Linear Algebra Examples

$v = [\begin{array}{c} - 1 \\ 4 \\ 11 \end{array}]$ , $S = {[\begin{array}{c} 1 \\ 2 \\ - 4 \end{array}], [\begin{array}{c} - 3 \\ - 5 \\ 13 \end{array}], [\begin{array}{c} 2 \\ - 1 \\ - 12 \end{array}]}$

Step 1

$S = {[\begin{array}{c} 1 \\ 2 \\ - 4 \end{array}], [\begin{array}{c} - 3 \\ - 5 \\ 13 \end{array}], [\begin{array}{c} 2 \\ - 1 \\ - 12 \end{array}]}$

$v = [\begin{array}{c} - 1 \\ 4 \\ 11 \end{array}]$

Assign the set the name $S$ and the vector the name $v$ .

Step 2

Set up a linear relation to see if there is a non-trivial solution to the system.

$a [\begin{array}{c} 1 \\ 2 \\ - 4 \end{array}] + b [\begin{array}{c} - 3 \\ - 5 \\ 13 \end{array}] + d [\begin{array}{c} 2 \\ - 1 \\ - 12 \end{array}] = [\begin{array}{c} - 1 \\ 4 \\ 11 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Write the vectors as a matrix.

$[\begin{array}{c} 1 & - 3 & 2 \\ 2 & - 5 & - 1 \\ - 4 & 13 & - 12 \end{array}]$

Step 3.2

Write as an augmented matrix for $A x = [\begin{array}{c} - 1 \\ 4 \\ 11 \end{array}]$ .

$[\begin{array}{c} 1 & - 3 & 2 & - 1 \\ 2 & - 5 & - 1 & 4 \\ - 4 & 13 & - 12 & 11 \end{array}]$

Step 3.3

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 3.3.1

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

$[\begin{array}{c} 1 & - 3 & 2 & - 1 \\ 2 - 2 \cdot 1 & - 5 - 2 \cdot - 3 & - 1 - 2 \cdot 2 & 4 - 2 \cdot - 1 \\ - 4 & 13 & - 12 & 11 \end{array}]$

Step 3.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 3 & 2 & - 1 \\ 0 & 1 & - 5 & 6 \\ - 4 & 13 & - 12 & 11 \end{array}]$

Step 3.4

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

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Step 3.4.1

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

$[\begin{array}{c} 1 & - 3 & 2 & - 1 \\ 0 & 1 & - 5 & 6 \\ - 4 + 4 \cdot 1 & 13 + 4 \cdot - 3 & - 12 + 4 \cdot 2 & 11 + 4 \cdot - 1 \end{array}]$

Step 3.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 3 & 2 & - 1 \\ 0 & 1 & - 5 & 6 \\ 0 & 1 & - 4 & 7 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & - 3 & 2 & - 1 \\ 0 & 1 & - 5 & 6 \\ 0 - 0 & 1 - 1 & - 4 + 5 & 7 - 6 \end{array}]$

Step 3.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 3 & 2 & - 1 \\ 0 & 1 & - 5 & 6 \\ 0 & 0 & 1 & 1 \end{array}]$

Step 3.6

Perform the row operation $R_{2} = R_{2} + 5 R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 3.6.1

Perform the row operation $R_{2} = R_{2} + 5 R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & - 3 & 2 & - 1 \\ 0 + 5 \cdot 0 & 1 + 5 \cdot 0 & - 5 + 5 \cdot 1 & 6 + 5 \cdot 1 \\ 0 & 0 & 1 & 1 \end{array}]$

Step 3.6.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 3 & 2 & - 1 \\ 0 & 1 & 0 & 11 \\ 0 & 0 & 1 & 1 \end{array}]$

Step 3.7

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

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Step 3.7.1

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

$[\begin{array}{c} 1 - 2 \cdot 0 & - 3 - 2 \cdot 0 & 2 - 2 \cdot 1 & - 1 - 2 \cdot 1 \\ 0 & 1 & 0 & 11 \\ 0 & 0 & 1 & 1 \end{array}]$

Step 3.7.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 3 & 0 & - 3 \\ 0 & 1 & 0 & 11 \\ 0 & 0 & 1 & 1 \end{array}]$

Step 3.8

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

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Step 3.8.1

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

$[\begin{array}{c} 1 + 3 \cdot 0 & - 3 + 3 \cdot 1 & 0 + 3 \cdot 0 & - 3 + 3 \cdot 11 \\ 0 & 1 & 0 & 11 \\ 0 & 0 & 1 & 1 \end{array}]$

Step 3.8.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 30 \\ 0 & 1 & 0 & 11 \\ 0 & 0 & 1 & 1 \end{array}]$

Step 4

Since the resulting system is consistent, the vector is an element of the set.

$v \in ⟨ S ⟩$

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