Find the Inverse [[7,-9,4],[0,3,-1],[2,-2,1]]

$[\begin{array}{c} 7 & - 9 & 4 \\ 0 & 3 & - 1 \\ 2 & - 2 & 1 \end{array}]$

Step 1

Set up a matrix that is broken into two pieces of equal size. On the left side, fill in the elements of the original matrix. On the right side, fill in elements of the identity matrix. In order to find the inverse matrix, use row operations to convert the left side into the identity matrix. After this is complete, the inverse of the original matrix will be on the right side of the double matrix.

$[\begin{array}{c} 7 & - 9 & 4 & 1 & 0 & 0 \\ 0 & 3 & - 1 & 0 & 1 & 0 \\ 2 & - 2 & 1 & 0 & 0 & 1 \end{array}]$

Step 2

Exchange row $3$ and row $2$ to organize the zeros into position.

$[\begin{array}{c} 7 & - 9 & 4 & 1 & 0 & 0 \\ 2 & - 2 & 1 & 0 & 0 & 1 \\ 0 & 3 & - 1 & 0 & 1 & 0 \end{array}]$

$R_{3} \leftrightarrow R$

Step 3

Perform the row operation $R_{1} = \frac{1}{7} R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $1$ .

Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = \frac{1}{7} R_{1}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} \frac{1}{7} R_{1} & \frac{1}{7} R_{1} & \frac{1}{7} R_{1} & \frac{1}{7} R_{1} & \frac{1}{7} R_{1} & \frac{1}{7} R_{1} \\ 2 & - 2 & 1 & 0 & 0 & 1 \\ 0 & 3 & - 1 & 0 & 1 & 0 \end{array}]$

$R_{1} = \frac{1}{7} R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = \frac{1}{7} R_{1}$ .

$[\begin{array}{c} (\frac{1}{7}) \cdot (7) & (\frac{1}{7}) \cdot (- 9) & (\frac{1}{7}) \cdot (4) & (\frac{1}{7}) \cdot (1) & (\frac{1}{7}) \cdot (0) & (\frac{1}{7}) \cdot (0) \\ 2 & - 2 & 1 & 0 & 0 & 1 \\ 0 & 3 & - 1 & 0 & 1 & 0 \end{array}]$

$R_{1} = \frac{1}{7} R_{1}$

Simplify $R_{1}$ (row $1$ ).

$[\begin{array}{c} 1 & - \frac{9}{7} & \frac{4}{7} & \frac{1}{7} & 0 & 0 \\ 2 & - 2 & 1 & 0 & 0 & 1 \\ 0 & 3 & - 1 & 0 & 1 & 0 \end{array}]$

Step 4

Perform the row operation $R_{2} = - 2 \cdot R_{1} + R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $0$ .

Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = - 2 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & - \frac{9}{7} & \frac{4}{7} & \frac{1}{7} & 0 & 0 \\ - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} \\ 0 & 3 & - 1 & 0 & 1 & 0 \end{array}]$

$R_{2} = - 2 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - 2 \cdot R_{1} + R_{2}$ .

$[\begin{array}{c} 1 & - \frac{9}{7} & \frac{4}{7} & \frac{1}{7} & 0 & 0 \\ (- 2) \cdot (1) + 2 & (- 2) \cdot (- \frac{9}{7}) - 2 & (- 2) \cdot (\frac{4}{7}) + 1 & (- 2) \cdot (\frac{1}{7}) + 0 & (- 2) \cdot (0) + 0 & (- 2) \cdot (0) + 1 \\ 0 & 3 & - 1 & 0 & 1 & 0 \end{array}]$

$R_{2} = - 2 \cdot R_{1} + R_{2}$

Simplify $R_{2}$ (row $2$ ).

$[\begin{array}{c} 1 & - \frac{9}{7} & \frac{4}{7} & \frac{1}{7} & 0 & 0 \\ 0 & \frac{4}{7} & - \frac{1}{7} & - \frac{2}{7} & 0 & 1 \\ 0 & 3 & - 1 & 0 & 1 & 0 \end{array}]$

Step 5

Perform the row operation $R_{2} = \frac{7}{4} R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $1$ .

Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = \frac{7}{4} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & - \frac{9}{7} & \frac{4}{7} & \frac{1}{7} & 0 & 0 \\ \frac{7}{4} R_{2} & \frac{7}{4} R_{2} & \frac{7}{4} R_{2} & \frac{7}{4} R_{2} & \frac{7}{4} R_{2} & \frac{7}{4} R_{2} \\ 0 & 3 & - 1 & 0 & 1 & 0 \end{array}]$

$R_{2} = \frac{7}{4} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{7}{4} R_{2}$ .

$[\begin{array}{c} 1 & - \frac{9}{7} & \frac{4}{7} & \frac{1}{7} & 0 & 0 \\ (\frac{7}{4}) \cdot (0) & (\frac{7}{4}) \cdot (\frac{4}{7}) & (\frac{7}{4}) \cdot (- \frac{1}{7}) & (\frac{7}{4}) \cdot (- \frac{2}{7}) & (\frac{7}{4}) \cdot (0) & (\frac{7}{4}) \cdot (1) \\ 0 & 3 & - 1 & 0 & 1 & 0 \end{array}]$

$R_{2} = \frac{7}{4} R_{2}$

Simplify $R_{2}$ (row $2$ ).

