Linear Algebra Examples

$\frac{1}{x} + \frac{2}{y} - \frac{4}{z} = 1$ , $\frac{2}{x} + \frac{3}{y} + \frac{8}{z} = 0$ , $- \frac{1}{x} + \frac{9}{y} + \frac{10}{z} = 5$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} \frac{1}{1} & \frac{2}{1} & - \frac{4}{1} & 1 \\ \frac{2}{1} & \frac{3}{1} & \frac{8}{1} & 0 \\ - \frac{1}{1} & \frac{9}{1} & \frac{10}{1} & 5 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

Divide $1$ by $1$ .

$[\begin{array}{c} 1 & \frac{2}{1} & - \frac{4}{1} & 1 \\ \frac{2}{1} & \frac{3}{1} & \frac{8}{1} & 0 \\ - \frac{1}{1} & \frac{9}{1} & \frac{10}{1} & 5 \end{array}]$

Step 2.2

Divide $2$ by $1$ .

$[\begin{array}{c} 1 & 2 & - \frac{4}{1} & 1 \\ \frac{2}{1} & \frac{3}{1} & \frac{8}{1} & 0 \\ - \frac{1}{1} & \frac{9}{1} & \frac{10}{1} & 5 \end{array}]$

Step 2.3

Divide $4$ by $1$ .

$[\begin{array}{c} 1 & 2 & - 1 \cdot 4 & 1 \\ \frac{2}{1} & \frac{3}{1} & \frac{8}{1} & 0 \\ - \frac{1}{1} & \frac{9}{1} & \frac{10}{1} & 5 \end{array}]$

Step 2.4

Multiply $- 1$ by $4$ .

$[\begin{array}{c} 1 & 2 & - 4 & 1 \\ \frac{2}{1} & \frac{3}{1} & \frac{8}{1} & 0 \\ - \frac{1}{1} & \frac{9}{1} & \frac{10}{1} & 5 \end{array}]$

Step 2.5

Divide $2$ by $1$ .

$[\begin{array}{c} 1 & 2 & - 4 & 1 \\ 2 & \frac{3}{1} & \frac{8}{1} & 0 \\ - \frac{1}{1} & \frac{9}{1} & \frac{10}{1} & 5 \end{array}]$

Step 2.6

Divide $3$ by $1$ .

$[\begin{array}{c} 1 & 2 & - 4 & 1 \\ 2 & 3 & \frac{8}{1} & 0 \\ - \frac{1}{1} & \frac{9}{1} & \frac{10}{1} & 5 \end{array}]$

Step 2.7

Divide $8$ by $1$ .

$[\begin{array}{c} 1 & 2 & - 4 & 1 \\ 2 & 3 & 8 & 0 \\ - \frac{1}{1} & \frac{9}{1} & \frac{10}{1} & 5 \end{array}]$

Step 2.8

Cancel the common factor of $1$ .

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Step 2.8.1

Cancel the common factor.

$[\begin{array}{c} 1 & 2 & - 4 & 1 \\ 2 & 3 & 8 & 0 \\ - \frac{1}{1} & \frac{9}{1} & \frac{10}{1} & 5 \end{array}]$

Step 2.8.2

Rewrite the expression.

$[\begin{array}{c} 1 & 2 & - 4 & 1 \\ 2 & 3 & 8 & 0 \\ - 1 \cdot 1 & \frac{9}{1} & \frac{10}{1} & 5 \end{array}]$

Step 2.9

Multiply $- 1$ by $1$ .

$[\begin{array}{c} 1 & 2 & - 4 & 1 \\ 2 & 3 & 8 & 0 \\ - 1 & \frac{9}{1} & \frac{10}{1} & 5 \end{array}]$

Step 2.10

Divide $9$ by $1$ .

$[\begin{array}{c} 1 & 2 & - 4 & 1 \\ 2 & 3 & 8 & 0 \\ - 1 & 9 & \frac{10}{1} & 5 \end{array}]$

Step 2.11

Divide $10$ by $1$ .

