Precalculus Examples

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

Step 1

Find the reduced row echelon form.

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Step 1.1

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

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Step 1.1.1

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

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

Step 1.1.2

Simplify $R_{1}$ .

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

Step 1.2

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

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Step 1.2.1

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

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

Step 1.2.2

Simplify $R_{3}$ .

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

Simplify $R_{2}$ .

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

Step 1.4

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

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Step 1.4.1

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

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

Step 1.4.2

Simplify $R_{3}$ .

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

Step 1.5

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

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Step 1.5.1

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

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

Step 1.5.2

Simplify $R_{1}$ .

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

Step 2

The pivot positions are the locations with the leading $1$ in each row. The pivot columns are the columns that have a pivot position.

Pivot Positions: $a_{11}$ and $a_{22}$

Pivot Columns: $1$ and $2$

Step 3

The basis for the column space of a matrix is formed by considering corresponding pivot columns in the original matrix. The dimension of $Col (A)$ is the number of vectors in a basis for $Col (A)$ .

Basis of $Col (A)$ : ${[\begin{array}{c} 3 \\ 0 \\ 1 \end{array}], [\begin{array}{c} - 1 \\ 2 \\ - 1 \end{array}]}$

Dimension of $Col (A)$ : $2$

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