Question

Find the inverse of the matrix using echelon reduction method .

x + 3y - z =0

x + y + z = 1

-x + 2y - z = 1

Answer 1

Solution.

$$A=\begin{pmatrix} 1 & 3 &-1 \\ 1&1&1\\ -1&2&-1 \end{pmatrix}$$

$A=\begin{pmatrix} 1 & 3 &-1|1&0&1 \\ 1&1 & 1 | 0&1&0\\ -1&2&-1|0&0&1 \end{pmatrix}= \begin{pmatrix} 1 & 3 &-1|1&0&1 \\ 0&-2&2|-1&1&0\\ 0&5&-2|1&0&1 \end{pmatrix}=\newline= \begin{pmatrix} 1 & 3 &-1|1&0&1 \\ 0&1&-1|\frac{1}{2}&-\frac{1}{2}&0\\ 0&0&3| -\frac{3}{2}&\frac{5}{2}&1 \end{pmatrix}= \begin{pmatrix} 1 & 3 &-1|1&0&1 \\ 0&1&-1|\frac{1}{2}&-\frac{1}{2}&0\\ 0&0&1| -\frac{1}{2}&\frac{5}{6}&\frac{1}{3} \end{pmatrix}=\newline =\begin{pmatrix} 1 & 3 &0|\frac{1}{2}&\frac{5}{6}&\frac{1}{3} \\ 0&1&0|0&\frac{1}{3}&\frac{1}{3}\\ 0&0&1| -\frac{1}{2}&\frac{5}{6}&\frac{1}{3} \end{pmatrix}= =\begin{pmatrix} 1 & 0 &0|\frac{1}{2}&-\frac{1}{6}&-\frac{2}{3} \\ 0&1&0|0&\frac{1}{3}&\frac{1}{3}\\ 0&0&1| -\frac{1}{2}&\frac{5}{6}&\frac{1}{3} \end{pmatrix}.$

So,

$A^{-1}=\begin{pmatrix} \frac{1}{2}&-\frac{1}{6}&-\frac{2}{3} \\ 0&\frac{1}{3}&\frac{1}{3}\\ -\frac{1}{2}&\frac{5}{6}&\frac{1}{3} \end{pmatrix}.$