Solve the Matrix Equation [[5,1,4],[2,-3,-5],[7,2,-6]]x=[[5],[2],[5]]

$[\begin{array}{c} 5 & 1 & 4 \\ 2 & - 3 & - 5 \\ 7 & 2 & - 6 \end{array}] x = [\begin{array}{c} 5 \\ 2 \\ 5 \end{array}]$

Find the inverse matrix.

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.

$[\begin{array}{c} 5 & 1 & 4 & 1 & 0 & 0 \\ 2 & - 3 & - 5 & 0 & 1 & 0 \\ 7 & 2 & - 6 & 0 & 0 & 1 \end{array}]$

Perform the row operation $R_{1} = \frac{1}{5} R_{1}$ 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 $R_{1} = \frac{1}{5} R_{1}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} \frac{1}{5} R_{1} & \frac{1}{5} R_{1} & \frac{1}{5} R_{1} & \frac{1}{5} R_{1} & \frac{1}{5} R_{1} & \frac{1}{5} R_{1} \\ 2 & - 3 & - 5 & 0 & 1 & 0 \\ 7 & 2 & - 6 & 0 & 0 & 1 \end{array}]$

$R_{1} = \frac{1}{5} R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = \frac{1}{5} R_{1}$ .

$[\begin{array}{c} (\frac{1}{5}) \cdot (5) & (\frac{1}{5}) \cdot (1) & (\frac{1}{5}) \cdot (4) & (\frac{1}{5}) \cdot (1) & (\frac{1}{5}) \cdot (0) & (\frac{1}{5}) \cdot (0) \\ 2 & - 3 & - 5 & 0 & 1 & 0 \\ 7 & 2 & - 6 & 0 & 0 & 1 \end{array}]$

$R_{1} = \frac{1}{5} R_{1}$

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

$[\begin{array}{c} 1 & \frac{1}{5} & \frac{4}{5} & \frac{1}{5} & 0 & 0 \\ 2 & - 3 & - 5 & 0 & 1 & 0 \\ 7 & 2 & - 6 & 0 & 0 & 1 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & \frac{1}{5} & \frac{4}{5} & \frac{1}{5} & 0 & 0 \\ - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} \\ 7 & 2 & - 6 & 0 & 0 & 1 \end{array}]$

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

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

$[\begin{array}{c} 1 & \frac{1}{5} & \frac{4}{5} & \frac{1}{5} & 0 & 0 \\ (- 2) \cdot (1) + 2 & (- 2) \cdot (\frac{1}{5}) - 3 & (- 2) \cdot (\frac{4}{5}) - 5 & (- 2) \cdot (\frac{1}{5}) + 0 & (- 2) \cdot (0) + 1 & (- 2) \cdot (0) + 0 \\ 7 & 2 & - 6 & 0 & 0 & 1 \end{array}]$

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

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

$[\begin{array}{c} 1 & \frac{1}{5} & \frac{4}{5} & \frac{1}{5} & 0 & 0 \\ 0 & - \frac{17}{5} & - \frac{33}{5} & - \frac{2}{5} & 1 & 0 \\ 7 & 2 & - 6 & 0 & 0 & 1 \end{array}]$

Perform the row operation $R_{3} = - 7 \cdot R_{1} + R_{3}$ 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 $R_{3} = - 7 \cdot R_{1} + R_{3}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & \frac{1}{5} & \frac{4}{5} & \frac{1}{5} & 0 & 0 \\ 0 & - \frac{17}{5} & - \frac{33}{5} & - \frac{2}{5} & 1 & 0 \\ - 7 \cdot R_{1} + R_{3} & - 7 \cdot R_{1} + R_{3} & - 7 \cdot R_{1} + R_{3} & - 7 \cdot R_{1} + R_{3} & - 7 \cdot R_{1} + R_{3} & - 7 \cdot R_{1} + R_{3} \end{array}]$

$R_{3} = - 7 \cdot R_{1} + R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = - 7 \cdot R_{1} + R_{3}$ .

