Linear Algebra Examples

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

Step 1

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

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

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.

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

Step 3

Find the reduced row echelon form.

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Step 3.1

Write the vectors as a matrix.

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

Step 3.2

Write as an augmented matrix for $A x = [\begin{array}{c} - 1 \\ 3 \end{array}]$ .

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

Step 3.3

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

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Step 3.3.1

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

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

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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