$[\begin{array}{c} 1 & 2 & - 4 & 1 \\ 2 & 3 & 8 & 0 \\ - 1 & 9 & 10 & 5 \end{array}]$

Step 2.12

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 2.12.1

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

$[\begin{array}{c} 1 & 2 & - 4 & 1 \\ 2 - 2 \cdot 1 & 3 - 2 \cdot 2 & 8 - 2 \cdot - 4 & 0 - 2 \cdot 1 \\ - 1 & 9 & 10 & 5 \end{array}]$

Step 2.12.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 2 & - 4 & 1 \\ 0 & - 1 & 16 & - 2 \\ - 1 & 9 & 10 & 5 \end{array}]$

Step 2.13

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

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Step 2.13.1

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

$[\begin{array}{c} 1 & 2 & - 4 & 1 \\ 0 & - 1 & 16 & - 2 \\ - 1 + 1 \cdot 1 & 9 + 1 \cdot 2 & 10 - 4 & 5 + 1 \cdot 1 \end{array}]$

Step 2.13.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 2 & - 4 & 1 \\ 0 & - 1 & 16 & - 2 \\ 0 & 11 & 6 & 6 \end{array}]$

Step 2.14

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

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Step 2.14.1

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

$[\begin{array}{c} 1 & 2 & - 4 & 1 \\ - 0 & - - 1 & - 1 \cdot 16 & - - 2 \\ 0 & 11 & 6 & 6 \end{array}]$

Step 2.14.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 2 & - 4 & 1 \\ 0 & 1 & - 16 & 2 \\ 0 & 11 & 6 & 6 \end{array}]$

Step 2.15

Perform the row operation $R_{3} = R_{3} - 11 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 2.15.1

Perform the row operation $R_{3} = R_{3} - 11 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & 2 & - 4 & 1 \\ 0 & 1 & - 16 & 2 \\ 0 - 11 \cdot 0 & 11 - 11 \cdot 1 & 6 - 11 \cdot - 16 & 6 - 11 \cdot 2 \end{array}]$

Step 2.15.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 2 & - 4 & 1 \\ 0 & 1 & - 16 & 2 \\ 0 & 0 & 182 & - 16 \end{array}]$

Step 2.16

Multiply each element of $R_{3}$ by $\frac{1}{182}$ to make the entry at $3, 3$ a $1$ .

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Step 2.16.1

Multiply each element of $R_{3}$ by $\frac{1}{182}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 2 & - 4 & 1 \\ 0 & 1 & - 16 & 2 \\ \frac{0}{182} & \frac{0}{182} & \frac{182}{182} & \frac{- 16}{182} \end{array}]$

Step 2.16.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 2 & - 4 & 1 \\ 0 & 1 & - 16 & 2 \\ 0 & 0 & 1 & - \frac{8}{91} \end{array}]$

Step 2.17

Perform the row operation $R_{2} = R_{2} + 16 R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 2.17.1

Perform the row operation $R_{2} = R_{2} + 16 R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & 2 & - 4 & 1 \\ 0 + 16 \cdot 0 & 1 + 16 \cdot 0 & - 16 + 16 \cdot 1 & 2 + 16 (- \frac{8}{91}) \\ 0 & 0 & 1 & - \frac{8}{91} \end{array}]$

Step 2.17.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 2 & - 4 & 1 \\ 0 & 1 & 0 & \frac{54}{91} \\ 0 & 0 & 1 & - \frac{8}{91} \end{array}]$

Step 2.18

Perform the row operation $R_{1} = R_{1} + 4 R_{3}$ to make the entry at $1, 3$ a $0$ .

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Step 2.18.1

Perform the row operation $R_{1} = R_{1} + 4 R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 + 4 \cdot 0 & 2 + 4 \cdot 0 & - 4 + 4 \cdot 1 & 1 + 4 (- \frac{8}{91}) \\ 0 & 1 & 0 & \frac{54}{91} \\ 0 & 0 & 1 & - \frac{8}{91} \end{array}]$

Step 2.18.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 2 & 0 & \frac{59}{91} \\ 0 & 1 & 0 & \frac{54}{91} \\ 0 & 0 & 1 & - \frac{8}{91} \end{array}]$

Step 2.19

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 2.19.1

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

$[\begin{array}{c} 1 - 2 \cdot 0 & 2 - 2 \cdot 1 & 0 - 2 \cdot 0 & \frac{59}{91} - 2 (\frac{54}{91}) \\ 0 & 1 & 0 & \frac{54}{91} \\ 0 & 0 & 1 & - \frac{8}{91} \end{array}]$

Step 2.19.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & - \frac{7}{13} \\ 0 & 1 & 0 & \frac{54}{91} \\ 0 & 0 & 1 & - \frac{8}{91} \end{array}]$

Step 3

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

$x = - \frac{7}{13}$

$y = \frac{54}{91}$

$z = - \frac{8}{91}$

Step 4

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

$(- \frac{7}{13}, \frac{54}{91}, - \frac{8}{91})$

Step 5

Decompose a solution vector by re-arranging each equation represented in the row-reduced form of the augmented matrix by solving for the dependent variable in each row yields the vector equality.

$X = [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} - \frac{7}{13} \\ \frac{54}{91} \\ - \frac{8}{91} \end{array}]$