Linear Algebra Examples

$- x + 3 y + 10 z = 8$ $4 x - 9 y - 34 z = - 17$ $3 x + 5 y - 2 z = 46$

Step 1

Write the system as a matrix.

$[\begin{array}{c} - 1 & 3 & 10 & 8 \\ 4 & - 9 & - 34 & - 17 \\ 3 & 5 & - 2 & 46 \end{array}]$

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 $1$ .

$[\begin{array}{c} - - 1 & - 1 \cdot 3 & - 1 \cdot 10 & - 1 \cdot 8 \\ 4 & - 9 & - 34 & - 17 \\ 3 & 5 & - 2 & 46 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 3 & - 10 & - 8 \\ 4 & - 9 & - 34 & - 17 \\ 3 & 5 & - 2 & 46 \end{array}]$

Step 2.2

Perform the row operation $R_{2} = R_{2} - 4 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 2.2.1

Perform the row operation $R_{2} = R_{2} - 4 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - 3 & - 10 & - 8 \\ 4 - 4 \cdot 1 & - 9 - 4 \cdot - 3 & - 34 - 4 \cdot - 10 & - 17 - 4 \cdot - 8 \\ 3 & 5 & - 2 & 46 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 3 & - 10 & - 8 \\ 0 & 3 & 6 & 15 \\ 3 & 5 & - 2 & 46 \end{array}]$

Step 2.3

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 2.3.1

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

$[\begin{array}{c} 1 & - 3 & - 10 & - 8 \\ 0 & 3 & 6 & 15 \\ 3 - 3 \cdot 1 & 5 - 3 \cdot - 3 & - 2 - 3 \cdot - 10 & 46 - 3 \cdot - 8 \end{array}]$

Step 2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 3 & - 10 & - 8 \\ 0 & 3 & 6 & 15 \\ 0 & 14 & 28 & 70 \end{array}]$

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 $1$ .

$[\begin{array}{c} 1 & - 3 & - 10 & - 8 \\ \frac{0}{3} & \frac{3}{3} & \frac{6}{3} & \frac{15}{3} \\ 0 & 14 & 28 & 70 \end{array}]$

Step 2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 3 & - 10 & - 8 \\ 0 & 1 & 2 & 5 \\ 0 & 14 & 28 & 70 \end{array}]$

Step 2.5

Perform the row operation $R_{3} = R_{3} - 14 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 2.5.1

Perform the row operation $R_{3} = R_{3} - 14 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - 3 & - 10 & - 8 \\ 0 & 1 & 2 & 5 \\ 0 - 14 \cdot 0 & 14 - 14 \cdot 1 & 28 - 14 \cdot 2 & 70 - 14 \cdot 5 \end{array}]$

Step 2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 3 & - 10 & - 8 \\ 0 & 1 & 2 & 5 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 2.6

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

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Step 2.6.1

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

$[\begin{array}{c} 1 + 3 \cdot 0 & - 3 + 3 \cdot 1 & - 10 + 3 \cdot 2 & - 8 + 3 \cdot 5 \\ 0 & 1 & 2 & 5 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 2.6.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - 4 & 7 \\ 0 & 1 & 2 & 5 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 3

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

$x - 4 z = 7$

$y + 2 z = 5$

$0 = 0$

Step 4

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

$(7 + 4 z, 5 - 2 z, z)$