Linear Algebra Examples

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

Step 1

Assign the set to the name $S$ to use throughout the problem.

$S = {[\begin{array}{c} 2 \\ - 2 \\ - 3 \end{array}], [\begin{array}{c} 1 \\ 3 \\ 5 \end{array}], [\begin{array}{c} 1 \\ 0 \\ 1 \end{array}]}$

Step 2

Create a matrix whose rows are the vectors in the spanning set.

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

Step 3

Find the reduced row echelon form of the matrix.

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Step 3.1

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

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Step 3.1.1

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

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

Step 3.1.2

Simplify $R_{1}$ .

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

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 $0$ .

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

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & - \frac{3}{2} \\ 0 & 4 & \frac{13}{2} \\ 1 & 0 & 1 \end{array}]$

Step 3.3

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

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Step 3.3.1

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

$[\begin{array}{c} 1 & - 1 & - \frac{3}{2} \\ 0 & 4 & \frac{13}{2} \\ 1 - 1 & 0 + 1 & 1 + \frac{3}{2} \end{array}]$

Step 3.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & - \frac{3}{2} \\ 0 & 4 & \frac{13}{2} \\ 0 & 1 & \frac{5}{2} \end{array}]$

Step 3.4

Multiply each element of $R_{2}$ by $\frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

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Step 3.4.1

Multiply each element of $R_{2}$ by $\frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 1 & - \frac{3}{2} \\ \frac{0}{4} & \frac{4}{4} & \frac{\frac{13}{2}}{4} \\ 0 & 1 & \frac{5}{2} \end{array}]$

Step 3.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & - \frac{3}{2} \\ 0 & 1 & \frac{13}{8} \\ 0 & 1 & \frac{5}{2} \end{array}]$

Step 3.5

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

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Step 3.5.1

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

$[\begin{array}{c} 1 & - 1 & - \frac{3}{2} \\ 0 & 1 & \frac{13}{8} \\ 0 - 0 & 1 - 1 & \frac{5}{2} - \frac{13}{8} \end{array}]$

Step 3.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & - \frac{3}{2} \\ 0 & 1 & \frac{13}{8} \\ 0 & 0 & \frac{7}{8} \end{array}]$

Step 3.6

Multiply each element of $R_{3}$ by $\frac{8}{7}$ to make the entry at $3, 3$ a $1$ .

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Step 3.6.1

Multiply each element of $R_{3}$ by $\frac{8}{7}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - 1 & - \frac{3}{2} \\ 0 & 1 & \frac{13}{8} \\ \frac{8}{7} \cdot 0 & \frac{8}{7} \cdot 0 & \frac{8}{7} \cdot \frac{7}{8} \end{array}]$

Step 3.6.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & - \frac{3}{2} \\ 0 & 1 & \frac{13}{8} \\ 0 & 0 & 1 \end{array}]$

Step 3.7

Perform the row operation $R_{2} = R_{2} - \frac{13}{8} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 3.7.1

Perform the row operation $R_{2} = R_{2} - \frac{13}{8} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & - 1 & - \frac{3}{2} \\ 0 - \frac{13}{8} \cdot 0 & 1 - \frac{13}{8} \cdot 0 & \frac{13}{8} - \frac{13}{8} \cdot 1 \\ 0 & 0 & 1 \end{array}]$

Step 3.7.2

Simplify $R_{2}$ .

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

Step 3.8

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

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Step 3.8.1

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

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

Step 3.8.2

Simplify $R_{1}$ .

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

Step 3.9

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

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Step 3.9.1

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

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

Step 3.9.2

Simplify $R_{1}$ .

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

Step 4

Convert the nonzero rows to column vectors to form the basis.

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

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