Linear Algebra Examples

S = {[\begin{matrix} 11 \end{matrix}], [\begin{matrix} 1 - 1 \end{matrix}]}

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

v = [\begin{matrix} - 1 3 \end{matrix}]

$v = [\begin{array}{c} - 1 \\ 3 \end{array}]$

Step 1

S = {[\begin{matrix} 11 \end{matrix}], [\begin{matrix} 1 - 1 \end{matrix}]}

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

v = [\begin{matrix} - 1 3 \end{matrix}]

$v = [\begin{array}{c} - 1 \\ 3 \end{array}]$

Assign the set the name

S

$S$ and the vector the name

v

$v$ .

Step 2

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

a [\begin{matrix} 11 \end{matrix}] + b [\begin{matrix} 1 - 1 \end{matrix}] = [\begin{matrix} - 1 3 \end{matrix}]

$a [\begin{array}{c} 1 \\ 1 \end{array}] + b [\begin{array}{c} 1 \\ - 1 \end{array}] = [\begin{array}{c} - 1 \\ 3 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Write the vectors as a matrix.

[\begin{matrix} 1 & 1 1 & - 1 \end{matrix}]

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

Step 3.2

Write as an augmented matrix for

A x = [\begin{matrix} - 1 3 \end{matrix}]

$A x = [\begin{array}{c} - 1 \\ 3 \end{array}]$ .

[\begin{matrix} 1 & 1 & - 1 1 & - 1 & 3 \end{matrix}]

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

Step 3.3

Perform the row operation

R_{2} = R_{2} - R_{1}

$R_{2} = R_{2} - R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

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Step 3.3.1

Perform the row operation

R_{2} = R_{2} - R_{1}

$R_{2} = R_{2} - R_{1}$ to make the entry at

$2, 1$ a

$0$ .

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

Step 3.3.2

Simplify

$R_{2}$ .

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

Step 3.4

Multiply each element of

$R_{2}$ by

$- \frac{1}{2}$ to make the entry at

$2, 2$ a

$1$ .

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Step 3.4.1

Multiply each element of

$R_{2}$ by

$- \frac{1}{2}$ to make the entry at

$2, 2$ a

$1$ .

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

Step 3.4.2

Simplify

$R_{2}$ .

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

Step 3.5

Perform the row operation

$R_{1} = R_{1} - R_{2}$ to make the entry at

$1, 2$ a

$0$ .

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Step 3.5.1

Perform the row operation

$R_{1} = R_{1} - R_{2}$ to make the entry at

$1, 2$ a

$0$ .

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

Step 3.5.2

Simplify

$R_{1}$ .

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

Step 4

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

$v \in ⟨ S ⟩$

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