Find the Inverse [[28,21,19],[17,24,15],[22,29,17]]

$[\begin{array}{c} 28 & 21 & 19 \\ 17 & 24 & 15 \\ 22 & 29 & 17 \end{array}]$

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.

$[\begin{array}{c} 28 & 21 & 19 & 1 & 0 & 0 \\ 17 & 24 & 15 & 0 & 1 & 0 \\ 22 & 29 & 17 & 0 & 0 & 1 \end{array}]$

Step 2

Find the reduced row echelon form.

Multiply each element of $R_{1}$ by $\frac{1}{28}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{28}{28} & \frac{21}{28} & \frac{19}{28} & \frac{1}{28} & \frac{0}{28} & \frac{0}{28} \\ 17 & 24 & 15 & 0 & 1 & 0 \\ 22 & 29 & 17 & 0 & 0 & 1 \end{array}]$

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{19}{28} & \frac{1}{28} & 0 & 0 \\ 17 & 24 & 15 & 0 & 1 & 0 \\ 22 & 29 & 17 & 0 & 0 & 1 \end{array}]$

Perform the row operation $R_{2} = R_{2} - 17 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{19}{28} & \frac{1}{28} & 0 & 0 \\ 17 - 17 \cdot 1 & 24 - 17 (\frac{3}{4}) & 15 - 17 (\frac{19}{28}) & 0 - 17 (\frac{1}{28}) & 1 - 17 \cdot 0 & 0 - 17 \cdot 0 \\ 22 & 29 & 17 & 0 & 0 & 1 \end{array}]$

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{19}{28} & \frac{1}{28} & 0 & 0 \\ 0 & \frac{45}{4} & \frac{97}{28} & - \frac{17}{28} & 1 & 0 \\ 22 & 29 & 17 & 0 & 0 & 1 \end{array}]$

Perform the row operation $R_{3} = R_{3} - 22 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{19}{28} & \frac{1}{28} & 0 & 0 \\ 0 & \frac{45}{4} & \frac{97}{28} & - \frac{17}{28} & 1 & 0 \\ 22 - 22 \cdot 1 & 29 - 22 (\frac{3}{4}) & 17 - 22 (\frac{19}{28}) & 0 - 22 (\frac{1}{28}) & 0 - 22 \cdot 0 & 1 - 22 \cdot 0 \end{array}]$

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{19}{28} & \frac{1}{28} & 0 & 0 \\ 0 & \frac{45}{4} & \frac{97}{28} & - \frac{17}{28} & 1 & 0 \\ 0 & \frac{25}{2} & \frac{29}{14} & - \frac{11}{14} & 0 & 1 \end{array}]$

Multiply each element of $R_{2}$ by $\frac{4}{45}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{19}{28} & \frac{1}{28} & 0 & 0 \\ \frac{4}{45} \cdot 0 & \frac{4}{45} \cdot \frac{45}{4} & \frac{4}{45} \cdot \frac{97}{28} & \frac{4}{45} (- \frac{17}{28}) & \frac{4}{45} \cdot 1 & \frac{4}{45} \cdot 0 \\ 0 & \frac{25}{2} & \frac{29}{14} & - \frac{11}{14} & 0 & 1 \end{array}]$

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{19}{28} & \frac{1}{28} & 0 & 0 \\ 0 & 1 & \frac{97}{315} & - \frac{17}{315} & \frac{4}{45} & 0 \\ 0 & \frac{25}{2} & \frac{29}{14} & - \frac{11}{14} & 0 & 1 \end{array}]$

Perform the row operation $R_{3} = R_{3} - \frac{25}{2} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{19}{28} & \frac{1}{28} & 0 & 0 \\ 0 & 1 & \frac{97}{315} & - \frac{17}{315} & \frac{4}{45} & 0 \\ 0 - \frac{25}{2} \cdot 0 & \frac{25}{2} - \frac{25}{2} \cdot 1 & \frac{29}{14} - \frac{25}{2} \cdot \frac{97}{315} & - \frac{11}{14} - \frac{25}{2} (- \frac{17}{315}) & 0 - \frac{25}{2} \cdot \frac{4}{45} & 1 - \frac{25}{2} \cdot 0 \end{array}]$

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{19}{28} & \frac{1}{28} & 0 & 0 \\ 0 & 1 & \frac{97}{315} & - \frac{17}{315} & \frac{4}{45} & 0 \\ 0 & 0 & - \frac{16}{9} & - \frac{1}{9} & - \frac{10}{9} & 1 \end{array}]$

