Find the Inverse [[1,-1,1],[0,2,-1],[-6,-7,0]]

$[\begin{array}{c} 1 & - 1 & 1 \\ 0 & 2 & - 1 \\ - 6 & - 7 & 0 \end{array}]$

Step 1

Set up a $3 \times 6$ matrix where the left half is the original matrix and the right half is its identity matrix.

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 \\ 0 & 2 & - 1 & 0 & 1 & 0 \\ - 6 & - 7 & 0 & 0 & 0 & 1 \end{array}]$

Step 2

Find the reduced row echelon form.

Perform the row operation $R_{3} = R_{3} + 6 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 \\ 0 & 2 & - 1 & 0 & 1 & 0 \\ - 6 + 6 \cdot 1 & - 7 + 6 \cdot - 1 & 0 + 6 \cdot 1 & 0 + 6 \cdot 1 & 0 + 6 \cdot 0 & 1 + 6 \cdot 0 \end{array}]$

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 \\ 0 & 2 & - 1 & 0 & 1 & 0 \\ 0 & - 13 & 6 & 6 & 0 & 1 \end{array}]$

Multiply each element of $R_{2}$ by $\frac{1}{2}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 \\ \frac{0}{2} & \frac{2}{2} & \frac{- 1}{2} & \frac{0}{2} & \frac{1}{2} & \frac{0}{2} \\ 0 & - 13 & 6 & 6 & 0 & 1 \end{array}]$

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & - \frac{1}{2} & 0 & \frac{1}{2} & 0 \\ 0 & - 13 & 6 & 6 & 0 & 1 \end{array}]$

Perform the row operation $R_{3} = R_{3} + 13 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & - \frac{1}{2} & 0 & \frac{1}{2} & 0 \\ 0 + 13 \cdot 0 & - 13 + 13 \cdot 1 & 6 + 13 (- \frac{1}{2}) & 6 + 13 \cdot 0 & 0 + 13 (\frac{1}{2}) & 1 + 13 \cdot 0 \end{array}]$

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & - \frac{1}{2} & 0 & \frac{1}{2} & 0 \\ 0 & 0 & - \frac{1}{2} & 6 & \frac{13}{2} & 1 \end{array}]$

Multiply each element of $R_{3}$ by $- 2$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & - \frac{1}{2} & 0 & \frac{1}{2} & 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 (- \frac{1}{2}) & - 2 \cdot 6 & - 2 (\frac{13}{2}) & - 2 \cdot 1 \end{array}]$

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & - \frac{1}{2} & 0 & \frac{1}{2} & 0 \\ 0 & 0 & 1 & - 12 & - 13 & - 2 \end{array}]$

Perform the row operation $R_{2} = R_{2} + \frac{1}{2} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 \\ 0 + \frac{1}{2} \cdot 0 & 1 + \frac{1}{2} \cdot 0 & - \frac{1}{2} + \frac{1}{2} \cdot 1 & 0 + \frac{1}{2} \cdot - 12 & \frac{1}{2} + \frac{1}{2} \cdot - 13 & 0 + \frac{1}{2} \cdot - 2 \\ 0 & 0 & 1 & - 12 & - 13 & - 2 \end{array}]$

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & - 6 & - 6 & - 1 \\ 0 & 0 & 1 & - 12 & - 13 & - 2 \end{array}]$

Perform the row operation $R_{1} = R_{1} - R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - 0 & - 1 - 0 & 1 - 1 & 1 + 12 & 0 + 13 & 0 + 2 \\ 0 & 1 & 0 & - 6 & - 6 & - 1 \\ 0 & 0 & 1 & - 12 & - 13 & - 2 \end{array}]$

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 1 & 0 & 13 & 13 & 2 \\ 0 & 1 & 0 & - 6 & - 6 & - 1 \\ 0 & 0 & 1 & - 12 & - 13 & - 2 \end{array}]$

Perform the row operation $R_{1} = R_{1} + R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + 0 & - 1 + 1 \cdot 1 & 0 + 0 & 13 - 6 & 13 - 6 & 2 - 1 \\ 0 & 1 & 0 & - 6 & - 6 & - 1 \\ 0 & 0 & 1 & - 12 & - 13 & - 2 \end{array}]$

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 7 & 7 & 1 \\ 0 & 1 & 0 & - 6 & - 6 & - 1 \\ 0 & 0 & 1 & - 12 & - 13 & - 2 \end{array}]$

Step 3

The right half of the reduced row echelon form is the inverse.

$[\begin{array}{c} 7 & 7 & 1 \\ - 6 & - 6 & - 1 \\ - 12 & - 13 & - 2 \end{array}]$

Find the Inverse [[1,-1,1],[0,2,-1],[-6,-7,0]]

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