Linear Algebra and Its Applications, Exercise 1.6.6

Exercise 1.6.6. Invert the following matrices using the Gauss-Jordan method:

A_1 = \begin{bmatrix} 1&0&0 \\ 1&1&1 \\ 0&0&1 \end{bmatrix} \quad A_2 = \begin{bmatrix} 2&-1&0 \\ -1&2&-1 \\ 0&-1&2 \end{bmatrix} \quad A_3 = \begin{bmatrix} 0&0&1 \\ 0&1&1 \\ 1&1&1 \end{bmatrix}

Answer: For the first matrix Gauss-Jordan elimination proceeds as follows: We first subtract 1 times the first row from the second row:

\begin{bmatrix} 1&0&0&1&0&0 \\ 1&1&1&0&1&0 \\ 0&0&1&0&0&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&0&0&1&0&0 \\ 0&1&1&-1&1&0 \\ 0&0&1&0&0&1 \end{bmatrix}

This completes the process of forward elimination. We then start reverse elimination by subtracting 1 times the third row from the second row:

\begin{bmatrix} 1&0&0&1&0&0 \\ 0&1&1&-1&1&0 \\ 0&0&1&0&0&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&0&0&1&0&0 \\ 1&1&0&-1&1&-1 \\ 0&0&1&0&0&1 \end{bmatrix}

This completes reverse elimination. Since the left matrix is now the identity matrix we are done, and we have

A_1^{-1} = \begin{bmatrix} 1&0&0 \\ -1&1&-1 \\ 0&0&1 \end{bmatrix}

For the second matrix we begin forward elimination by subtracting -1/2 times the first row from the second row:

\begin{bmatrix} 2&-1&0&1&0&0 \\ -1&2&-1&0&1&0 \\ 0&-1&2&0&0&1 \end{bmatrix} \rightarrow \begin{bmatrix} 2&-1&0&1&0&0 \\ 0&\frac{3}{2}&-1&\frac{1}{2}&1&0 \\ 0&-1&2&0&0&1 \end{bmatrix}

and then multiply -2/3 by the second row and subtract it from the third:

\begin{bmatrix} 2&-1&0&1&0&0 \\ 0&\frac{3}{2}&-1&\frac{1}{2}&1&0 \\ 0&-1&2&0&0&1 \end{bmatrix} \rightarrow \begin{bmatrix} 2&-1&0&1&0&0 \\ 0&\frac{3}{2}&-1&\frac{1}{2}&1&0 \\ 0&0&\frac{4}{3}&\frac{1}{3}&\frac{2}{3}&1 \end{bmatrix}

This completes forward elimination, so we begin reverse elimination by multiplying -3/4 by the third row and subtracting it from the second:

\begin{bmatrix} 2&-1&0&1&0&0 \\ 0&\frac{3}{2}&-1&\frac{1}{2}&1&0 \\ 0&0&\frac{4}{3}&\frac{1}{3}&\frac{2}{3}&1 \end{bmatrix} \rightarrow \begin{bmatrix} 2&-1&0&1&0&0 \\ 0&\frac{3}{2}&0&\frac{3}{4}&\frac{3}{2}&\frac{3}{4} \\ 0&0&\frac{4}{3}&\frac{1}{3}&\frac{2}{3}&1 \end{bmatrix}

We continue by multiplying the second row by -2/3 and subtracting it from the first row:

\begin{bmatrix} 2&-1&0&1&0&0 \\ 0&\frac{3}{2}&0&\frac{3}{4}&\frac{3}{2}&\frac{3}{4} \\ 0&0&\frac{4}{3}&\frac{1}{3}&\frac{2}{3}&1 \end{bmatrix} \rightarrow \begin{bmatrix} 2&0&0&\frac{3}{2}&1&\frac{1}{2} \\ 0&\frac{3}{2}&0&\frac{3}{4}&\frac{3}{2}&\frac{3}{4} \\ 0&0&\frac{4}{3}&\frac{1}{3}&\frac{2}{3}&1 \end{bmatrix}

In the final step we multiply the first row by 1/2, the second row by 2/3, and the third row by 3/4:

\begin{bmatrix} 2&0&0&\frac{3}{2}&1&\frac{1}{2} \\ 0&\frac{3}{2}&0&\frac{3}{4}&\frac{3}{2}&\frac{3}{4} \\ 0&0&\frac{4}{3}&\frac{1}{3}&\frac{2}{3}&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&0&0&\frac{3}{4}&\frac{1}{2}&\frac{1}{4} \\ 0&1&0&\frac{1}{2}&1&\frac{1}{2} \\ 0&0&1&\frac{1}{4}&\frac{1}{2}&\frac{3}{4} \end{bmatrix}

to obtain

A_2^{-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}

For the third matrix we must first do a row exchange to get a nonzero pivot in the first row. We exchange the first row with the third:

\begin{bmatrix} 0&0&1&1&0&0 \\ 0&1&1&0&1&0 \\ 1&1&1&0&0&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&1&1&0&0&1 \\ 0&1&1&0&1&0 \\ 0&0&1&1&0&0\end{bmatrix}

This completes forward elimination, so we begin reverse elimination by subtracting 1 times the third row from the second row and 1 times the third row from the first row:

\begin{bmatrix} 1&1&1&0&0&1 \\ 0&1&1&0&1&0 \\ 0&0&1&1&0&0\end{bmatrix} \rightarrow \begin{bmatrix} 1&1&0&-1&0&1 \\ 0&1&0&-1&1&0 \\ 0&0&1&1&0&0\end{bmatrix}

Finally we subtract 1 times the second row from the first row:

\begin{bmatrix} 1&1&0&-1&0&1 \\ 0&1&0&-1&1&0 \\ 0&0&1&1&0&0\end{bmatrix} \rightarrow \begin{bmatrix} 1&0&0&0&-1&1 \\ 0&1&0&-1&1&0 \\ 0&0&1&1&0&0\end{bmatrix}

This completes elimination and we have

A_3^{-1} = \begin{bmatrix} 0&-1&1 \\ -1&1&0 \\ 1&0&0\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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