Precalculus Examples

$x + y + z = 6$ , $x + y - z = 0$ , $3 x + 2 y + z = 10$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 1 & 1 & 1 & 6 \\ 1 & 1 & - 1 & 0 \\ 3 & 2 & 1 & 10 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} 1 & 1 & 1 & 6 \\ 1 - 1 & 1 - 1 & - 1 - 1 & 0 - 6 \\ 3 & 2 & 1 & 10 \end{array}]$

Step 2.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 6 \\ 0 & 0 & - 2 & - 6 \\ 3 & 2 & 1 & 10 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & 1 & 1 & 6 \\ 0 & 0 & - 2 & - 6 \\ 3 - 3 \cdot 1 & 2 - 3 \cdot 1 & 1 - 3 \cdot 1 & 10 - 3 \cdot 6 \end{array}]$

Step 2.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 6 \\ 0 & 0 & - 2 & - 6 \\ 0 & - 1 & - 2 & - 8 \end{array}]$

Step 2.3

Swap $R_{3}$ with $R_{2}$ to put a nonzero entry at $2, 2$ .

$[\begin{array}{c} 1 & 1 & 1 & 6 \\ 0 & - 1 & - 2 & - 8 \\ 0 & 0 & - 2 & - 6 \end{array}]$

Step 2.4

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

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Step 2.4.1

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

$[\begin{array}{c} 1 & 1 & 1 & 6 \\ - 0 & - - 1 & - - 2 & - - 8 \\ 0 & 0 & - 2 & - 6 \end{array}]$

Step 2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 6 \\ 0 & 1 & 2 & 8 \\ 0 & 0 & - 2 & - 6 \end{array}]$

Step 2.5

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

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Step 2.5.1

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

$[\begin{array}{c} 1 & 1 & 1 & 6 \\ 0 & 1 & 2 & 8 \\ - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot - 2 & - \frac{1}{2} \cdot - 6 \end{array}]$

Step 2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 6 \\ 0 & 1 & 2 & 8 \\ 0 & 0 & 1 & 3 \end{array}]$

Step 2.6

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

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Step 2.6.1

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

$[\begin{array}{c} 1 & 1 & 1 & 6 \\ 0 - 2 \cdot 0 & 1 - 2 \cdot 0 & 2 - 2 \cdot 1 & 8 - 2 \cdot 3 \\ 0 & 0 & 1 & 3 \end{array}]$

Step 2.6.2

Simplify $R_{2}$ .

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

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 $0$ .

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

Step 2.7.2

Simplify $R_{1}$ .

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

Step 2.8

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

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Step 2.8.1

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

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

Step 2.8.2

Simplify $R_{1}$ .

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

Step 3

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

$x = 1$

$y = 2$

$z = 3$

Step 4

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

$(1, 2, 3)$