$[\begin{array}{c} 1 & - \frac{9}{7} & \frac{4}{7} & \frac{1}{7} & 0 & 0 \\ 0 & 1 & - \frac{1}{4} & - \frac{1}{2} & 0 & \frac{7}{4} \\ 0 & 3 & - 1 & 0 & 1 & 0 \end{array}]$

Step 6

Perform the row operation $R_{1} = \frac{9}{7} R_{2} + R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $0$ .

Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = \frac{9}{7} R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} \frac{9}{7} R_{2} + R_{1} & \frac{9}{7} R_{2} + R_{1} & \frac{9}{7} R_{2} + R_{1} & \frac{9}{7} R_{2} + R_{1} & \frac{9}{7} R_{2} + R_{1} & \frac{9}{7} R_{2} + R_{1} \\ 0 & 1 & - \frac{1}{4} & - \frac{1}{2} & 0 & \frac{7}{4} \\ 0 & 3 & - 1 & 0 & 1 & 0 \end{array}]$

$R_{1} = \frac{9}{7} R_{2} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = \frac{9}{7} R_{2} + R_{1}$ .

$[\begin{array}{c} (\frac{9}{7}) \cdot (0) + 1 & (\frac{9}{7}) \cdot (1) - \frac{9}{7} & (\frac{9}{7}) \cdot (- \frac{1}{4}) + \frac{4}{7} & (\frac{9}{7}) \cdot (- \frac{1}{2}) + \frac{1}{7} & (\frac{9}{7}) \cdot (0) + 0 & (\frac{9}{7}) \cdot (\frac{7}{4}) + 0 \\ 0 & 1 & - \frac{1}{4} & - \frac{1}{2} & 0 & \frac{7}{4} \\ 0 & 3 & - 1 & 0 & 1 & 0 \end{array}]$

$R_{1} = \frac{9}{7} R_{2} + R_{1}$

Simplify $R_{1}$ (row $1$ ).

$[\begin{array}{c} 1 & 0 & \frac{1}{4} & - \frac{1}{2} & 0 & \frac{9}{4} \\ 0 & 1 & - \frac{1}{4} & - \frac{1}{2} & 0 & \frac{7}{4} \\ 0 & 3 & - 1 & 0 & 1 & 0 \end{array}]$

Step 7

Perform the row operation $R_{3} = - 3 \cdot R_{2} + R_{3}$ on $R_{3}$ (row $3$ ) in order to convert some elements in the row to $0$ .

Replace $R_{3}$ (row $3$ ) with the row operation $R_{3} = - 3 \cdot R_{2} + R_{3}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 0 & \frac{1}{4} & - \frac{1}{2} & 0 & \frac{9}{4} \\ 0 & 1 & - \frac{1}{4} & - \frac{1}{2} & 0 & \frac{7}{4} \\ - 3 \cdot R_{2} + R_{3} & - 3 \cdot R_{2} + R_{3} & - 3 \cdot R_{2} + R_{3} & - 3 \cdot R_{2} + R_{3} & - 3 \cdot R_{2} + R_{3} & - 3 \cdot R_{2} + R_{3} \end{array}]$

$R_{3} = - 3 \cdot R_{2} + R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = - 3 \cdot R_{2} + R_{3}$ .

$[\begin{array}{c} 1 & 0 & \frac{1}{4} & - \frac{1}{2} & 0 & \frac{9}{4} \\ 0 & 1 & - \frac{1}{4} & - \frac{1}{2} & 0 & \frac{7}{4} \\ (- 3) \cdot (0) + 0 & (- 3) \cdot (1) + 3 & (- 3) \cdot (- \frac{1}{4}) - 1 & (- 3) \cdot (- \frac{1}{2}) + 0 & (- 3) \cdot (0) + 1 & (- 3) \cdot (\frac{7}{4}) + 0 \end{array}]$

$R_{3} = - 3 \cdot R_{2} + R_{3}$

Simplify $R_{3}$ (row $3$ ).

$[\begin{array}{c} 1 & 0 & \frac{1}{4} & - \frac{1}{2} & 0 & \frac{9}{4} \\ 0 & 1 & - \frac{1}{4} & - \frac{1}{2} & 0 & \frac{7}{4} \\ 0 & 0 & - \frac{1}{4} & \frac{3}{2} & 1 & - \frac{21}{4} \end{array}]$

Step 8

Perform the row operation $R_{3} = - 4 \cdot R_{3}$ on $R_{3}$ (row $3$ ) in order to convert some elements in the row to $1$ .

Replace $R_{3}$ (row $3$ ) with the row operation $R_{3} = - 4 \cdot R_{3}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & 0 & \frac{1}{4} & - \frac{1}{2} & 0 & \frac{9}{4} \\ 0 & 1 & - \frac{1}{4} & - \frac{1}{2} & 0 & \frac{7}{4} \\ - 4 \cdot R_{3} & - 4 \cdot R_{3} & - 4 \cdot R_{3} & - 4 \cdot R_{3} & - 4 \cdot R_{3} & - 4 \cdot R_{3} \end{array}]$

$R_{3} = - 4 \cdot R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = - 4 \cdot R_{3}$ .

