Linear Algebra Examples

[\begin{matrix} - 9 & - 6 1 & - 2 \end{matrix}]

$[\begin{array}{c} - 9 & - 6 \\ 1 & - 2 \end{array}]$

Step 1

Find the reduced row echelon form.

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Step 1.1

Multiply each element of

$R_{1}$ by

$- \frac{1}{9}$ to make the entry at

$1, 1$ a

$1$ .

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Step 1.1.1

Multiply each element of

$R_{1}$ by

$- \frac{1}{9}$ to make the entry at

$1, 1$ a

$1$ .

$[\begin{array}{c} - \frac{1}{9} \cdot - 9 & - \frac{1}{9} \cdot - 6 \\ 1 & - 2 \end{array}]$

Step 1.1.2

Simplify

$R_{1}$ .

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

Step 1.2

Perform the row operation

$R_{2} = R_{2} - R_{1}$ to make the entry at

$2, 1$ a

$0$ .

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Step 1.2.1

Perform the row operation

$R_{2} = R_{2} - R_{1}$ to make the entry at

$2, 1$ a

$0$ .

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

Step 1.2.2

Simplify

$R_{2}$ .

$[\begin{array}{c} 1 & \frac{2}{3} \\ 0 & - \frac{8}{3} \end{array}]$

Step 1.3

Multiply each element of

$R_{2}$ by

$- \frac{3}{8}$ to make the entry at

$2, 2$ a

$1$ .

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Step 1.3.1

Multiply each element of

$R_{2}$ by

$- \frac{3}{8}$ to make the entry at

$2, 2$ a

$1$ .

$[\begin{array}{c} 1 & \frac{2}{3} \\ - \frac{3}{8} \cdot 0 & - \frac{3}{8} (- \frac{8}{3}) \end{array}]$

Step 1.3.2

Simplify

$R_{2}$ .

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

Step 1.4

Perform the row operation

$R_{1} = R_{1} - \frac{2}{3} R_{2}$ to make the entry at

$1, 2$ a

$0$ .

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Step 1.4.1

Perform the row operation

$R_{1} = R_{1} - \frac{2}{3} R_{2}$ to make the entry at

$1, 2$ a

$0$ .

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

Step 1.4.2

Simplify

$R_{1}$ .

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

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

$2$

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