Algebra Examples

$[\begin{array}{c} 1 & 4 & 3 \\ 3 & 7 & - 1 \\ - 2 & 1 & 12 \end{array}]$

Step 1

Find the reduced row echelon form.

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Step 1.1

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

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Step 1.1.1

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

$[\begin{array}{c} 1 & 4 & 3 \\ 3 - 3 \cdot 1 & 7 - 3 \cdot 4 & - 1 - 3 \cdot 3 \\ - 2 & 1 & 12 \end{array}]$

Step 1.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 4 & 3 \\ 0 & - 5 & - 10 \\ - 2 & 1 & 12 \end{array}]$

Step 1.2

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

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Step 1.2.1

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

$[\begin{array}{c} 1 & 4 & 3 \\ 0 & - 5 & - 10 \\ - 2 + 2 \cdot 1 & 1 + 2 \cdot 4 & 12 + 2 \cdot 3 \end{array}]$

Step 1.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 4 & 3 \\ 0 & - 5 & - 10 \\ 0 & 9 & 18 \end{array}]$

Step 1.3

Multiply each element of $R_{2}$ by $- \frac{1}{5}$ to make the entry at $2, 2$ a $1$ .

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Step 1.3.1

Multiply each element of $R_{2}$ by $- \frac{1}{5}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 4 & 3 \\ - \frac{1}{5} \cdot 0 & - \frac{1}{5} \cdot - 5 & - \frac{1}{5} \cdot - 10 \\ 0 & 9 & 18 \end{array}]$

Step 1.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 4 & 3 \\ 0 & 1 & 2 \\ 0 & 9 & 18 \end{array}]$

Step 1.4

Perform the row operation $R_{3} = R_{3} - 9 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 1.4.1

Perform the row operation $R_{3} = R_{3} - 9 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & 4 & 3 \\ 0 & 1 & 2 \\ 0 - 9 \cdot 0 & 9 - 9 \cdot 1 & 18 - 9 \cdot 2 \end{array}]$

Step 1.4.2

Simplify $R_{3}$ .

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

Step 1.5

Perform the row operation $R_{1} = R_{1} - 4 R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 1.5.1

Perform the row operation $R_{1} = R_{1} - 4 R_{2}$ to make the entry at $1, 2$ a $0$ .

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

Step 1.5.2

Simplify $R_{1}$ .

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

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 $Col (A)$ : ${[\begin{array}{c} 1 \\ 3 \\ - 2 \end{array}], [\begin{array}{c} 4 \\ 7 \\ 1 \end{array}]}$

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

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