$[\begin{array}{c} 1 & \frac{1}{5} & \frac{4}{5} & \frac{1}{5} & 0 & 0 \\ 0 & - \frac{17}{5} & - \frac{33}{5} & - \frac{2}{5} & 1 & 0 \\ (- 7) \cdot (1) + 7 & (- 7) \cdot (\frac{1}{5}) + 2 & (- 7) \cdot (\frac{4}{5}) - 6 & (- 7) \cdot (\frac{1}{5}) + 0 & (- 7) \cdot (0) + 0 & (- 7) \cdot (0) + 1 \end{array}]$

$R_{3} = - 7 \cdot R_{1} + R_{3}$

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

$[\begin{array}{c} 1 & \frac{1}{5} & \frac{4}{5} & \frac{1}{5} & 0 & 0 \\ 0 & - \frac{17}{5} & - \frac{33}{5} & - \frac{2}{5} & 1 & 0 \\ 0 & \frac{3}{5} & - \frac{58}{5} & - \frac{7}{5} & 0 & 1 \end{array}]$

Perform the row operation $R_{2} = - \frac{5}{17} R_{2}$ 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 $R_{2} = - \frac{5}{17} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & \frac{1}{5} & \frac{4}{5} & \frac{1}{5} & 0 & 0 \\ - \frac{5}{17} R_{2} & - \frac{5}{17} R_{2} & - \frac{5}{17} R_{2} & - \frac{5}{17} R_{2} & - \frac{5}{17} R_{2} & - \frac{5}{17} R_{2} \\ 0 & \frac{3}{5} & - \frac{58}{5} & - \frac{7}{5} & 0 & 1 \end{array}]$

$R_{2} = - \frac{5}{17} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - \frac{5}{17} R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{5} & \frac{4}{5} & \frac{1}{5} & 0 & 0 \\ (- \frac{5}{17}) \cdot (0) & (- \frac{5}{17}) \cdot (- \frac{17}{5}) & (- \frac{5}{17}) \cdot (- \frac{33}{5}) & (- \frac{5}{17}) \cdot (- \frac{2}{5}) & (- \frac{5}{17}) \cdot (1) & (- \frac{5}{17}) \cdot (0) \\ 0 & \frac{3}{5} & - \frac{58}{5} & - \frac{7}{5} & 0 & 1 \end{array}]$

$R_{2} = - \frac{5}{17} R_{2}$

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

$[\begin{array}{c} 1 & \frac{1}{5} & \frac{4}{5} & \frac{1}{5} & 0 & 0 \\ 0 & 1 & \frac{33}{17} & \frac{2}{17} & - \frac{5}{17} & 0 \\ 0 & \frac{3}{5} & - \frac{58}{5} & - \frac{7}{5} & 0 & 1 \end{array}]$

Perform the row operation $R_{1} = - \frac{1}{5} R_{2} + R_{1}$ 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 $R_{1} = - \frac{1}{5} R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} - \frac{1}{5} R_{2} + R_{1} & - \frac{1}{5} R_{2} + R_{1} & - \frac{1}{5} R_{2} + R_{1} & - \frac{1}{5} R_{2} + R_{1} & - \frac{1}{5} R_{2} + R_{1} & - \frac{1}{5} R_{2} + R_{1} \\ 0 & 1 & \frac{33}{17} & \frac{2}{17} & - \frac{5}{17} & 0 \\ 0 & \frac{3}{5} & - \frac{58}{5} & - \frac{7}{5} & 0 & 1 \end{array}]$

$R_{1} = - \frac{1}{5} R_{2} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = - \frac{1}{5} R_{2} + R_{1}$ .

$[\begin{array}{c} (- \frac{1}{5}) \cdot (0) + 1 & (- \frac{1}{5}) \cdot (1) + \frac{1}{5} & (- \frac{1}{5}) \cdot (\frac{33}{17}) + \frac{4}{5} & (- \frac{1}{5}) \cdot (\frac{2}{17}) + \frac{1}{5} & (- \frac{1}{5}) \cdot (- \frac{5}{17}) + 0 & (- \frac{1}{5}) \cdot (0) + 0 \\ 0 & 1 & \frac{33}{17} & \frac{2}{17} & - \frac{5}{17} & 0 \\ 0 & \frac{3}{5} & - \frac{58}{5} & - \frac{7}{5} & 0 & 1 \end{array}]$

$R_{1} = - \frac{1}{5} R_{2} + R_{1}$

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

$[\begin{array}{c} 1 & 0 & \frac{7}{17} & \frac{3}{17} & \frac{1}{17} & 0 \\ 0 & 1 & \frac{33}{17} & \frac{2}{17} & - \frac{5}{17} & 0 \\ 0 & \frac{3}{5} & - \frac{58}{5} & - \frac{7}{5} & 0 & 1 \end{array}]$

