Solve Using a Matrix by Elimination x-3y-z=9 , -x+8y-4z=-24 , 2x-15y+7z=45

$x - 3 y - z = 9$ , $- x + 8 y - 4 z = - 24$ , $2 x - 15 y + 7 z = 45$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & - 3 & - 1 & 9 \\ - 1 & 8 & - 4 & - 24 \\ 2 & - 15 & 7 & 45 \end{array}]$

Step 2

Find the reduced row echelon form of the matrix.

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

$[\begin{array}{c} 1 & - 3 & - 1 & 9 \\ R_{1} + R_{2} & R_{1} + R_{2} & R_{1} + R_{2} & R_{1} + R_{2} \\ 2 & - 15 & 7 & 45 \end{array}]$

$R_{2} = R_{1} + R_{2}$

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

$[\begin{array}{c} 1 & - 3 & - 1 & 9 \\ 1 - 1 & - 3 + 8 & - 1 - 4 & 9 - 24 \\ 2 & - 15 & 7 & 45 \end{array}]$

$R_{2} = R_{1} + R_{2}$

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

$[\begin{array}{c} 1 & - 3 & - 1 & 9 \\ 0 & 5 & - 5 & - 15 \\ 2 & - 15 & 7 & 45 \end{array}]$

Perform the row operation $R_{3} = - 2 \cdot R_{1} + R_{3}$ 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 $R_{3} = - 2 \cdot R_{1} + R_{3}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & - 3 & - 1 & 9 \\ 0 & 5 & - 5 & - 15 \\ - 2 \cdot R_{1} + R_{3} & - 2 \cdot R_{1} + R_{3} & - 2 \cdot R_{1} + R_{3} & - 2 \cdot R_{1} + R_{3} \end{array}]$

$R_{3} = - 2 \cdot R_{1} + R_{3}$

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

$[\begin{array}{c} 1 & - 3 & - 1 & 9 \\ 0 & 5 & - 5 & - 15 \\ (- 2) \cdot (1) + 2 & (- 2) \cdot (- 3) - 15 & (- 2) \cdot (- 1) + 7 & (- 2) \cdot (9) + 45 \end{array}]$

$R_{3} = - 2 \cdot R_{1} + R_{3}$

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

$[\begin{array}{c} 1 & - 3 & - 1 & 9 \\ 0 & 5 & - 5 & - 15 \\ 0 & - 9 & 9 & 27 \end{array}]$

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

$[\begin{array}{c} 1 & - 3 & - 1 & 9 \\ \frac{1}{5} R_{2} & \frac{1}{5} R_{2} & \frac{1}{5} R_{2} & \frac{1}{5} R_{2} \\ 0 & - 9 & 9 & 27 \end{array}]$

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

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

$[\begin{array}{c} 1 & - 3 & - 1 & 9 \\ (\frac{1}{5}) \cdot (0) & (\frac{1}{5}) \cdot (5) & (\frac{1}{5}) \cdot (- 5) & (\frac{1}{5}) \cdot (- 15) \\ 0 & - 9 & 9 & 27 \end{array}]$

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

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

$[\begin{array}{c} 1 & - 3 & - 1 & 9 \\ 0 & 1 & - 1 & - 3 \\ 0 & - 9 & 9 & 27 \end{array}]$

Perform the row operation $R_{1} = 3 \cdot R_{2} + R_{1}$ 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 $R_{1} = 3 \cdot R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 3 \cdot R_{2} + R_{1} & 3 \cdot R_{2} + R_{1} & 3 \cdot R_{2} + R_{1} & 3 \cdot R_{2} + R_{1} \\ 0 & 1 & - 1 & - 3 \\ 0 & - 9 & 9 & 27 \end{array}]$

$R_{1} = 3 \cdot R_{2} + R_{1}$

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

$[\begin{array}{c} (3) \cdot (0) + 1 & (3) \cdot (1) - 3 & (3) \cdot (- 1) - 1 & (3) \cdot (- 3) + 9 \\ 0 & 1 & - 1 & - 3 \\ 0 & - 9 & 9 & 27 \end{array}]$

$R_{1} = 3 \cdot R_{2} + R_{1}$

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

$[\begin{array}{c} 1 & 0 & - 4 & 0 \\ 0 & 1 & - 1 & - 3 \\ 0 & - 9 & 9 & 27 \end{array}]$

Perform the row operation $R_{3} = 9 \cdot R_{2} + R_{3}$ 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 $R_{3} = 9 \cdot R_{2} + R_{3}$ in order to convert some elements in the row to the desired value $0$ .

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

$R_{3} = 9 \cdot R_{2} + R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = 9 \cdot R_{2} + R_{3}$ .

$[\begin{array}{c} 1 & 0 & - 4 & 0 \\ 0 & 1 & - 1 & - 3 \\ (9) \cdot (0) + 0 & (9) \cdot (1) - 9 & (9) \cdot (- 1) + 9 & (9) \cdot (- 3) + 27 \end{array}]$

$R_{3} = 9 \cdot R_{2} + R_{3}$

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

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

Step 3

Use the result matrix to declare the final solutions to the system of equations.

$x - 4 z = 0$

$y - z = - 3$

Step 4

Add $4 z$ to both sides of the equation.

$x = 4 z$

$y - z = - 3$

Step 5

Add $z$ to both sides of the equation.

$y = - 3 + z$

$x = 4 z$

Step 6

The solution is the set of ordered pairs that makes the system true.

$(4 z, - 3 + z, z)$

Solve Using a Matrix by Elimination x-3y-z=9 , -x+8y-4z=-24 , 2x-15y+7z=45

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