Linear Algebra Examples

$S = {[\begin{array}{c} - 7 \\ 5 \\ 1 \end{array}], [\begin{array}{c} - 6 \\ 5 \\ 0 \end{array}]}$

Step 1

Write as an augmented matrix for $A x = 0$ .

$[\begin{array}{c} - 7 & - 6 & 0 \\ 5 & 5 & 0 \\ 1 & 0 & 0 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

Multiply each element of $R_{1}$ by $- \frac{1}{7}$ to make the entry at $1, 1$ a $1$ .

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Step 2.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{7}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{7} \cdot - 7 & - \frac{1}{7} \cdot - 6 & - \frac{1}{7} \cdot 0 \\ 5 & 5 & 0 \\ 1 & 0 & 0 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{6}{7} & 0 \\ 5 & 5 & 0 \\ 1 & 0 & 0 \end{array}]$

Step 2.2

Perform the row operation $R_{2} = R_{2} - 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 2.2.1

Perform the row operation $R_{2} = R_{2} - 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{6}{7} & 0 \\ 5 - 5 \cdot 1 & 5 - 5 (\frac{6}{7}) & 0 - 5 \cdot 0 \\ 1 & 0 & 0 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{6}{7} & 0 \\ 0 & \frac{5}{7} & 0 \\ 1 & 0 & 0 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & \frac{6}{7} & 0 \\ 0 & \frac{5}{7} & 0 \\ 1 - 1 & 0 - \frac{6}{7} & 0 - 0 \end{array}]$

Step 2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{6}{7} & 0 \\ 0 & \frac{5}{7} & 0 \\ 0 & - \frac{6}{7} & 0 \end{array}]$

Step 2.4

Multiply each element of $R_{2}$ by $\frac{7}{5}$ to make the entry at $2, 2$ a $1$ .

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Step 2.4.1

Multiply each element of $R_{2}$ by $\frac{7}{5}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{6}{7} & 0 \\ \frac{7}{5} \cdot 0 & \frac{7}{5} \cdot \frac{5}{7} & \frac{7}{5} \cdot 0 \\ 0 & - \frac{6}{7} & 0 \end{array}]$

Step 2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{6}{7} & 0 \\ 0 & 1 & 0 \\ 0 & - \frac{6}{7} & 0 \end{array}]$

Step 2.5

Perform the row operation $R_{3} = R_{3} + \frac{6}{7} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 2.5.1

Perform the row operation $R_{3} = R_{3} + \frac{6}{7} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{6}{7} & 0 \\ 0 & 1 & 0 \\ 0 + \frac{6}{7} \cdot 0 & - \frac{6}{7} + \frac{6}{7} \cdot 1 & 0 + \frac{6}{7} \cdot 0 \end{array}]$

Step 2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{6}{7} & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}]$

Step 2.6

Perform the row operation $R_{1} = R_{1} - \frac{6}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 2.6.1

Perform the row operation $R_{1} = R_{1} - \frac{6}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{6}{7} \cdot 0 & \frac{6}{7} - \frac{6}{7} \cdot 1 & 0 - \frac{6}{7} \cdot 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}]$

Step 2.6.2

Simplify $R_{1}$ .

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

Step 3

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

$x = 0$

$y = 0$

$0 = 0$

Step 4

Write a solution vector by solving in terms of the free variables in each row.

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 0 \\ 0 \end{array}]$

Step 5

Write as a solution set.

${[\begin{array}{c} 0 \\ 0 \end{array}]}$

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