Linear Algebra Examples

⎡ ⎢ ⎣ \begin{matrix} - 6 & 7 2 & 6 - 4 & 1 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} - 6 & 7 \\ 2 & 6 \\ - 4 & 1 \end{array}]$

Step 1

Multiply each element of

R_{1}

$R_{1}$ by

- \frac{1}{6}

$- \frac{1}{6}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

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Step 1.1

Multiply each element of

R_{1}

$R_{1}$ by

- \frac{1}{6}

$- \frac{1}{6}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} - \frac{1}{6} \cdot - 6 & - \frac{1}{6} \cdot 7 2 & 6 - 4 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} - \frac{1}{6} \cdot - 6 & - \frac{1}{6} \cdot 7 \\ 2 & 6 \\ - 4 & 1 \end{array}]$

Step 1.2

Simplify

R_{1}

$R_{1}$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - \frac{7}{6} 2 & 6 - 4 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & - \frac{7}{6} \\ 2 & 6 \\ - 4 & 1 \end{array}]$

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - \frac{7}{6} 2 & 6 - 4 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & - \frac{7}{6} \\ 2 & 6 \\ - 4 & 1 \end{array}]$

Step 2

Perform the row operation

R_{2} = R_{2} - 2 R_{1}

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

2, 1

$2, 1$ a

0

$0$ .

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Step 2.1

Perform the row operation

R_{2} = R_{2} - 2 R_{1}

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

2, 1

$2, 1$ a

0

$0$ .

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & - \frac{7}{6} 2 - 2 \cdot 1 & 6 - 2 (- \frac{7}{6}) - 4 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & - \frac{7}{6} \\ 2 - 2 \cdot 1 & 6 - 2 (- \frac{7}{6}) \\ - 4 & 1 \end{array}]$

Step 2.2

Simplify

R_{2}

$R_{2}$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - \frac{7}{6} 0 & \frac{25}{3} - 4 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & - \frac{7}{6} \\ 0 & \frac{25}{3} \\ - 4 & 1 \end{array}]$

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - \frac{7}{6} 0 & \frac{25}{3} - 4 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & - \frac{7}{6} \\ 0 & \frac{25}{3} \\ - 4 & 1 \end{array}]$

Step 3

Perform the row operation

R_{3} = R_{3} + 4 R_{1}

$R_{3} = R_{3} + 4 R_{1}$ to make the entry at

3, 1

$3, 1$ a

0

$0$ .

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Step 3.1

Perform the row operation

R_{3} = R_{3} + 4 R_{1}

$R_{3} = R_{3} + 4 R_{1}$ to make the entry at

3, 1

$3, 1$ a

0

$0$ .

⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & - \frac{7}{6} 0 & \frac{25}{3} - 4 + 4 \cdot 1 & 1 + 4 (- \frac{7}{6}) \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & - \frac{7}{6} \\ 0 & \frac{25}{3} \\ - 4 + 4 \cdot 1 & 1 + 4 (- \frac{7}{6}) \end{array}]$

Step 3.2

Simplify

R_{3}

$R_{3}$ .

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & - \frac{7}{6} 0 & \frac{25}{3} 0 & - \frac{11}{3} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & - \frac{7}{6} \\ 0 & \frac{25}{3} \\ 0 & - \frac{11}{3} \end{array}]$

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & - \frac{7}{6} 0 & \frac{25}{3} 0 & - \frac{11}{3} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & - \frac{7}{6} \\ 0 & \frac{25}{3} \\ 0 & - \frac{11}{3} \end{array}]$

Step 4

Multiply each element of

R_{2}

$R_{2}$ by

\frac{3}{25}

$\frac{3}{25}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

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Step 4.1

Multiply each element of

R_{2}

$R_{2}$ by

\frac{3}{25}

$\frac{3}{25}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & - \frac{7}{6} \frac{3}{25} \cdot 0 & \frac{3}{25} \cdot \frac{25}{3} 0 & - \frac{11}{3} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & - \frac{7}{6} \\ \frac{3}{25} \cdot 0 & \frac{3}{25} \cdot \frac{25}{3} \\ 0 & - \frac{11}{3} \end{array}]$

Step 4.2

Simplify

R_{2}

$R_{2}$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - \frac{7}{6} 0 & 1 0 & - \frac{11}{3} \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & - \frac{7}{6} \\ 0 & 1 \\ 0 & - \frac{11}{3} \end{array}]$

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - \frac{7}{6} 0 & 1 0 & - \frac{11}{3} \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & - \frac{7}{6} \\ 0 & 1 \\ 0 & - \frac{11}{3} \end{array}]$

Step 5

Perform the row operation

R_{3} = R_{3} + \frac{11}{3} R_{2}

$R_{3} = R_{3} + \frac{11}{3} R_{2}$ to make the entry at

3, 2

$3, 2$ a

0

$0$ .

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Step 5.1

Perform the row operation

R_{3} = R_{3} + \frac{11}{3} R_{2}

$R_{3} = R_{3} + \frac{11}{3} R_{2}$ to make the entry at

3, 2

$3, 2$ a

0

$0$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - \frac{7}{6} 0 & 1 0 + \frac{11}{3} \cdot 0 & - \frac{11}{3} + \frac{11}{3} \cdot 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & - \frac{7}{6} \\ 0 & 1 \\ 0 + \frac{11}{3} \cdot 0 & - \frac{11}{3} + \frac{11}{3} \cdot 1 \end{array}]$

Step 5.2

Simplify

R_{3}

$R_{3}$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - \frac{7}{6} 0 & 1 0 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & - \frac{7}{6} \\ 0 & 1 \\ 0 & 0 \end{array}]$

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - \frac{7}{6} 0 & 1 0 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & - \frac{7}{6} \\ 0 & 1 \\ 0 & 0 \end{array}]$

Step 6

Perform the row operation

R_{1} = R_{1} + \frac{7}{6} R_{2}

$R_{1} = R_{1} + \frac{7}{6} R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

Tap for more steps...

Step 6.1

Perform the row operation

R_{1} = R_{1} + \frac{7}{6} R_{2}

$R_{1} = R_{1} + \frac{7}{6} R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 + \frac{7}{6} \cdot 0 & - \frac{7}{6} + \frac{7}{6} \cdot 1 0 & 1 0 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 + \frac{7}{6} \cdot 0 & - \frac{7}{6} + \frac{7}{6} \cdot 1 \\ 0 & 1 \\ 0 & 0 \end{array}]$

Step 6.2

Simplify

$R_{1}$ .

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