Precalculus Examples

$y - 5 z = 1$ , $4 x + 4 z = 4$ , $5 x - 4 y - 5 z = 1$

Step 1

Write the system as a matrix.

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

Step 2

Find the reduced row echelon form.

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Step 2.1

Swap $R_{2}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

Simplify $R_{1}$ .

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

Step 2.3

Perform the row operation $R_{3} = R_{3} - 5 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 2.3.1

Perform the row operation $R_{3} = R_{3} - 5 R_{1}$ to make the entry at $3, 1$ a $0$ .

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

Step 2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 0 & 1 & 1 \\ 0 & 1 & - 5 & 1 \\ 0 & - 4 & - 10 & - 4 \end{array}]$

Step 2.4

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 2.4.1

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

$[\begin{array}{c} 1 & 0 & 1 & 1 \\ 0 & 1 & - 5 & 1 \\ 0 + 4 \cdot 0 & - 4 + 4 \cdot 1 & - 10 + 4 \cdot - 5 & - 4 + 4 \cdot 1 \end{array}]$

Step 2.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 0 & 1 & 1 \\ 0 & 1 & - 5 & 1 \\ 0 & 0 & - 30 & 0 \end{array}]$

Step 2.5

Multiply each element of $R_{3}$ by $- \frac{1}{30}$ to make the entry at $3, 3$ a $1$ .

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Step 2.5.1

Multiply each element of $R_{3}$ by $- \frac{1}{30}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 0 & 1 & 1 \\ 0 & 1 & - 5 & 1 \\ - \frac{1}{30} \cdot 0 & - \frac{1}{30} \cdot 0 & - \frac{1}{30} \cdot - 30 & - \frac{1}{30} \cdot 0 \end{array}]$

Step 2.5.2

Simplify $R_{3}$ .

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

Step 2.6

Perform the row operation $R_{2} = R_{2} + 5 R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 2.6.1

Perform the row operation $R_{2} = R_{2} + 5 R_{3}$ to make the entry at $2, 3$ a $0$ .

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

Step 2.6.2

Simplify $R_{2}$ .

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

Step 2.7

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

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Step 2.7.1

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

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

Step 2.7.2

Simplify $R_{1}$ .

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

Step 3

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

$x = 1$

$y = 1$

$z = 0$

Step 4

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

$(1, 1, 0)$