 1&0&1&\vline&1&0&0 \\ 0&1&-1&\vline&-1&1&0 \\ 0&0&2&\vline&1&-1&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&0&0&\vline&\frac{1}{2}&\frac{1}{2}&-\frac{1}{2} \\ 0&1&0&\vline&-\frac{1}{2}&\frac{1}{2}&\frac{1}{2} \\ 0&0&2&\vline&1&-1&1 \end{bmatrix}$

Finally we divide the third row by 2: $\begin{bmatrix} 1&0&0&\vline&\frac{1}{2}&\frac{1}{2}&-\frac{1}{2} \\ 0&1&0&\vline&-\frac{1}{2}&\frac{1}{2}&\frac{1}{2} \\ 0&0&2&\vline&1&-1&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&0&0&\vline&\frac{1}{2}&\frac{1}{2}&-\frac{1}{2} \\ 0&1&0&\vline&-\frac{1}{2}&\frac{1}{2}&\frac{1}{2} \\ 0&0&1&\vline&\frac{1}{2}&-\frac{1}{2}&\frac{1}{2} \end{bmatrix}$

We then have $A = \begin{bmatrix} 1&0&1 \\ 1&1&0 \\ 0&1&1 \end{bmatrix} \quad \rm and \quad A^{-1} = \begin{bmatrix} \frac{1}{2}&\frac{1}{2}&-\frac{1}{2} \\ -\frac{1}{2}&\frac{1}{2}&\frac{1}{2} \\ \frac{1}{2}&-\frac{1}{2}&\frac{1}{2} \end{bmatrix}$

We also use Gauss-Jordan elimination on the second matrix, starting by multiplying the first row by $\frac{1}{2}$ and subtracting it from the second row: $\begin{bmatrix} 2&1&0&\vline&1&0&0 \\ 1&2&1&\vline&0&1&0 \\ 0&1&2&\vline&0&0&1 \end{bmatrix} \rightarrow \begin{bmatrix} 2&1&0&\vline&1&0&0 \\ 0&\frac{3}{2}&1&\vline&-\frac{1}{2}&1&0 \\ 0&1&2&\vline&0&0&1 \end{bmatrix}$

We next multiply the second row by $\frac{2}{3}$ and subtract it from the third row: $\begin{bmatrix} 2&1&0&\vline&1&0&0 \\ 0&\frac{3}{2}&1&\vline&-\frac{1}{2}&1&0 \\ 0&1&2&\vline&0&0&1 \end{bmatrix} \rightarrow \begin{bmatrix} 2&1&0&\vline&1&0&0 \\ 0&\frac{3}{2}&1&\vline&-\frac{1}{2}&1&0 \\ 0&0&\frac{4}{3}&\vline&\frac{1}{3}&-\frac{2}{3}&1 \end{bmatrix}$

This completes forward elimination. We start backward elimination by multiplying the third row by $\frac{3}{4}$ and subtracting it from the second row: $\begin{bmatrix} 2&1&0&\vline&1&0&0 \\ 0&\frac{3}{2}&1&\vline&-\frac{1}{2}&1&0 \\ 0&0&\frac{4}{3}&\vline&\frac{1}{3}&-\frac{2}{3}&1 \end{bmatrix} \rightarrow \begin{bmatrix} 2&1&0&\vline&1&0&0 \\ 0&\frac{3}{2}&0&\vline&-\frac{3}{4}&\frac{3}{2}&-\frac{3}{4} \\ 0&0&\frac{4}{3}&\vline&\frac{1}{3}&-\frac{2}{3}&1 \end{bmatrix}$

We next multiply the second row by $\frac{2}{3}$ and subtract it from the first: $\begin{bmatrix} 2&1&0&\vline&1&0&0 \\ 0&\frac{3}{2}&0&\vline&-\frac{3}{4}&\frac{3}{2}&-\frac{3}{4} \\ 0&0&\frac{4}{3}&\vline&\frac{1}{3}&-\frac{2}{3}&1 \end{bmatrix} \rightarrow \begin{bmatrix} 2&0&0&\vline&\frac{3}{2}&-1&\frac{1}{2} \\ 0&\frac{3}{2}&0&\vline&-\frac{3}{4}&\frac{3}{2}&-\frac{3}{4} \\ 0&0&\frac{4}{3}&\vline&\frac{1}{3}&-\frac{2}{3}&1 \end{bmatrix}$

This completes backward elimination. We then divide the first row by 2, the second row by $\frac{3}{2}$, and the third row by $\frac{4}{3}$: $\begin{bmatrix} 2&0&0&\vline&\frac{3}{2}&-1&\frac{1}{2} \\ 0&\frac{3}{2}&0&\vline&-\frac{3}{4}&\frac{3}{2}&-\frac{3}{4} \\ 0&0&\frac{4}{3}&\vline&\frac{1}{3}&-\frac{2}{3}&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&0&0&\vline&\frac{3}{4}&-\frac{1}{2}&\frac{1}{4} \\ 0&1&0&\vline&-\frac{1}{2}&1&-\frac{1}{2} \\ 0&0&1&\vline&\frac{1}{4}&-\frac{1}{2}&\frac{3}{4} \end{bmatrix}$

We then have $A = \begin{bmatrix} 2&1&0 \\ 1&2&1 \\ 0&1&2 \end{bmatrix} \quad A^{-1} = \begin{bmatrix} \frac{3}{4}&-\frac{1}{2}&\frac{1}{4} \\ -\frac{1}{2}&1&-\frac{1}{2} \\ \frac{1}{4}&-\frac{1}{2}&\frac{3}{4} \end{bmatrix}$

Finally we use Gauss-Jordan elimination on the third matrix, starting by multiplying the first row by 1 and subtracting it from the second row, and multiplying the first row by -2 and subtracting it from the third: $\begin{bmatrix} 1&1&-2&\vline&1&0&0 \\ 1&-2&1&\vline&0&1&0 \\ -2&1&1&\vline&0&0&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&1&-2&\vline&1&0&0 \\ 0&-3&3&\vline&-1&1&0 \\ 0&3&-3&\vline&2&0&1 \end{bmatrix}$

For the next step we multiply the second row by -1 and subtract it from the third row: $\begin{bmatrix} 1&1&-2&\vline&1&0&0 \\ 0&-3&3&\vline&-1&1&0 \\ 0&3&-3&\vline&2&0&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&1&-2&\vline&1&0&0 \\ 0&-3&3&\vline&-1&1&0 \\ 0&0&0&\vline&1&1&1 \end{bmatrix}$

This leaves us with no pivot for the third row, so elimination fails and the third matrix $A$ is not invertible.

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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