Linear Algebra Examples

$x - 2 y + 3 x = 9$ , $- x + 3 y = - 4$ , $2 x - 5 y + 5 z = 17$

Step 1

Add $x$ and $3 x$ .

$4 x - 2 y = 9$

$- x + 3 y = - 4$

$2 x - 5 y + 5 z = 17$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 4 & - 2 & 0 & 9 \\ - 1 & 3 & 0 & - 4 \\ 2 & - 5 & 5 & 17 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Multiply each element of $R_{1}$ by $\frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

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Step 3.1.1

Multiply each element of $R_{1}$ by $\frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{4}{4} & \frac{- 2}{4} & \frac{0}{4} & \frac{9}{4} \\ - 1 & 3 & 0 & - 4 \\ 2 & - 5 & 5 & 17 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 0 & \frac{9}{4} \\ - 1 & 3 & 0 & - 4 \\ 2 & - 5 & 5 & 17 \end{array}]$

Step 3.2

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

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Step 3.2.1

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

$[\begin{array}{c} 1 & - \frac{1}{2} & 0 & \frac{9}{4} \\ - 1 + 1 \cdot 1 & 3 - \frac{1}{2} & 0 + 0 & - 4 + \frac{9}{4} \\ 2 & - 5 & 5 & 17 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 0 & \frac{9}{4} \\ 0 & \frac{5}{2} & 0 & - \frac{7}{4} \\ 2 & - 5 & 5 & 17 \end{array}]$

Step 3.3

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

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Step 3.3.1

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

$[\begin{array}{c} 1 & - \frac{1}{2} & 0 & \frac{9}{4} \\ 0 & \frac{5}{2} & 0 & - \frac{7}{4} \\ 2 - 2 \cdot 1 & - 5 - 2 (- \frac{1}{2}) & 5 - 2 \cdot 0 & 17 - 2 (\frac{9}{4}) \end{array}]$

Step 3.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 0 & \frac{9}{4} \\ 0 & \frac{5}{2} & 0 & - \frac{7}{4} \\ 0 & - 4 & 5 & \frac{25}{2} \end{array}]$

Step 3.4

Multiply each element of $R_{2}$ by $\frac{2}{5}$ to make the entry at $2, 2$ a $1$ .

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Step 3.4.1

Multiply each element of $R_{2}$ by $\frac{2}{5}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 0 & \frac{9}{4} \\ \frac{2}{5} \cdot 0 & \frac{2}{5} \cdot \frac{5}{2} & \frac{2}{5} \cdot 0 & \frac{2}{5} (- \frac{7}{4}) \\ 0 & - 4 & 5 & \frac{25}{2} \end{array}]$

Step 3.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 0 & \frac{9}{4} \\ 0 & 1 & 0 & - \frac{7}{10} \\ 0 & - 4 & 5 & \frac{25}{2} \end{array}]$

Step 3.5

Perform the row operation $R_{3} = R_{3} + 4 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 3.5.1

Perform the row operation $R_{3} = R_{3} + 4 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 0 & \frac{9}{4} \\ 0 & 1 & 0 & - \frac{7}{10} \\ 0 + 4 \cdot 0 & - 4 + 4 \cdot 1 & 5 + 4 \cdot 0 & \frac{25}{2} + 4 (- \frac{7}{10}) \end{array}]$

Step 3.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 0 & \frac{9}{4} \\ 0 & 1 & 0 & - \frac{7}{10} \\ 0 & 0 & 5 & \frac{97}{10} \end{array}]$

Step 3.6

Multiply each element of $R_{3}$ by $\frac{1}{5}$ to make the entry at $3, 3$ a $1$ .

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Step 3.6.1

Multiply each element of $R_{3}$ by $\frac{1}{5}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 0 & \frac{9}{4} \\ 0 & 1 & 0 & - \frac{7}{10} \\ \frac{0}{5} & \frac{0}{5} & \frac{5}{5} & \frac{\frac{97}{10}}{5} \end{array}]$

Step 3.6.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 0 & \frac{9}{4} \\ 0 & 1 & 0 & - \frac{7}{10} \\ 0 & 0 & 1 & \frac{97}{50} \end{array}]$

Step 3.7

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

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Step 3.7.1

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

$[\begin{array}{c} 1 + \frac{1}{2} \cdot 0 & - \frac{1}{2} + \frac{1}{2} \cdot 1 & 0 + \frac{1}{2} \cdot 0 & \frac{9}{4} + \frac{1}{2} (- \frac{7}{10}) \\ 0 & 1 & 0 & - \frac{7}{10} \\ 0 & 0 & 1 & \frac{97}{50} \end{array}]$

Step 3.7.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{19}{10} \\ 0 & 1 & 0 & - \frac{7}{10} \\ 0 & 0 & 1 & \frac{97}{50} \end{array}]$

Step 4

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

$x = \frac{19}{10}$

$y = - \frac{7}{10}$

$z = \frac{97}{50}$

Step 5

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

$(\frac{19}{10}, - \frac{7}{10}, \frac{97}{50})$