Linear Algebra Examples

$[\begin{array}{c} 0 & 1 & 2 & 3 \\ 10 & 11 & 12 & 13 \\ 20 & 21 & 22 & 23 \\ 30 & 31 & 32 & 33 \end{array}]$

Step 1

Find the reduced row echelon form.

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Step 1.1

Swap $R_{2}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

$[\begin{array}{c} 10 & 11 & 12 & 13 \\ 0 & 1 & 2 & 3 \\ 20 & 21 & 22 & 23 \\ 30 & 31 & 32 & 33 \end{array}]$

Step 1.2

Multiply each element of $R_{1}$ by $\frac{1}{10}$ to make the entry at $1, 1$ a $1$ .

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Step 1.2.1

Multiply each element of $R_{1}$ by $\frac{1}{10}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{10}{10} & \frac{11}{10} & \frac{12}{10} & \frac{13}{10} \\ 0 & 1 & 2 & 3 \\ 20 & 21 & 22 & 23 \\ 30 & 31 & 32 & 33 \end{array}]$

Step 1.2.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{11}{10} & \frac{6}{5} & \frac{13}{10} \\ 0 & 1 & 2 & 3 \\ 20 & 21 & 22 & 23 \\ 30 & 31 & 32 & 33 \end{array}]$

Step 1.3

Perform the row operation $R_{3} = R_{3} - 20 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 1.3.1

Perform the row operation $R_{3} = R_{3} - 20 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{11}{10} & \frac{6}{5} & \frac{13}{10} \\ 0 & 1 & 2 & 3 \\ 20 - 20 \cdot 1 & 21 - 20 (\frac{11}{10}) & 22 - 20 (\frac{6}{5}) & 23 - 20 (\frac{13}{10}) \\ 30 & 31 & 32 & 33 \end{array}]$

Step 1.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{11}{10} & \frac{6}{5} & \frac{13}{10} \\ 0 & 1 & 2 & 3 \\ 0 & - 1 & - 2 & - 3 \\ 30 & 31 & 32 & 33 \end{array}]$

Step 1.4

Perform the row operation $R_{4} = R_{4} - 30 R_{1}$ to make the entry at $4, 1$ a $0$ .

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Step 1.4.1

Perform the row operation $R_{4} = R_{4} - 30 R_{1}$ to make the entry at $4, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{11}{10} & \frac{6}{5} & \frac{13}{10} \\ 0 & 1 & 2 & 3 \\ 0 & - 1 & - 2 & - 3 \\ 30 - 30 \cdot 1 & 31 - 30 (\frac{11}{10}) & 32 - 30 (\frac{6}{5}) & 33 - 30 (\frac{13}{10}) \end{array}]$

Step 1.4.2

Simplify $R_{4}$ .

$[\begin{array}{c} 1 & \frac{11}{10} & \frac{6}{5} & \frac{13}{10} \\ 0 & 1 & 2 & 3 \\ 0 & - 1 & - 2 & - 3 \\ 0 & - 2 & - 4 & - 6 \end{array}]$

Step 1.5

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

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Step 1.5.1

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

$[\begin{array}{c} 1 & \frac{11}{10} & \frac{6}{5} & \frac{13}{10} \\ 0 & 1 & 2 & 3 \\ 0 + 0 & - 1 + 1 \cdot 1 & - 2 + 1 \cdot 2 & - 3 + 1 \cdot 3 \\ 0 & - 2 & - 4 & - 6 \end{array}]$

Step 1.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{11}{10} & \frac{6}{5} & \frac{13}{10} \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \\ 0 & - 2 & - 4 & - 6 \end{array}]$

Step 1.6

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

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Step 1.6.1

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

$[\begin{array}{c} 1 & \frac{11}{10} & \frac{6}{5} & \frac{13}{10} \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \\ 0 + 2 \cdot 0 & - 2 + 2 \cdot 1 & - 4 + 2 \cdot 2 & - 6 + 2 \cdot 3 \end{array}]$

Step 1.6.2

Simplify $R_{4}$ .

$[\begin{array}{c} 1 & \frac{11}{10} & \frac{6}{5} & \frac{13}{10} \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 1.7

Perform the row operation $R_{1} = R_{1} - \frac{11}{10} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 1.7.1

Perform the row operation $R_{1} = R_{1} - \frac{11}{10} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{11}{10} \cdot 0 & \frac{11}{10} - \frac{11}{10} \cdot 1 & \frac{6}{5} - \frac{11}{10} \cdot 2 & \frac{13}{10} - \frac{11}{10} \cdot 3 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 1.7.2

Simplify $R_{1}$ .

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

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 rank is the number of pivot columns.

$2$