Find the Inverse [[2,0,-3],[0,-1,0],[3,-2,-5]]

$[\begin{array}{c} 2 & 0 & - 3 \\ 0 & - 1 & 0 \\ 3 & - 2 & - 5 \end{array}]$

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.

$[\begin{array}{c} 2 & 0 & - 3 & 1 & 0 & 0 \\ 0 & - 1 & 0 & 0 & 1 & 0 \\ 3 & - 2 & - 5 & 0 & 0 & 1 \end{array}]$

Step 2

Find the reduced row echelon form.

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

$[\begin{array}{c} \frac{2}{2} & \frac{0}{2} & \frac{- 3}{2} & \frac{1}{2} & \frac{0}{2} & \frac{0}{2} \\ 0 & - 1 & 0 & 0 & 1 & 0 \\ 3 & - 2 & - 5 & 0 & 0 & 1 \end{array}]$

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - \frac{3}{2} & \frac{1}{2} & 0 & 0 \\ 0 & - 1 & 0 & 0 & 1 & 0 \\ 3 & - 2 & - 5 & 0 & 0 & 1 \end{array}]$

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

$[\begin{array}{c} 1 & 0 & - \frac{3}{2} & \frac{1}{2} & 0 & 0 \\ 0 & - 1 & 0 & 0 & 1 & 0 \\ 3 - 3 \cdot 1 & - 2 - 3 \cdot 0 & - 5 - 3 (- \frac{3}{2}) & 0 - 3 (\frac{1}{2}) & 0 - 3 \cdot 0 & 1 - 3 \cdot 0 \end{array}]$

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 0 & - \frac{3}{2} & \frac{1}{2} & 0 & 0 \\ 0 & - 1 & 0 & 0 & 1 & 0 \\ 0 & - 2 & - \frac{1}{2} & - \frac{3}{2} & 0 & 1 \end{array}]$

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

$[\begin{array}{c} 1 & 0 & - \frac{3}{2} & \frac{1}{2} & 0 & 0 \\ - 0 & - - 1 & - 0 & - 0 & - 1 \cdot 1 & - 0 \\ 0 & - 2 & - \frac{1}{2} & - \frac{3}{2} & 0 & 1 \end{array}]$

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & - \frac{3}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 & 0 & - 1 & 0 \\ 0 & - 2 & - \frac{1}{2} & - \frac{3}{2} & 0 & 1 \end{array}]$

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

$[\begin{array}{c} 1 & 0 & - \frac{3}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 & 0 & - 1 & 0 \\ 0 + 2 \cdot 0 & - 2 + 2 \cdot 1 & - \frac{1}{2} + 2 \cdot 0 & - \frac{3}{2} + 2 \cdot 0 & 0 + 2 \cdot - 1 & 1 + 2 \cdot 0 \end{array}]$

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 0 & - \frac{3}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 & 0 & - 1 & 0 \\ 0 & 0 & - \frac{1}{2} & - \frac{3}{2} & - 2 & 1 \end{array}]$

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

$[\begin{array}{c} 1 & 0 & - \frac{3}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 & 0 & - 1 & 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 (- \frac{1}{2}) & - 2 (- \frac{3}{2}) & - 2 \cdot - 2 & - 2 \cdot 1 \end{array}]$

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 0 & - \frac{3}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 & 0 & - 1 & 0 \\ 0 & 0 & 1 & 3 & 4 & - 2 \end{array}]$

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

$[\begin{array}{c} 1 + \frac{3}{2} \cdot 0 & 0 + \frac{3}{2} \cdot 0 & - \frac{3}{2} + \frac{3}{2} \cdot 1 & \frac{1}{2} + \frac{3}{2} \cdot 3 & 0 + \frac{3}{2} \cdot 4 & 0 + \frac{3}{2} \cdot - 2 \\ 0 & 1 & 0 & 0 & - 1 & 0 \\ 0 & 0 & 1 & 3 & 4 & - 2 \end{array}]$

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 5 & 6 & - 3 \\ 0 & 1 & 0 & 0 & - 1 & 0 \\ 0 & 0 & 1 & 3 & 4 & - 2 \end{array}]$

Step 3

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

$[\begin{array}{c} 5 & 6 & - 3 \\ 0 & - 1 & 0 \\ 3 & 4 & - 2 \end{array}]$

Find the Inverse [[2,0,-3],[0,-1,0],[3,-2,-5]]

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