Linear Algebra Examples

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

Step 1

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

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

Step 2

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

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

Step 3

Find the reduced row echelon form of the matrix.

Tap for more steps...

Step 3.1

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

Tap for more steps...

Step 3.1.1

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

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

Step 3.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 2 & 3 \\ 0 & 1 & 6 \\ 1 & 0 & - 1 \end{array}]$

Step 3.2

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

Tap for more steps...

Step 3.2.1

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

$[\begin{array}{c} 1 & 2 & 3 \\ 0 & 1 & 6 \\ 1 - 1 & 0 - 2 & - 1 - 3 \end{array}]$

Step 3.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & - 2 & - 4 \end{array}]$

Step 3.3

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

Tap for more steps...

Step 3.3.1

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

$[\begin{array}{c} 1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 + 2 \cdot 0 & - 2 + 2 \cdot 1 & - 4 + 2 \cdot 6 \end{array}]$

Step 3.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & 8 \end{array}]$

Step 3.4

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

Tap for more steps...

Step 3.4.1

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

$[\begin{array}{c} 1 & 2 & 3 \\ 0 & 1 & 6 \\ \frac{0}{8} & \frac{0}{8} & \frac{8}{8} \end{array}]$

Step 3.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & 1 \end{array}]$

Step 3.5

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

Tap for more steps...

Step 3.5.1

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

$[\begin{array}{c} 1 & 2 & 3 \\ 0 - 6 \cdot 0 & 1 - 6 \cdot 0 & 6 - 6 \cdot 1 \\ 0 & 0 & 1 \end{array}]$

Step 3.5.2

Simplify $R_{2}$ .

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

Step 3.6

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

Tap for more steps...

Step 3.6.1

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

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

Step 3.6.2

Simplify $R_{1}$ .

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

Step 3.7

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

Tap for more steps...

Step 3.7.1

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

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

Step 3.7.2

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.

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

Enter YOUR Problem