Linear Algebra Examples

$- x - 2 y - 2 z = - 3$ , $- x - 5 y - 2 z = - 9$ , $- x - 5 y - 5 z = - 6$

Step 1

Write the system as a matrix.

$[\begin{array}{c} - 1 & - 2 & - 2 & - 3 \\ - 1 & - 5 & - 2 & - 9 \\ - 1 & - 5 & - 5 & - 6 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} - - 1 & - - 2 & - - 2 & - - 3 \\ - 1 & - 5 & - 2 & - 9 \\ - 1 & - 5 & - 5 & - 6 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 2 & 2 & 3 \\ - 1 & - 5 & - 2 & - 9 \\ - 1 & - 5 & - 5 & - 6 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} 1 & 2 & 2 & 3 \\ - 1 + 1 \cdot 1 & - 5 + 1 \cdot 2 & - 2 + 1 \cdot 2 & - 9 + 1 \cdot 3 \\ - 1 & - 5 & - 5 & - 6 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 2 & 2 & 3 \\ 0 & - 3 & 0 & - 6 \\ - 1 & - 5 & - 5 & - 6 \end{array}]$

Step 2.3

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

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Step 2.3.1

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

$[\begin{array}{c} 1 & 2 & 2 & 3 \\ 0 & - 3 & 0 & - 6 \\ - 1 + 1 \cdot 1 & - 5 + 1 \cdot 2 & - 5 + 1 \cdot 2 & - 6 + 1 \cdot 3 \end{array}]$

Step 2.3.2

Simplify $R_{3}$ .

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

Step 2.4

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

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Step 2.4.1

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

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

Step 2.4.2

Simplify $R_{2}$ .

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

Step 2.5

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

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Step 2.5.1

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

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

Step 2.5.2

Simplify $R_{3}$ .

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

Step 2.6

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

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Step 2.6.1

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

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

Step 2.6.2

Simplify $R_{3}$ .

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

Step 2.7

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

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Step 2.7.1

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

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

Step 2.7.2

Simplify $R_{1}$ .

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

Step 2.8

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 2.8.1

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

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

Step 2.8.2

Simplify $R_{1}$ .

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

Step 3

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

$x = 1$

$y = 2$

$z = - 1$

Step 4

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

$(1, 2, - 1)$