Perform the row operation $R_{3} = - \frac{3}{5} R_{2} + R_{3}$ 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 $R_{3} = - \frac{3}{5} R_{2} + R_{3}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 0 & \frac{7}{17} & \frac{3}{17} & \frac{1}{17} & 0 \\ 0 & 1 & \frac{33}{17} & \frac{2}{17} & - \frac{5}{17} & 0 \\ - \frac{3}{5} R_{2} + R_{3} & - \frac{3}{5} R_{2} + R_{3} & - \frac{3}{5} R_{2} + R_{3} & - \frac{3}{5} R_{2} + R_{3} & - \frac{3}{5} R_{2} + R_{3} & - \frac{3}{5} R_{2} + R_{3} \end{array}]$

$R_{3} = - \frac{3}{5} R_{2} + R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = - \frac{3}{5} R_{2} + R_{3}$ .

$[\begin{array}{c} 1 & 0 & \frac{7}{17} & \frac{3}{17} & \frac{1}{17} & 0 \\ 0 & 1 & \frac{33}{17} & \frac{2}{17} & - \frac{5}{17} & 0 \\ (- \frac{3}{5}) \cdot (0) + 0 & (- \frac{3}{5}) \cdot (1) + \frac{3}{5} & (- \frac{3}{5}) \cdot (\frac{33}{17}) - \frac{58}{5} & (- \frac{3}{5}) \cdot (\frac{2}{17}) - \frac{7}{5} & (- \frac{3}{5}) \cdot (- \frac{5}{17}) + 0 & (- \frac{3}{5}) \cdot (0) + 1 \end{array}]$

$R_{3} = - \frac{3}{5} R_{2} + R_{3}$

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

$[\begin{array}{c} 1 & 0 & \frac{7}{17} & \frac{3}{17} & \frac{1}{17} & 0 \\ 0 & 1 & \frac{33}{17} & \frac{2}{17} & - \frac{5}{17} & 0 \\ 0 & 0 & - \frac{217}{17} & - \frac{25}{17} & \frac{3}{17} & 1 \end{array}]$

Perform the row operation $R_{3} = - \frac{17}{217} R_{3}$ 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 $R_{3} = - \frac{17}{217} R_{3}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & 0 & \frac{7}{17} & \frac{3}{17} & \frac{1}{17} & 0 \\ 0 & 1 & \frac{33}{17} & \frac{2}{17} & - \frac{5}{17} & 0 \\ - \frac{17}{217} R_{3} & - \frac{17}{217} R_{3} & - \frac{17}{217} R_{3} & - \frac{17}{217} R_{3} & - \frac{17}{217} R_{3} & - \frac{17}{217} R_{3} \end{array}]$

$R_{3} = - \frac{17}{217} R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = - \frac{17}{217} R_{3}$ .

$[\begin{array}{c} 1 & 0 & \frac{7}{17} & \frac{3}{17} & \frac{1}{17} & 0 \\ 0 & 1 & \frac{33}{17} & \frac{2}{17} & - \frac{5}{17} & 0 \\ (- \frac{17}{217}) \cdot (0) & (- \frac{17}{217}) \cdot (0) & (- \frac{17}{217}) \cdot (- \frac{217}{17}) & (- \frac{17}{217}) \cdot (- \frac{25}{17}) & (- \frac{17}{217}) \cdot (\frac{3}{17}) & (- \frac{17}{217}) \cdot (1) \end{array}]$

$R_{3} = - \frac{17}{217} R_{3}$

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

$[\begin{array}{c} 1 & 0 & \frac{7}{17} & \frac{3}{17} & \frac{1}{17} & 0 \\ 0 & 1 & \frac{33}{17} & \frac{2}{17} & - \frac{5}{17} & 0 \\ 0 & 0 & 1 & \frac{25}{217} & - \frac{3}{217} & - \frac{17}{217} \end{array}]$

Perform the row operation $R_{1} = - \frac{7}{17} R_{3} + R_{1}$ 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 $R_{1} = - \frac{7}{17} R_{3} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} - \frac{7}{17} R_{3} + R_{1} & - \frac{7}{17} R_{3} + R_{1} & - \frac{7}{17} R_{3} + R_{1} & - \frac{7}{17} R_{3} + R_{1} & - \frac{7}{17} R_{3} + R_{1} & - \frac{7}{17} R_{3} + R_{1} \\ 0 & 1 & \frac{33}{17} & \frac{2}{17} & - \frac{5}{17} & 0 \\ 0 & 0 & 1 & \frac{25}{217} & - \frac{3}{217} & - \frac{17}{217} \end{array}]$

$R_{1} = - \frac{7}{17} R_{3} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = - \frac{7}{17} R_{3} + R_{1}$ .

