Linear Algebra Examples

$3 x - 3 y - 6 z = 3$ $2 x - 3 y - 5 z = 8$ $x + 4 y + 2 z = 6$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 3 & - 3 & - 6 & 3 \\ 2 & - 3 & - 5 & 8 \\ 1 & 4 & 2 & 6 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} \frac{3}{3} & \frac{- 3}{3} & \frac{- 6}{3} & \frac{3}{3} \\ 2 & - 3 & - 5 & 8 \\ 1 & 4 & 2 & 6 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 1 & - 2 & 1 \\ 2 & - 3 & - 5 & 8 \\ 1 & 4 & 2 & 6 \end{array}]$

Step 2.2

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 2.2.1

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

$[\begin{array}{c} 1 & - 1 & - 2 & 1 \\ 2 - 2 \cdot 1 & - 3 - 2 \cdot - 1 & - 5 - 2 \cdot - 2 & 8 - 2 \cdot 1 \\ 1 & 4 & 2 & 6 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

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

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 $0$ .

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

Step 2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & - 2 & 1 \\ 0 & - 1 & - 1 & 6 \\ 0 & 5 & 4 & 5 \end{array}]$

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 $1$ .

$[\begin{array}{c} 1 & - 1 & - 2 & 1 \\ - 0 & - - 1 & - - 1 & - 1 \cdot 6 \\ 0 & 5 & 4 & 5 \end{array}]$

Step 2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & - 2 & 1 \\ 0 & 1 & 1 & - 6 \\ 0 & 5 & 4 & 5 \end{array}]$

Step 2.5

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

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Step 2.5.1

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

$[\begin{array}{c} 1 & - 1 & - 2 & 1 \\ 0 & 1 & 1 & - 6 \\ 0 - 5 \cdot 0 & 5 - 5 \cdot 1 & 4 - 5 \cdot 1 & 5 - 5 \cdot - 6 \end{array}]$

Step 2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & - 2 & 1 \\ 0 & 1 & 1 & - 6 \\ 0 & 0 & - 1 & 35 \end{array}]$

Step 2.6

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

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Step 2.6.1

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

$[\begin{array}{c} 1 & - 1 & - 2 & 1 \\ 0 & 1 & 1 & - 6 \\ - 0 & - 0 & - - 1 & - 1 \cdot 35 \end{array}]$

Step 2.6.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & - 2 & 1 \\ 0 & 1 & 1 & - 6 \\ 0 & 0 & 1 & - 35 \end{array}]$

Step 2.7

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

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Step 2.7.1

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

$[\begin{array}{c} 1 & - 1 & - 2 & 1 \\ 0 - 0 & 1 - 0 & 1 - 1 & - 6 + 35 \\ 0 & 0 & 1 & - 35 \end{array}]$

Step 2.7.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & - 2 & 1 \\ 0 & 1 & 0 & 29 \\ 0 & 0 & 1 & - 35 \end{array}]$

Step 2.8

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 2.8.1

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

$[\begin{array}{c} 1 + 2 \cdot 0 & - 1 + 2 \cdot 0 & - 2 + 2 \cdot 1 & 1 + 2 \cdot - 35 \\ 0 & 1 & 0 & 29 \\ 0 & 0 & 1 & - 35 \end{array}]$

Step 2.8.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 1 & 0 & - 69 \\ 0 & 1 & 0 & 29 \\ 0 & 0 & 1 & - 35 \end{array}]$

Step 2.9

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

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Step 2.9.1

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

$[\begin{array}{c} 1 + 0 & - 1 + 1 \cdot 1 & 0 + 0 & - 69 + 1 \cdot 29 \\ 0 & 1 & 0 & 29 \\ 0 & 0 & 1 & - 35 \end{array}]$

Step 2.9.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & - 40 \\ 0 & 1 & 0 & 29 \\ 0 & 0 & 1 & - 35 \end{array}]$

Step 3

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

$x = - 40$

$y = 29$

$z = - 35$

Step 4

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

$(- 40, 29, - 35)$