Linear Algebra Examples

B = [\begin{array}{c} 9 & - 6 \\ - 5 & 2 \end{array}]

$B = [\begin{array}{c} 9 & - 6 \\ - 5 & 2 \end{array}]$

Step 1

Find the reduced row echelon form.

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Step 1.1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{9}

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

1, 1

$1, 1$ a

1

$1$ .

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Step 1.1.1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{9}

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

1, 1

$1, 1$ a

1

$1$ .

[\begin{array}{c} \frac{9}{9} & \frac{- 6}{9} \\ - 5 & 2 \end{array}]

$[\begin{array}{c} \frac{9}{9} & \frac{- 6}{9} \\ - 5 & 2 \end{array}]$

Step 1.1.2

Simplify

R_{1}

$R_{1}$ .

[\begin{array}{c} 1 & - \frac{2}{3} \\ - 5 & 2 \end{array}]

$[\begin{array}{c} 1 & - \frac{2}{3} \\ - 5 & 2 \end{array}]$

[\begin{array}{c} 1 & - \frac{2}{3} \\ - 5 & 2 \end{array}]

$[\begin{array}{c} 1 & - \frac{2}{3} \\ - 5 & 2 \end{array}]$

Step 1.2

Perform the row operation

R_{2} = R_{2} + 5 R_{1}

$R_{2} = R_{2} + 5 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

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Step 1.2.1

Perform the row operation

R_{2} = R_{2} + 5 R_{1}

$R_{2} = R_{2} + 5 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

[\begin{array}{c} 1 & - \frac{2}{3} \\ - 5 + 5 \cdot 1 & 2 + 5 (- \frac{2}{3}) \end{array}]

$[\begin{array}{c} 1 & - \frac{2}{3} \\ - 5 + 5 \cdot 1 & 2 + 5 (- \frac{2}{3}) \end{array}]$

Step 1.2.2

Simplify

R_{2}

$R_{2}$ .

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

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

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

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

Step 1.3

Multiply each element of

R_{2}

$R_{2}$ by

- \frac{3}{4}

$- \frac{3}{4}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

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Step 1.3.1

Multiply each element of

R_{2}

$R_{2}$ by

- \frac{3}{4}

$- \frac{3}{4}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

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

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

Step 1.3.2

Simplify

R_{2}

$R_{2}$ .

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

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

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

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

Step 1.4

Perform the row operation

R_{1} = R_{1} + \frac{2}{3} R_{2}

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

1, 2

$1, 2$ a

0

$0$ .

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Step 1.4.1

Perform the row operation

R_{1} = R_{1} + \frac{2}{3} R_{2}

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

1, 2

$1, 2$ a

0

$0$ .

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

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

Step 1.4.2

Simplify

R_{1}

$R_{1}$ .

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

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

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

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

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

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

Step 2

The pivot positions are the locations with the leading

1

$1$ in each row. The pivot columns are the columns that have a pivot position.

Pivot Positions:

b_{11}

$b_{11}$ and

b_{22}

$b_{22}$

Pivot Columns:

1

$1$ and

2

$2$

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