$[\begin{array}{c} (- \frac{7}{17}) \cdot (0) + 1 & (- \frac{7}{17}) \cdot (0) + 0 & (- \frac{7}{17}) \cdot (1) + \frac{7}{17} & (- \frac{7}{17}) \cdot (\frac{25}{217}) + \frac{3}{17} & (- \frac{7}{17}) \cdot (- \frac{3}{217}) + \frac{1}{17} & (- \frac{7}{17}) \cdot (- \frac{17}{217}) + 0 \\ 0 & 1 & \frac{33}{17} & \frac{2}{17} & - \frac{5}{17} & 0 \\ 0 & 0 & 1 & \frac{25}{217} & - \frac{3}{217} & - \frac{17}{217} \end{array}]$

$R_{1} = - \frac{7}{17} R_{3} + R_{1}$

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

$[\begin{array}{c} 1 & 0 & 0 & \frac{4}{31} & \frac{2}{31} & \frac{1}{31} \\ 0 & 1 & \frac{33}{17} & \frac{2}{17} & - \frac{5}{17} & 0 \\ 0 & 0 & 1 & \frac{25}{217} & - \frac{3}{217} & - \frac{17}{217} \end{array}]$

Perform the row operation $R_{2} = - \frac{33}{17} R_{3} + R_{2}$ 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 $R_{2} = - \frac{33}{17} R_{3} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{4}{31} & \frac{2}{31} & \frac{1}{31} \\ - \frac{33}{17} R_{3} + R_{2} & - \frac{33}{17} R_{3} + R_{2} & - \frac{33}{17} R_{3} + R_{2} & - \frac{33}{17} R_{3} + R_{2} & - \frac{33}{17} R_{3} + R_{2} & - \frac{33}{17} R_{3} + R_{2} \\ 0 & 0 & 1 & \frac{25}{217} & - \frac{3}{217} & - \frac{17}{217} \end{array}]$

$R_{2} = - \frac{33}{17} R_{3} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - \frac{33}{17} R_{3} + R_{2}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{4}{31} & \frac{2}{31} & \frac{1}{31} \\ (- \frac{33}{17}) \cdot (0) + 0 & (- \frac{33}{17}) \cdot (0) + 1 & (- \frac{33}{17}) \cdot (1) + \frac{33}{17} & (- \frac{33}{17}) \cdot (\frac{25}{217}) + \frac{2}{17} & (- \frac{33}{17}) \cdot (- \frac{3}{217}) - \frac{5}{17} & (- \frac{33}{17}) \cdot (- \frac{17}{217}) + 0 \\ 0 & 0 & 1 & \frac{25}{217} & - \frac{3}{217} & - \frac{17}{217} \end{array}]$

$R_{2} = - \frac{33}{17} R_{3} + R_{2}$

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

$[\begin{array}{c} 1 & 0 & 0 & \frac{4}{31} & \frac{2}{31} & \frac{1}{31} \\ 0 & 1 & 0 & - \frac{23}{217} & - \frac{58}{217} & \frac{33}{217} \\ 0 & 0 & 1 & \frac{25}{217} & - \frac{3}{217} & - \frac{17}{217} \end{array}]$

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

$[\begin{array}{c} \frac{4}{31} & \frac{2}{31} & \frac{1}{31} \\ - \frac{23}{217} & - \frac{58}{217} & \frac{33}{217} \\ \frac{25}{217} & - \frac{3}{217} & - \frac{17}{217} \end{array}]$

Assuming that $X$ is the matrix to solve for, multiply the inverse matrix by both sides of the equation.

