Linear Algebra Examples

⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 1 3 & - 1 & 0 6 & - 2 & 0 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 2 & 1 \\ 3 & - 1 & 0 \\ 6 & - 2 & 0 \end{array}]$

Step 1

To determine if the columns in the matrix are linearly dependent, determine if the equation

A x = 0

$A x = 0$ has any non-trivial solutions.

Step 2

Write as an augmented matrix for

A x = 0

$A x = 0$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & 1 & 0 3 & - 1 & 0 & 0 6 & - 2 & 0 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 2 & 1 & 0 \\ 3 & - 1 & 0 & 0 \\ 6 & - 2 & 0 & 0 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Perform the row operation

R_{2} = R_{2} - 3 R_{1}

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

2, 1

$2, 1$ a

0

$0$ .

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Step 3.1.1

Perform the row operation

R_{2} = R_{2} - 3 R_{1}

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

2, 1

$2, 1$ a

0

$0$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & 1 & 0 3 - 3 \cdot 1 & - 1 - 3 \cdot 2 & 0 - 3 \cdot 1 & 0 - 3 \cdot 0 6 & - 2 & 0 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 2 & 1 & 0 \\ 3 - 3 \cdot 1 & - 1 - 3 \cdot 2 & 0 - 3 \cdot 1 & 0 - 3 \cdot 0 \\ 6 & - 2 & 0 & 0 \end{array}]$

Step 3.1.2

Simplify

R_{2}

$R_{2}$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & 1 & 0 0 & - 7 & - 3 & 0 6 & - 2 & 0 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 2 & 1 & 0 \\ 0 & - 7 & - 3 & 0 \\ 6 & - 2 & 0 & 0 \end{array}]$

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & 1 & 0 0 & - 7 & - 3 & 0 6 & - 2 & 0 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 2 & 1 & 0 \\ 0 & - 7 & - 3 & 0 \\ 6 & - 2 & 0 & 0 \end{array}]$

Step 3.2

Perform the row operation

R_{3} = R_{3} - 6 R_{1}

$R_{3} = R_{3} - 6 R_{1}$ to make the entry at

3, 1

$3, 1$ a

0

$0$ .

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Step 3.2.1

Perform the row operation

R_{3} = R_{3} - 6 R_{1}

$R_{3} = R_{3} - 6 R_{1}$ to make the entry at

3, 1

$3, 1$ a

0

$0$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & 1 & 0 0 & - 7 & - 3 & 0 6 - 6 \cdot 1 & - 2 - 6 \cdot 2 & 0 - 6 \cdot 1 & 0 - 6 \cdot 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 2 & 1 & 0 \\ 0 & - 7 & - 3 & 0 \\ 6 - 6 \cdot 1 & - 2 - 6 \cdot 2 & 0 - 6 \cdot 1 & 0 - 6 \cdot 0 \end{array}]$

Step 3.2.2

Simplify

R_{3}

$R_{3}$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & 1 & 0 0 & - 7 & - 3 & 0 0 & - 14 & - 6 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 2 & 1 & 0 \\ 0 & - 7 & - 3 & 0 \\ 0 & - 14 & - 6 & 0 \end{array}]$

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & 1 & 0 0 & - 7 & - 3 & 0 0 & - 14 & - 6 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 2 & 1 & 0 \\ 0 & - 7 & - 3 & 0 \\ 0 & - 14 & - 6 & 0 \end{array}]$

Step 3.3

Multiply each element of

R_{2}

$R_{2}$ by

- \frac{1}{7}

$- \frac{1}{7}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

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Step 3.3.1

Multiply each element of

R_{2}

$R_{2}$ by

$- \frac{1}{7}$ to make the entry at

$2, 2$ a

$1$ .

$[\begin{array}{c} 1 & 2 & 1 & 0 \\ - \frac{1}{7} \cdot 0 & - \frac{1}{7} \cdot - 7 & - \frac{1}{7} \cdot - 3 & - \frac{1}{7} \cdot 0 \\ 0 & - 14 & - 6 & 0 \end{array}]$

Step 3.3.2

Simplify

$R_{2}$ .

$[\begin{array}{c} 1 & 2 & 1 & 0 \\ 0 & 1 & \frac{3}{7} & 0 \\ 0 & - 14 & - 6 & 0 \end{array}]$

Step 3.4

Perform the row operation

$R_{3} = R_{3} + 14 R_{2}$ to make the entry at

$3, 2$ a

$0$ .

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Step 3.4.1

Perform the row operation

$R_{3} = R_{3} + 14 R_{2}$ to make the entry at

$3, 2$ a

$0$ .

$[\begin{array}{c} 1 & 2 & 1 & 0 \\ 0 & 1 & \frac{3}{7} & 0 \\ 0 + 14 \cdot 0 & - 14 + 14 \cdot 1 & - 6 + 14 (\frac{3}{7}) & 0 + 14 \cdot 0 \end{array}]$

Step 3.4.2

Simplify

$R_{3}$ .

$[\begin{array}{c} 1 & 2 & 1 & 0 \\ 0 & 1 & \frac{3}{7} & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 3.5

Perform the row operation

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

$1, 2$ a

$0$ .

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Step 3.5.1

Perform the row operation

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

$1, 2$ a

$0$ .

$[\begin{array}{c} 1 - 2 \cdot 0 & 2 - 2 \cdot 1 & 1 - 2 (\frac{3}{7}) & 0 - 2 \cdot 0 \\ 0 & 1 & \frac{3}{7} & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 3.5.2

Simplify

$R_{1}$ .

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

Step 4

Remove rows that are all zeros.

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

Step 5

Write the matrix as a system of linear equations.

$x + \frac{1}{7} z = 0$

$y + \frac{3}{7} z = 0$

Step 6

Since there are non-trivial solutions to

$A x = 0$ , the vectors are linearly dependent.

Linearly Dependent