Linear Algebra and Its Applications, Exercise 1.6.22

Exercise 1.6.22. Find the inverse of the following lower triangular matrix:

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

Answer: We use Gauss-Jordan elimination on the matrix A, starting by multiplying the first row by the multiplier \frac{1}{4} and subtracting it from the second row, multiplying the first row by the multiplier \frac{1}{3} and subtracting it from the third row, and multiplying the first row by the multiplier \frac{1}{2} and subtracting it from the fourth row:

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

We then multiply the second row by the multiplier \frac{1}{3} and subtract it from the third row, and multiply the second row by the multiplier \frac{1}{2} and subtract it from the fourth row:

\begin{bmatrix} 1&0&0&0&\vline&1&0&0&0 \\ 0&1&0&0&\vline&-\frac{1}{4}&1&0&0 \\ 0&\frac{1}{3}&1&0&\vline&-\frac{1}{3}&0&1&0 \\ 0&\frac{1}{2}&\frac{1}{2}&1&\vline&-\frac{1}{2}&0&0&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&0&0&0&\vline&1&0&0&0 \\ 0&1&0&0&\vline&-\frac{1}{4}&1&0&0 \\ 0&0&1&0&\vline&-\frac{1}{4}&-\frac{1}{3}&1&0 \\ 0&0&\frac{1}{2}&1&\vline&-\frac{3}{8}&-\frac{1}{2}&0&1 \end{bmatrix}

Finally we multiply the third row by the multiplier \frac{1}{2} and subtract it from the fourth row:

\begin{bmatrix} 1&0&0&0&\vline&1&0&0&0 \\ 0&1&0&0&\vline&-\frac{1}{4}&1&0&0 \\ 0&0&1&0&\vline&-\frac{1}{4}&-\frac{1}{3}&1&0 \\ 0&0&\frac{1}{2}&1&\vline&-\frac{3}{8}&-\frac{1}{2}&0&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&0&0&0&\vline&1&0&0&0 \\ 0&1&0&0&\vline&-\frac{1}{4}&1&0&0 \\ 0&0&1&0&\vline&-\frac{1}{4}&-\frac{1}{3}&1&0 \\ 0&0&0&1&\vline&-\frac{1}{4}&-\frac{1}{3}&-\frac{1}{2}&1 \end{bmatrix}

Since the original matrix A is lower triangular and has ones on the diagonal we do not need to do reverse elimination or any other steps. We thus have

A^{-1} = \begin{bmatrix} 1&0&0&0 \\ -\frac{1}{4}&1&0&0 \\ -\frac{1}{4}&-\frac{1}{3}&1&0 \\ -\frac{1}{4}&-\frac{1}{3}&-\frac{1}{2}&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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