$[\begin{array}{c} \frac{4}{31} & \frac{2}{31} & \frac{1}{31} \\ - \frac{23}{217} & - \frac{58}{217} & \frac{33}{217} \\ \frac{25}{217} & - \frac{3}{217} & - \frac{17}{217} \end{array}] \cdot ([\begin{array}{c} 5 & 1 & 4 \\ 2 & - 3 & - 5 \\ 7 & 2 & - 6 \end{array}] x) = [\begin{array}{c} \frac{4}{31} & \frac{2}{31} & \frac{1}{31} \\ - \frac{23}{217} & - \frac{58}{217} & \frac{33}{217} \\ \frac{25}{217} & - \frac{3}{217} & - \frac{17}{217} \end{array}] \cdot [\begin{array}{c} 5 \\ 2 \\ 5 \end{array}]$

Simplify the left side of the equation.

Multiply each row in the first matrix by each column in the second matrix.

$[\begin{array}{c} \frac{4}{31} \cdot 5 + \frac{2}{31} \cdot 2 + \frac{1}{31} \cdot 7 & \frac{4}{31} \cdot 1 + \frac{2}{31} \cdot - 3 + \frac{1}{31} \cdot 2 & \frac{4}{31} \cdot 4 + \frac{2}{31} \cdot - 5 + \frac{1}{31} \cdot - 6 \\ - \frac{23}{217} \cdot 5 - \frac{58}{217} \cdot 2 + \frac{33}{217} \cdot 7 & - \frac{23}{217} \cdot 1 - \frac{58}{217} \cdot - 3 + \frac{33}{217} \cdot 2 & - \frac{23}{217} \cdot 4 - \frac{58}{217} \cdot - 5 + \frac{33}{217} \cdot - 6 \\ \frac{25}{217} \cdot 5 - \frac{3}{217} \cdot 2 - \frac{17}{217} \cdot 7 & \frac{25}{217} \cdot 1 - \frac{3}{217} \cdot - 3 - \frac{17}{217} \cdot 2 & \frac{25}{217} \cdot 4 - \frac{3}{217} \cdot - 5 - \frac{17}{217} \cdot - 6 \end{array}] x = [\begin{array}{c} \frac{4}{31} & \frac{2}{31} & \frac{1}{31} \\ - \frac{23}{217} & - \frac{58}{217} & \frac{33}{217} \\ \frac{25}{217} & - \frac{3}{217} & - \frac{17}{217} \end{array}] \cdot [\begin{array}{c} 5 \\ 2 \\ 5 \end{array}]$

Simplify each element of the matrix by multiplying out all the expressions.

$[\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] x = [\begin{array}{c} \frac{4}{31} & \frac{2}{31} & \frac{1}{31} \\ - \frac{23}{217} & - \frac{58}{217} & \frac{33}{217} \\ \frac{25}{217} & - \frac{3}{217} & - \frac{17}{217} \end{array}] \cdot [\begin{array}{c} 5 \\ 2 \\ 5 \end{array}]$

Multiplying the identity matrix by any matrix $x$ is matrix $x$ .

$x = [\begin{array}{c} \frac{4}{31} & \frac{2}{31} & \frac{1}{31} \\ - \frac{23}{217} & - \frac{58}{217} & \frac{33}{217} \\ \frac{25}{217} & - \frac{3}{217} & - \frac{17}{217} \end{array}] \cdot [\begin{array}{c} 5 \\ 2 \\ 5 \end{array}]$

Simplify the right side of the equation.

Multiply each row in the first matrix by each column in the second matrix.

$x = [\begin{array}{c} \frac{4}{31} \cdot 5 + \frac{2}{31} \cdot 2 + \frac{1}{31} \cdot 5 \\ - \frac{23}{217} \cdot 5 - \frac{58}{217} \cdot 2 + \frac{33}{217} \cdot 5 \\ \frac{25}{217} \cdot 5 - \frac{3}{217} \cdot 2 - \frac{17}{217} \cdot 5 \end{array}]$

Simplify each element of the matrix by multiplying out all the expressions.

$x = [\begin{array}{c} \frac{29}{31} \\ - \frac{66}{217} \\ \frac{34}{217} \end{array}]$

The matrix is in the most simplified form.

$x = [\begin{array}{c} \frac{29}{31} \\ - \frac{66}{217} \\ \frac{34}{217} \end{array}]$

Solve the Matrix Equation [[5,1,4],[2,-3,-5],[7,2,-6]]x=[[5],[2],[5]]

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