## 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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