Multiply each element of $R_{3}$ by $- \frac{9}{16}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{19}{28} & \frac{1}{28} & 0 & 0 \\ 0 & 1 & \frac{97}{315} & - \frac{17}{315} & \frac{4}{45} & 0 \\ - \frac{9}{16} \cdot 0 & - \frac{9}{16} \cdot 0 & - \frac{9}{16} (- \frac{16}{9}) & - \frac{9}{16} (- \frac{1}{9}) & - \frac{9}{16} (- \frac{10}{9}) & - \frac{9}{16} \cdot 1 \end{array}]$

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{19}{28} & \frac{1}{28} & 0 & 0 \\ 0 & 1 & \frac{97}{315} & - \frac{17}{315} & \frac{4}{45} & 0 \\ 0 & 0 & 1 & \frac{1}{16} & \frac{5}{8} & - \frac{9}{16} \end{array}]$

Perform the row operation $R_{2} = R_{2} - \frac{97}{315} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{19}{28} & \frac{1}{28} & 0 & 0 \\ 0 - \frac{97}{315} \cdot 0 & 1 - \frac{97}{315} \cdot 0 & \frac{97}{315} - \frac{97}{315} \cdot 1 & - \frac{17}{315} - \frac{97}{315} \cdot \frac{1}{16} & \frac{4}{45} - \frac{97}{315} \cdot \frac{5}{8} & 0 - \frac{97}{315} (- \frac{9}{16}) \\ 0 & 0 & 1 & \frac{1}{16} & \frac{5}{8} & - \frac{9}{16} \end{array}]$

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{19}{28} & \frac{1}{28} & 0 & 0 \\ 0 & 1 & 0 & - \frac{41}{560} & - \frac{29}{280} & \frac{97}{560} \\ 0 & 0 & 1 & \frac{1}{16} & \frac{5}{8} & - \frac{9}{16} \end{array}]$

Perform the row operation $R_{1} = R_{1} - \frac{19}{28} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - \frac{19}{28} \cdot 0 & \frac{3}{4} - \frac{19}{28} \cdot 0 & \frac{19}{28} - \frac{19}{28} \cdot 1 & \frac{1}{28} - \frac{19}{28} \cdot \frac{1}{16} & 0 - \frac{19}{28} \cdot \frac{5}{8} & 0 - \frac{19}{28} (- \frac{9}{16}) \\ 0 & 1 & 0 & - \frac{41}{560} & - \frac{29}{280} & \frac{97}{560} \\ 0 & 0 & 1 & \frac{1}{16} & \frac{5}{8} & - \frac{9}{16} \end{array}]$

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3}{4} & 0 & - \frac{3}{448} & - \frac{95}{224} & \frac{171}{448} \\ 0 & 1 & 0 & - \frac{41}{560} & - \frac{29}{280} & \frac{97}{560} \\ 0 & 0 & 1 & \frac{1}{16} & \frac{5}{8} & - \frac{9}{16} \end{array}]$

Perform the row operation $R_{1} = R_{1} - \frac{3}{4} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{3}{4} \cdot 0 & \frac{3}{4} - \frac{3}{4} \cdot 1 & 0 - \frac{3}{4} \cdot 0 & - \frac{3}{448} - \frac{3}{4} (- \frac{41}{560}) & - \frac{95}{224} - \frac{3}{4} (- \frac{29}{280}) & \frac{171}{448} - \frac{3}{4} \cdot \frac{97}{560} \\ 0 & 1 & 0 & - \frac{41}{560} & - \frac{29}{280} & \frac{97}{560} \\ 0 & 0 & 1 & \frac{1}{16} & \frac{5}{8} & - \frac{9}{16} \end{array}]$

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{27}{560} & - \frac{97}{280} & \frac{141}{560} \\ 0 & 1 & 0 & - \frac{41}{560} & - \frac{29}{280} & \frac{97}{560} \\ 0 & 0 & 1 & \frac{1}{16} & \frac{5}{8} & - \frac{9}{16} \end{array}]$

Step 3

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

$[\begin{array}{c} \frac{27}{560} & - \frac{97}{280} & \frac{141}{560} \\ - \frac{41}{560} & - \frac{29}{280} & \frac{97}{560} \\ \frac{1}{16} & \frac{5}{8} & - \frac{9}{16} \end{array}]$

Find the Inverse [[28,21,19],[17,24,15],[22,29,17]]

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