Algebra Examples

[\begin{matrix} - 2 & - 9 - 1 & 0 \end{matrix}]

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

Step 1

Find the reduced row echelon form.

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Step 1.1

Multiply each element of

$R_{1}$ by

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

$1, 1$ a

$1$ .

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Step 1.1.1

Multiply each element of

$R_{1}$ by

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

$1, 1$ a

$1$ .

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

Step 1.1.2

Simplify

$R_{1}$ .

$[\begin{array}{c} 1 & \frac{9}{2} \\ - 1 & 0 \end{array}]$

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

$0$ .

$[\begin{array}{c} 1 & \frac{9}{2} \\ - 1 + 1 \cdot 1 & 0 + \frac{9}{2} \end{array}]$

Step 1.2.2

Simplify

$R_{2}$ .

$[\begin{array}{c} 1 & \frac{9}{2} \\ 0 & \frac{9}{2} \end{array}]$

Step 1.3

Multiply each element of

$R_{2}$ by

$\frac{2}{9}$ to make the entry at

$2, 2$ a

$1$ .

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Step 1.3.1

Multiply each element of

$R_{2}$ by

$\frac{2}{9}$ to make the entry at

$2, 2$ a

$1$ .

$[\begin{array}{c} 1 & \frac{9}{2} \\ \frac{2}{9} \cdot 0 & \frac{2}{9} \cdot \frac{9}{2} \end{array}]$

Step 1.3.2

Simplify

$R_{2}$ .

$[\begin{array}{c} 1 & \frac{9}{2} \\ 0 & 1 \end{array}]$

Step 1.4

Perform the row operation

$R_{1} = R_{1} - \frac{9}{2} R_{2}$ to make the entry at

$1, 2$ a

$0$ .

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Step 1.4.1

Perform the row operation

$R_{1} = R_{1} - \frac{9}{2} R_{2}$ to make the entry at

$1, 2$ a

$0$ .

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

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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