Find the Inverse [[1,-1,-1],[0,-2,-1],[-7,-8,0]]

$[\begin{array}{c} 1 & - 1 & - 1 \\ 0 & - 2 & - 1 \\ - 7 & - 8 & 0 \end{array}]$

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.

$[\begin{array}{c} 1 & - 1 & - 1 & 1 & 0 & 0 \\ 0 & - 2 & - 1 & 0 & 1 & 0 \\ - 7 & - 8 & 0 & 0 & 0 & 1 \end{array}]$

Step 2

Find the reduced row echelon form.

Perform the row operation $R_{3} = R_{3} + 7 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 1 & 0 & 0 \\ 0 & - 2 & - 1 & 0 & 1 & 0 \\ - 7 + 7 \cdot 1 & - 8 + 7 \cdot - 1 & 0 + 7 \cdot - 1 & 0 + 7 \cdot 1 & 0 + 7 \cdot 0 & 1 + 7 \cdot 0 \end{array}]$

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 1 & 0 & 0 \\ 0 & - 2 & - 1 & 0 & 1 & 0 \\ 0 & - 15 & - 7 & 7 & 0 & 1 \end{array}]$

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

$[\begin{array}{c} 1 & - 1 & - 1 & 1 & 0 & 0 \\ - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot - 2 & - \frac{1}{2} \cdot - 1 & - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot 1 & - \frac{1}{2} \cdot 0 \\ 0 & - 15 & - 7 & 7 & 0 & 1 \end{array}]$

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 1 & 0 & 0 \\ 0 & 1 & \frac{1}{2} & 0 & - \frac{1}{2} & 0 \\ 0 & - 15 & - 7 & 7 & 0 & 1 \end{array}]$

Perform the row operation $R_{3} = R_{3} + 15 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 1 & 0 & 0 \\ 0 & 1 & \frac{1}{2} & 0 & - \frac{1}{2} & 0 \\ 0 + 15 \cdot 0 & - 15 + 15 \cdot 1 & - 7 + 15 (\frac{1}{2}) & 7 + 15 \cdot 0 & 0 + 15 (- \frac{1}{2}) & 1 + 15 \cdot 0 \end{array}]$

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 1 & 0 & 0 \\ 0 & 1 & \frac{1}{2} & 0 & - \frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{2} & 7 & - \frac{15}{2} & 1 \end{array}]$

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

$[\begin{array}{c} 1 & - 1 & - 1 & 1 & 0 & 0 \\ 0 & 1 & \frac{1}{2} & 0 & - \frac{1}{2} & 0 \\ 2 \cdot 0 & 2 \cdot 0 & 2 (\frac{1}{2}) & 2 \cdot 7 & 2 (- \frac{15}{2}) & 2 \cdot 1 \end{array}]$

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 1 & 0 & 0 \\ 0 & 1 & \frac{1}{2} & 0 & - \frac{1}{2} & 0 \\ 0 & 0 & 1 & 14 & - 15 & 2 \end{array}]$

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

$[\begin{array}{c} 1 & - 1 & - 1 & 1 & 0 & 0 \\ 0 - \frac{1}{2} \cdot 0 & 1 - \frac{1}{2} \cdot 0 & \frac{1}{2} - \frac{1}{2} \cdot 1 & 0 - \frac{1}{2} \cdot 14 & - \frac{1}{2} - \frac{1}{2} \cdot - 15 & 0 - \frac{1}{2} \cdot 2 \\ 0 & 0 & 1 & 14 & - 15 & 2 \end{array}]$

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & - 7 & 7 & - 1 \\ 0 & 0 & 1 & 14 & - 15 & 2 \end{array}]$

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

$[\begin{array}{c} 1 + 0 & - 1 + 0 & - 1 + 1 \cdot 1 & 1 + 1 \cdot 14 & 0 - 15 & 0 + 1 \cdot 2 \\ 0 & 1 & 0 & - 7 & 7 & - 1 \\ 0 & 0 & 1 & 14 & - 15 & 2 \end{array}]$

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 1 & 0 & 15 & - 15 & 2 \\ 0 & 1 & 0 & - 7 & 7 & - 1 \\ 0 & 0 & 1 & 14 & - 15 & 2 \end{array}]$

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

$[\begin{array}{c} 1 + 0 & - 1 + 1 \cdot 1 & 0 + 0 & 15 - 7 & - 15 + 1 \cdot 7 & 2 - 1 \\ 0 & 1 & 0 & - 7 & 7 & - 1 \\ 0 & 0 & 1 & 14 & - 15 & 2 \end{array}]$

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 8 & - 8 & 1 \\ 0 & 1 & 0 & - 7 & 7 & - 1 \\ 0 & 0 & 1 & 14 & - 15 & 2 \end{array}]$

Step 3

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

$[\begin{array}{c} 8 & - 8 & 1 \\ - 7 & 7 & - 1 \\ 14 & - 15 & 2 \end{array}]$

Find the Inverse [[1,-1,-1],[0,-2,-1],[-7,-8,0]]

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