$[\begin{array}{c} 1 & 0 & \frac{1}{4} & - \frac{1}{2} & 0 & \frac{9}{4} \\ 0 & 1 & - \frac{1}{4} & - \frac{1}{2} & 0 & \frac{7}{4} \\ (- 4) \cdot (0) & (- 4) \cdot (0) & (- 4) \cdot (- \frac{1}{4}) & (- 4) \cdot (\frac{3}{2}) & (- 4) \cdot (1) & (- 4) \cdot (- \frac{21}{4}) \end{array}]$

$R_{3} = - 4 \cdot R_{3}$

Simplify $R_{3}$ (row $3$ ).

$[\begin{array}{c} 1 & 0 & \frac{1}{4} & - \frac{1}{2} & 0 & \frac{9}{4} \\ 0 & 1 & - \frac{1}{4} & - \frac{1}{2} & 0 & \frac{7}{4} \\ 0 & 0 & 1 & - 6 & - 4 & 21 \end{array}]$

Step 9

Perform the row operation $R_{1} = - \frac{1}{4} R_{3} + R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $0$ .

Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = - \frac{1}{4} R_{3} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} - \frac{1}{4} R_{3} + R_{1} & - \frac{1}{4} R_{3} + R_{1} & - \frac{1}{4} R_{3} + R_{1} & - \frac{1}{4} R_{3} + R_{1} & - \frac{1}{4} R_{3} + R_{1} & - \frac{1}{4} R_{3} + R_{1} \\ 0 & 1 & - \frac{1}{4} & - \frac{1}{2} & 0 & \frac{7}{4} \\ 0 & 0 & 1 & - 6 & - 4 & 21 \end{array}]$

$R_{1} = - \frac{1}{4} R_{3} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = - \frac{1}{4} R_{3} + R_{1}$ .

$[\begin{array}{c} (- \frac{1}{4}) \cdot (0) + 1 & (- \frac{1}{4}) \cdot (0) + 0 & (- \frac{1}{4}) \cdot (1) + \frac{1}{4} & (- \frac{1}{4}) \cdot (- 6) - \frac{1}{2} & (- \frac{1}{4}) \cdot (- 4) + 0 & (- \frac{1}{4}) \cdot (21) + \frac{9}{4} \\ 0 & 1 & - \frac{1}{4} & - \frac{1}{2} & 0 & \frac{7}{4} \\ 0 & 0 & 1 & - 6 & - 4 & 21 \end{array}]$

$R_{1} = - \frac{1}{4} R_{3} + R_{1}$

Simplify $R_{1}$ (row $1$ ).

$[\begin{array}{c} 1 & 0 & 0 & 1 & 1 & - 3 \\ 0 & 1 & - \frac{1}{4} & - \frac{1}{2} & 0 & \frac{7}{4} \\ 0 & 0 & 1 & - 6 & - 4 & 21 \end{array}]$

Step 10

Perform the row operation $R_{2} = \frac{1}{4} R_{3} + R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $0$ .

Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = \frac{1}{4} R_{3} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 1 & 1 & - 3 \\ \frac{1}{4} R_{3} + R_{2} & \frac{1}{4} R_{3} + R_{2} & \frac{1}{4} R_{3} + R_{2} & \frac{1}{4} R_{3} + R_{2} & \frac{1}{4} R_{3} + R_{2} & \frac{1}{4} R_{3} + R_{2} \\ 0 & 0 & 1 & - 6 & - 4 & 21 \end{array}]$

$R_{2} = \frac{1}{4} R_{3} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{1}{4} R_{3} + R_{2}$ .

$[\begin{array}{c} 1 & 0 & 0 & 1 & 1 & - 3 \\ (\frac{1}{4}) \cdot (0) + 0 & (\frac{1}{4}) \cdot (0) + 1 & (\frac{1}{4}) \cdot (1) - \frac{1}{4} & (\frac{1}{4}) \cdot (- 6) - \frac{1}{2} & (\frac{1}{4}) \cdot (- 4) + 0 & (\frac{1}{4}) \cdot (21) + \frac{7}{4} \\ 0 & 0 & 1 & - 6 & - 4 & 21 \end{array}]$

$R_{2} = \frac{1}{4} R_{3} + R_{2}$

Simplify $R_{2}$ (row $2$ ).

$[\begin{array}{c} 1 & 0 & 0 & 1 & 1 & - 3 \\ 0 & 1 & 0 & - 2 & - 1 & 7 \\ 0 & 0 & 1 & - 6 & - 4 & 21 \end{array}]$

Step 11

The right side of the double matrix is now the inverse of the original matrix.

$[\begin{array}{c} 1 & 1 & - 3 \\ - 2 & - 1 & 7 \\ - 6 & - 4 & 21 \end{array}]$

Find the Inverse [[7,-9,4],[0,3,-1],[2,-2,1]]

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