Question

Find the inverse of the matrix
1 −1 0
2 −1 1
1 1 −1

using row reduction.

Answer 1

Answer on Question #42865 – Math - Linear Algebra

Find the inverse of the matrix

1 - 1 0

2 - 1 1

1 1 - 1

using row reduction.

Solution

Form the augmented matrix $(A|E)$: $\begin{pmatrix} 1 & -1 & 0 & 1 & 0 & 0 \\ 2 & -1 & 1 & 0 & 1 & 0 \\ 1 & 1 & -1 & 0 & 0 & 1 \end{pmatrix}$.

$$R_2 - 2R_1 \rightarrow R_2: \begin{pmatrix} 1 & -1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & -2 & 1 & 0 \\ 1 & 1 & -1 & 0 & 0 & 1 \end{pmatrix}$$ $$R_3 - R_1 \rightarrow R_3: \begin{pmatrix} 1 & -1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & -2 & 1 & 0 \\ 0 & 2 & -1 & -1 & 0 & 1 \end{pmatrix}$$ $$R_3 - 2R_2 \rightarrow R_3: \begin{pmatrix} 1 & -1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & -2 & 1 & 0 \\ 0 & 0 & -3 & 3 & -2 & 1 \end{pmatrix}$$ $$\frac{R_3}{3} \rightarrow R_3: \begin{pmatrix} 1 & -1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & -2 & 1 & 0 \\ 0 & 0 & -1 & 1 & -\frac{2}{3} & \frac{1}{3} \end{pmatrix}$$ $$R_2 + R_3 \rightarrow R_2: \begin{pmatrix} 1 & -1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & -1 & \frac{1}{3} & \frac{1}{3} \\ 0 & 0 & -1 & 1 & -\frac{2}{3} & \frac{1}{3} \end{pmatrix}$$ $$R_2 + R_1 \rightarrow R_1: \begin{pmatrix} 1 & 0 & 0 & 0 & \frac{1}{3} & \frac{1}{3} \\ 0 & 1 & 0 & -1 & \frac{1}{3} & \frac{1}{3} \\ 0 & 0 & -1 & 1 & -2 & 1 \\ 0 & 0 & 1 & -1 & \frac{2}{3} & \frac{1}{3} \end{pmatrix}$$ $$-R_3 \rightarrow R_3: \begin{pmatrix} 1 & 0 & 0 & 0 & \frac{1}{3} & \frac{1}{3} \\ 0 & 1 & 0 & -1 & \frac{1}{3} & \frac{1}{3} \\ 0 & 0 & 1 & -1 & \frac{2}{3} & -\frac{1}{3} \end{pmatrix}$$

Answer: $\begin{pmatrix} 0 & \frac{1}{3} & \frac{1}{3} \\ -1 & \frac{1}{3} & \frac{1}{3} \\ -1 & \frac{2}{3} & -\frac{1}{3} \end{pmatrix}$.

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