Linear Algebra Examples

$A = [\begin{array}{c} 1 & 2 & 10 \\ 8 & 1 & 10 \\ - 6 & - 2 & - 5 \end{array}]$

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 $A x = 0$ .

$[\begin{array}{c} 1 & 2 & 10 & 0 \\ 8 & 1 & 10 & 0 \\ - 6 & - 2 & - 5 & 0 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Perform the row operation $R_{2} = R_{2} - 8 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.1.1

Perform the row operation $R_{2} = R_{2} - 8 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 2 & 10 & 0 \\ 8 - 8 \cdot 1 & 1 - 8 \cdot 2 & 10 - 8 \cdot 10 & 0 - 8 \cdot 0 \\ - 6 & - 2 & - 5 & 0 \end{array}]$

Step 3.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 2 & 10 & 0 \\ 0 & - 15 & - 70 & 0 \\ - 6 & - 2 & - 5 & 0 \end{array}]$

Step 3.2

Perform the row operation $R_{3} = R_{3} + 6 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 3.2.1

Perform the row operation $R_{3} = R_{3} + 6 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & 2 & 10 & 0 \\ 0 & - 15 & - 70 & 0 \\ - 6 + 6 \cdot 1 & - 2 + 6 \cdot 2 & - 5 + 6 \cdot 10 & 0 + 6 \cdot 0 \end{array}]$

Step 3.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 2 & 10 & 0 \\ 0 & - 15 & - 70 & 0 \\ 0 & 10 & 55 & 0 \end{array}]$

Step 3.3

Multiply each element of $R_{2}$ by $- \frac{1}{15}$ to make the entry at $2, 2$ a $1$ .

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Step 3.3.1

Multiply each element of $R_{2}$ by $- \frac{1}{15}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 2 & 10 & 0 \\ - \frac{1}{15} \cdot 0 & - \frac{1}{15} \cdot - 15 & - \frac{1}{15} \cdot - 70 & - \frac{1}{15} \cdot 0 \\ 0 & 10 & 55 & 0 \end{array}]$

Step 3.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 2 & 10 & 0 \\ 0 & 1 & \frac{14}{3} & 0 \\ 0 & 10 & 55 & 0 \end{array}]$

Step 3.4

Perform the row operation $R_{3} = R_{3} - 10 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 3.4.1

Perform the row operation $R_{3} = R_{3} - 10 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & 2 & 10 & 0 \\ 0 & 1 & \frac{14}{3} & 0 \\ 0 - 10 \cdot 0 & 10 - 10 \cdot 1 & 55 - 10 (\frac{14}{3}) & 0 - 10 \cdot 0 \end{array}]$

Step 3.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 2 & 10 & 0 \\ 0 & 1 & \frac{14}{3} & 0 \\ 0 & 0 & \frac{25}{3} & 0 \end{array}]$

Step 3.5

Multiply each element of $R_{3}$ by $\frac{3}{25}$ to make the entry at $3, 3$ a $1$ .

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Step 3.5.1

Multiply each element of $R_{3}$ by $\frac{3}{25}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 2 & 10 & 0 \\ 0 & 1 & \frac{14}{3} & 0 \\ \frac{3}{25} \cdot 0 & \frac{3}{25} \cdot 0 & \frac{3}{25} \cdot \frac{25}{3} & \frac{3}{25} \cdot 0 \end{array}]$

Step 3.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 2 & 10 & 0 \\ 0 & 1 & \frac{14}{3} & 0 \\ 0 & 0 & 1 & 0 \end{array}]$

Step 3.6

Perform the row operation $R_{2} = R_{2} - \frac{14}{3} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 3.6.1

Perform the row operation $R_{2} = R_{2} - \frac{14}{3} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & 2 & 10 & 0 \\ 0 - \frac{14}{3} \cdot 0 & 1 - \frac{14}{3} \cdot 0 & \frac{14}{3} - \frac{14}{3} \cdot 1 & 0 - \frac{14}{3} \cdot 0 \\ 0 & 0 & 1 & 0 \end{array}]$

Step 3.6.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 2 & 10 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array}]$

Step 3.7

Perform the row operation $R_{1} = R_{1} - 10 R_{3}$ to make the entry at $1, 3$ a $0$ .

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Step 3.7.1

Perform the row operation $R_{1} = R_{1} - 10 R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - 10 \cdot 0 & 2 - 10 \cdot 0 & 10 - 10 \cdot 1 & 0 - 10 \cdot 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array}]$

Step 3.7.2

Simplify $R_{1}$ .

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

Step 3.8

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 3.8.1

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

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

Step 3.8.2

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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