Linear Algebra Examples

$x + 3 y - z = 4$ , $3 y - z = 0$ , $x - y + 5 z = 0$

Step 1

Write the system as a matrix.

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

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

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

Step 2.1.2

Simplify $R_{3}$ .

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

Simplify $R_{2}$ .

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 3 & - 1 & 4 \\ 0 & 1 & - \frac{1}{3} & 0 \\ 0 & 0 & \frac{14}{3} & - 4 \end{array}]$

Step 2.4

Multiply each element of $R_{3}$ by $\frac{3}{14}$ to make the entry at $3, 3$ a $1$ .

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Step 2.4.1

Multiply each element of $R_{3}$ by $\frac{3}{14}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 3 & - 1 & 4 \\ 0 & 1 & - \frac{1}{3} & 0 \\ \frac{3}{14} \cdot 0 & \frac{3}{14} \cdot 0 & \frac{3}{14} \cdot \frac{14}{3} & \frac{3}{14} \cdot - 4 \end{array}]$

Step 2.4.2

Simplify $R_{3}$ .

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

Step 2.5

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

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Step 2.5.1

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

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

Step 2.5.2

Simplify $R_{2}$ .

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

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 $0$ .

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

Step 2.6.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 3 & 0 & \frac{22}{7} \\ 0 & 1 & 0 & - \frac{2}{7} \\ 0 & 0 & 1 & - \frac{6}{7} \end{array}]$

Step 2.7

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

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Step 2.7.1

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

$[\begin{array}{c} 1 - 3 \cdot 0 & 3 - 3 \cdot 1 & 0 - 3 \cdot 0 & \frac{22}{7} - 3 (- \frac{2}{7}) \\ 0 & 1 & 0 & - \frac{2}{7} \\ 0 & 0 & 1 & - \frac{6}{7} \end{array}]$

Step 2.7.2

Simplify $R_{1}$ .

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

Step 3

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

$x = 4$

$y = - \frac{2}{7}$

$z = - \frac{6}{7}$

Step 4

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

$(4, - \frac{2}{7}, - \frac{6}{7})$