Finite Math Examples Systems of Linear Equations

Solve Using a Matrix by Row Operations

$x + 2 y - z = 4$ , $2 x + y + z = - 2$ , $x + 2 y + z = 2$

Step 1

Write the system as a matrix.

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

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

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

Step 2.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 2 & - 1 & 4 \\ 0 & - 3 & 3 & - 10 \\ 1 & 2 & 1 & 2 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} 1 & 2 & - 1 & 4 \\ 0 & - 3 & 3 & - 10 \\ 1 - 1 & 2 - 2 & 1 + 1 & 2 - 4 \end{array}]$

Step 2.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 2 & - 1 & 4 \\ 0 & - 3 & 3 & - 10 \\ 0 & 0 & 2 & - 2 \end{array}]$

Step 2.3

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

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Step 2.3.1

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

$[\begin{array}{c} 1 & 2 & - 1 & 4 \\ - \frac{1}{3} \cdot 0 & - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot 3 & - \frac{1}{3} \cdot - 10 \\ 0 & 0 & 2 & - 2 \end{array}]$

Step 2.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 2 & - 1 & 4 \\ 0 & 1 & - 1 & \frac{10}{3} \\ 0 & 0 & 2 & - 2 \end{array}]$

Step 2.4

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

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Step 2.4.1

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

$[\begin{array}{c} 1 & 2 & - 1 & 4 \\ 0 & 1 & - 1 & \frac{10}{3} \\ \frac{0}{2} & \frac{0}{2} & \frac{2}{2} & \frac{- 2}{2} \end{array}]$

Step 2.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 2 & - 1 & 4 \\ 0 & 1 & - 1 & \frac{10}{3} \\ 0 & 0 & 1 & - 1 \end{array}]$

Step 2.5

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

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Step 2.5.1

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

$[\begin{array}{c} 1 & 2 & - 1 & 4 \\ 0 + 0 & 1 + 0 & - 1 + 1 \cdot 1 & \frac{10}{3} - 1 \\ 0 & 0 & 1 & - 1 \end{array}]$

Step 2.5.2

Simplify $R_{2}$ .

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

Step 2.6

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

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Step 2.6.1

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

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

Step 2.6.2

Simplify $R_{1}$ .

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

Step 2.7

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

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Step 2.7.1

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

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

Step 2.7.2

Simplify $R_{1}$ .

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

Step 3

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

$x = - \frac{5}{3}$

$y = \frac{7}{3}$

$z = - 1$

Step 4

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

$(- \frac{5}{3}, \frac{7}{3}, - 1)$

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