Linear Algebra Examples

$B = {[\begin{array}{c} - 1 \\ 4 \\ 7 \end{array}], [\begin{array}{c} 6 \\ - 5 \\ 8 \end{array}], [\begin{array}{c} 1 \\ 5 \\ 9 \end{array}]}$

Step 1

To determine if the columns in the matrix are linearly dependent, determine if the equation $A x = 0$ has any non-trivial solutions.

Step 2

Write as an augmented matrix for $A x = 0$ .

$[\begin{array}{c} - 1 & 6 & 1 & 0 \\ 4 & - 5 & 5 & 0 \\ 7 & 8 & 9 & 0 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Multiply each element of $R_{1}$ by $- 1$ to make the entry at $1, 1$ a $1$ .

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Step 3.1.1

Multiply each element of $R_{1}$ by $- 1$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - - 1 & - 1 \cdot 6 & - 1 \cdot 1 & - 0 \\ 4 & - 5 & 5 & 0 \\ 7 & 8 & 9 & 0 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 6 & - 1 & 0 \\ 4 & - 5 & 5 & 0 \\ 7 & 8 & 9 & 0 \end{array}]$

Step 3.2

Perform the row operation $R_{2} = R_{2} - 4 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.2.1

Perform the row operation $R_{2} = R_{2} - 4 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - 6 & - 1 & 0 \\ 4 - 4 \cdot 1 & - 5 - 4 \cdot - 6 & 5 - 4 \cdot - 1 & 0 - 4 \cdot 0 \\ 7 & 8 & 9 & 0 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 6 & - 1 & 0 \\ 0 & 19 & 9 & 0 \\ 7 & 8 & 9 & 0 \end{array}]$

Step 3.3

Perform the row operation $R_{3} = R_{3} - 7 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 3.3.1

Perform the row operation $R_{3} = R_{3} - 7 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - 6 & - 1 & 0 \\ 0 & 19 & 9 & 0 \\ 7 - 7 \cdot 1 & 8 - 7 \cdot - 6 & 9 - 7 \cdot - 1 & 0 - 7 \cdot 0 \end{array}]$

Step 3.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 6 & - 1 & 0 \\ 0 & 19 & 9 & 0 \\ 0 & 50 & 16 & 0 \end{array}]$

Step 3.4

Multiply each element of $R_{2}$ by $\frac{1}{19}$ to make the entry at $2, 2$ a $1$ .

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Step 3.4.1

Multiply each element of $R_{2}$ by $\frac{1}{19}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 6 & - 1 & 0 \\ \frac{0}{19} & \frac{19}{19} & \frac{9}{19} & \frac{0}{19} \\ 0 & 50 & 16 & 0 \end{array}]$

Step 3.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 6 & - 1 & 0 \\ 0 & 1 & \frac{9}{19} & 0 \\ 0 & 50 & 16 & 0 \end{array}]$

Step 3.5

Perform the row operation $R_{3} = R_{3} - 50 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 3.5.1

Perform the row operation $R_{3} = R_{3} - 50 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - 6 & - 1 & 0 \\ 0 & 1 & \frac{9}{19} & 0 \\ 0 - 50 \cdot 0 & 50 - 50 \cdot 1 & 16 - 50 (\frac{9}{19}) & 0 - 50 \cdot 0 \end{array}]$

Step 3.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 6 & - 1 & 0 \\ 0 & 1 & \frac{9}{19} & 0 \\ 0 & 0 & - \frac{146}{19} & 0 \end{array}]$

Step 3.6

Multiply each element of $R_{3}$ by $- \frac{19}{146}$ to make the entry at $3, 3$ a $1$ .

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Step 3.6.1

Multiply each element of $R_{3}$ by $- \frac{19}{146}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - 6 & - 1 & 0 \\ 0 & 1 & \frac{9}{19} & 0 \\ - \frac{19}{146} \cdot 0 & - \frac{19}{146} \cdot 0 & - \frac{19}{146} (- \frac{146}{19}) & - \frac{19}{146} \cdot 0 \end{array}]$

Step 3.6.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 6 & - 1 & 0 \\ 0 & 1 & \frac{9}{19} & 0 \\ 0 & 0 & 1 & 0 \end{array}]$

Step 3.7

Perform the row operation $R_{2} = R_{2} - \frac{9}{19} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 3.7.1

Perform the row operation $R_{2} = R_{2} - \frac{9}{19} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & - 6 & - 1 & 0 \\ 0 - \frac{9}{19} \cdot 0 & 1 - \frac{9}{19} \cdot 0 & \frac{9}{19} - \frac{9}{19} \cdot 1 & 0 - \frac{9}{19} \cdot 0 \\ 0 & 0 & 1 & 0 \end{array}]$

Step 3.7.2

Simplify $R_{2}$ .

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

Step 3.8

Perform the row operation $R_{1} = R_{1} + R_{3}$ to make the entry at $1, 3$ a $0$ .

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Step 3.8.1

Perform the row operation $R_{1} = R_{1} + R_{3}$ to make the entry at $1, 3$ a $0$ .

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

Step 3.8.2

Simplify $R_{1}$ .

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

Step 3.9

Perform the row operation $R_{1} = R_{1} + 6 R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 3.9.1

Perform the row operation $R_{1} = R_{1} + 6 R_{2}$ to make the entry at $1, 2$ a $0$ .

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

Step 3.9.2

Simplify $R_{1}$ .

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

Step 4

Write the matrix as a system of linear equations.

$x = 0$

$y = 0$

$z = 0$

Step 5

Since the only solution to $A x = 0$ is the trivial solution, the vectors are linearly independent.

Linearly Independent

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