Precalculus Examples

$x - 2 y + 5 z = 3$ , $- 2 x + 6 y - 11 z = 1$ , $3 x - 16 y + 20 z = - 26$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 1 & - 2 & 5 & 3 \\ - 2 & 6 & - 11 & 1 \\ 3 & - 16 & 20 & - 26 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & - 2 & 5 & 3 \\ - 2 + 2 \cdot 1 & 6 + 2 \cdot - 2 & - 11 + 2 \cdot 5 & 1 + 2 \cdot 3 \\ 3 & - 16 & 20 & - 26 \end{array}]$

Step 2.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 2 & 5 & 3 \\ 0 & 2 & - 1 & 7 \\ 3 & - 16 & 20 & - 26 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} 1 & - 2 & 5 & 3 \\ 0 & 2 & - 1 & 7 \\ 3 - 3 \cdot 1 & - 16 - 3 \cdot - 2 & 20 - 3 \cdot 5 & - 26 - 3 \cdot 3 \end{array}]$

Step 2.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 2 & 5 & 3 \\ 0 & 2 & - 1 & 7 \\ 0 & - 10 & 5 & - 35 \end{array}]$

Step 2.3

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

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Step 2.3.1

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

$[\begin{array}{c} 1 & - 2 & 5 & 3 \\ \frac{0}{2} & \frac{2}{2} & \frac{- 1}{2} & \frac{7}{2} \\ 0 & - 10 & 5 & - 35 \end{array}]$

Step 2.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 2 & 5 & 3 \\ 0 & 1 & - \frac{1}{2} & \frac{7}{2} \\ 0 & - 10 & 5 & - 35 \end{array}]$

Step 2.4

Perform the row operation $R_{3} = R_{3} + 10 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 2.4.1

Perform the row operation $R_{3} = R_{3} + 10 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - 2 & 5 & 3 \\ 0 & 1 & - \frac{1}{2} & \frac{7}{2} \\ 0 + 10 \cdot 0 & - 10 + 10 \cdot 1 & 5 + 10 (- \frac{1}{2}) & - 35 + 10 (\frac{7}{2}) \end{array}]$

Step 2.4.2

Simplify $R_{3}$ .

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

Step 2.5

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 2.5.1

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

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

Step 2.5.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 4 & 10 \\ 0 & 1 & - \frac{1}{2} & \frac{7}{2} \\ 0 & 0 & 0 & 0 \end{array}]$

Step 3

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

$x + 4 z = 10$

$y - \frac{1}{2} z = \frac{7}{2}$

$0 = 0$

Step 4

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

$(10 - 4 z, \frac{7}{2} + \frac{z}{2}, z)$