Linear Algebra Examples

$[\begin{array}{c} 17 & 22 & 4 \\ 2 & 9 & 11 \\ 4 & 6 & 13 \end{array}]$

Step 1

Nullity is the dimension of the null space, which is the same as the number of free variables in the system after row reducing. The free variables are the columns without pivot positions.

Step 2

Find the reduced row echelon form.

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Step 2.1

Multiply each element of $R_{1}$ by $\frac{1}{17}$ to make the entry at $1, 1$ a $1$ .

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Step 2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{17}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{17}{17} & \frac{22}{17} & \frac{4}{17} \\ 2 & 9 & 11 \\ 4 & 6 & 13 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{22}{17} & \frac{4}{17} \\ 2 & 9 & 11 \\ 4 & 6 & 13 \end{array}]$

Step 2.2

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 2.2.1

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

$[\begin{array}{c} 1 & \frac{22}{17} & \frac{4}{17} \\ 2 - 2 \cdot 1 & 9 - 2 (\frac{22}{17}) & 11 - 2 (\frac{4}{17}) \\ 4 & 6 & 13 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{22}{17} & \frac{4}{17} \\ 0 & \frac{109}{17} & \frac{179}{17} \\ 4 & 6 & 13 \end{array}]$

Step 2.3

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 2.3.1

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

$[\begin{array}{c} 1 & \frac{22}{17} & \frac{4}{17} \\ 0 & \frac{109}{17} & \frac{179}{17} \\ 4 - 4 \cdot 1 & 6 - 4 (\frac{22}{17}) & 13 - 4 (\frac{4}{17}) \end{array}]$

Step 2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{22}{17} & \frac{4}{17} \\ 0 & \frac{109}{17} & \frac{179}{17} \\ 0 & \frac{14}{17} & \frac{205}{17} \end{array}]$

Step 2.4

Multiply each element of $R_{2}$ by $\frac{17}{109}$ to make the entry at $2, 2$ a $1$ .

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Step 2.4.1

Multiply each element of $R_{2}$ by $\frac{17}{109}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{22}{17} & \frac{4}{17} \\ \frac{17}{109} \cdot 0 & \frac{17}{109} \cdot \frac{109}{17} & \frac{17}{109} \cdot \frac{179}{17} \\ 0 & \frac{14}{17} & \frac{205}{17} \end{array}]$

Step 2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{22}{17} & \frac{4}{17} \\ 0 & 1 & \frac{179}{109} \\ 0 & \frac{14}{17} & \frac{205}{17} \end{array}]$

Step 2.5

Perform the row operation $R_{3} = R_{3} - \frac{14}{17} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 2.5.1

Perform the row operation $R_{3} = R_{3} - \frac{14}{17} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{22}{17} & \frac{4}{17} \\ 0 & 1 & \frac{179}{109} \\ 0 - \frac{14}{17} \cdot 0 & \frac{14}{17} - \frac{14}{17} \cdot 1 & \frac{205}{17} - \frac{14}{17} \cdot \frac{179}{109} \end{array}]$

Step 2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{22}{17} & \frac{4}{17} \\ 0 & 1 & \frac{179}{109} \\ 0 & 0 & \frac{1167}{109} \end{array}]$

Step 2.6

Multiply each element of $R_{3}$ by $\frac{109}{1167}$ to make the entry at $3, 3$ a $1$ .

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Step 2.6.1

Multiply each element of $R_{3}$ by $\frac{109}{1167}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{22}{17} & \frac{4}{17} \\ 0 & 1 & \frac{179}{109} \\ \frac{109}{1167} \cdot 0 & \frac{109}{1167} \cdot 0 & \frac{109}{1167} \cdot \frac{1167}{109} \end{array}]$

Step 2.6.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{22}{17} & \frac{4}{17} \\ 0 & 1 & \frac{179}{109} \\ 0 & 0 & 1 \end{array}]$

Step 2.7

Perform the row operation $R_{2} = R_{2} - \frac{179}{109} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 2.7.1

Perform the row operation $R_{2} = R_{2} - \frac{179}{109} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & \frac{22}{17} & \frac{4}{17} \\ 0 - \frac{179}{109} \cdot 0 & 1 - \frac{179}{109} \cdot 0 & \frac{179}{109} - \frac{179}{109} \cdot 1 \\ 0 & 0 & 1 \end{array}]$

Step 2.7.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{22}{17} & \frac{4}{17} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$

Step 2.8

Perform the row operation $R_{1} = R_{1} - \frac{4}{17} R_{3}$ to make the entry at $1, 3$ a $0$ .

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Step 2.8.1

Perform the row operation $R_{1} = R_{1} - \frac{4}{17} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - \frac{4}{17} \cdot 0 & \frac{22}{17} - \frac{4}{17} \cdot 0 & \frac{4}{17} - \frac{4}{17} \cdot 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$

Step 2.8.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{22}{17} & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$

Step 2.9

Perform the row operation $R_{1} = R_{1} - \frac{22}{17} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 2.9.1

Perform the row operation $R_{1} = R_{1} - \frac{22}{17} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{22}{17} \cdot 0 & \frac{22}{17} - \frac{22}{17} \cdot 1 & 0 - \frac{22}{17} \cdot 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$

Step 2.9.2

Simplify $R_{1}$ .

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

Step 3

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: $a_{11}, a_{22},$ and $a_{33}$

Pivot Columns: $1, 2,$ and $3$

Step 4

The nullity is the number of columns without a pivot position in the row reduced matrix.

$0$