Linear Algebra Examples

$[\begin{array}{c} - 3 & 6 & - 1 & 1 & - 7 \\ 1 & - 2 & 2 & 3 & - 1 \\ 2 & - 4 & 5 & 8 & - 4 \end{array}]$

Step 1

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

$[\begin{array}{c} - 3 & 6 & - 1 & 1 & - 7 & 0 \\ 1 & - 2 & 2 & 3 & - 1 & 0 \\ 2 & - 4 & 5 & 8 & - 4 & 0 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot 6 & - \frac{1}{3} \cdot - 1 & - \frac{1}{3} \cdot 1 & - \frac{1}{3} \cdot - 7 & - \frac{1}{3} \cdot 0 \\ 1 & - 2 & 2 & 3 & - 1 & 0 \\ 2 & - 4 & 5 & 8 & - 4 & 0 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 2 & \frac{1}{3} & - \frac{1}{3} & \frac{7}{3} & 0 \\ 1 & - 2 & 2 & 3 & - 1 & 0 \\ 2 & - 4 & 5 & 8 & - 4 & 0 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} 1 & - 2 & \frac{1}{3} & - \frac{1}{3} & \frac{7}{3} & 0 \\ 1 - 1 & - 2 + 2 & 2 - \frac{1}{3} & 3 + \frac{1}{3} & - 1 - \frac{7}{3} & 0 - 0 \\ 2 & - 4 & 5 & 8 & - 4 & 0 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 2 & \frac{1}{3} & - \frac{1}{3} & \frac{7}{3} & 0 \\ 0 & 0 & \frac{5}{3} & \frac{10}{3} & - \frac{10}{3} & 0 \\ 2 & - 4 & 5 & 8 & - 4 & 0 \end{array}]$

Step 2.3

Perform the row operation $R_{3} = R_{3} - 2 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 2.3.1

Perform the row operation $R_{3} = R_{3} - 2 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - 2 & \frac{1}{3} & - \frac{1}{3} & \frac{7}{3} & 0 \\ 0 & 0 & \frac{5}{3} & \frac{10}{3} & - \frac{10}{3} & 0 \\ 2 - 2 \cdot 1 & - 4 - 2 \cdot - 2 & 5 - 2 (\frac{1}{3}) & 8 - 2 (- \frac{1}{3}) & - 4 - 2 (\frac{7}{3}) & 0 - 2 \cdot 0 \end{array}]$

Step 2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 2 & \frac{1}{3} & - \frac{1}{3} & \frac{7}{3} & 0 \\ 0 & 0 & \frac{5}{3} & \frac{10}{3} & - \frac{10}{3} & 0 \\ 0 & 0 & \frac{13}{3} & \frac{26}{3} & - \frac{26}{3} & 0 \end{array}]$

Step 2.4

Multiply each element of $R_{2}$ by $\frac{3}{5}$ to make the entry at $2, 3$ a $1$ .

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Step 2.4.1

Multiply each element of $R_{2}$ by $\frac{3}{5}$ to make the entry at $2, 3$ a $1$ .

$[\begin{array}{c} 1 & - 2 & \frac{1}{3} & - \frac{1}{3} & \frac{7}{3} & 0 \\ \frac{3}{5} \cdot 0 & \frac{3}{5} \cdot 0 & \frac{3}{5} \cdot \frac{5}{3} & \frac{3}{5} \cdot \frac{10}{3} & \frac{3}{5} (- \frac{10}{3}) & \frac{3}{5} \cdot 0 \\ 0 & 0 & \frac{13}{3} & \frac{26}{3} & - \frac{26}{3} & 0 \end{array}]$

Step 2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 2 & \frac{1}{3} & - \frac{1}{3} & \frac{7}{3} & 0 \\ 0 & 0 & 1 & 2 & - 2 & 0 \\ 0 & 0 & \frac{13}{3} & \frac{26}{3} & - \frac{26}{3} & 0 \end{array}]$

Step 2.5

Perform the row operation $R_{3} = R_{3} - \frac{13}{3} R_{2}$ to make the entry at $3, 3$ a $0$ .

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Step 2.5.1

Perform the row operation $R_{3} = R_{3} - \frac{13}{3} R_{2}$ to make the entry at $3, 3$ a $0$ .

$[\begin{array}{c} 1 & - 2 & \frac{1}{3} & - \frac{1}{3} & \frac{7}{3} & 0 \\ 0 & 0 & 1 & 2 & - 2 & 0 \\ 0 - \frac{13}{3} \cdot 0 & 0 - \frac{13}{3} \cdot 0 & \frac{13}{3} - \frac{13}{3} \cdot 1 & \frac{26}{3} - \frac{13}{3} \cdot 2 & - \frac{26}{3} - \frac{13}{3} \cdot - 2 & 0 - \frac{13}{3} \cdot 0 \end{array}]$

Step 2.5.2

Simplify $R_{3}$ .

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

Step 2.6

Perform the row operation $R_{1} = R_{1} - \frac{1}{3} R_{2}$ to make the entry at $1, 3$ a $0$ .

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Step 2.6.1

Perform the row operation $R_{1} = R_{1} - \frac{1}{3} R_{2}$ to make the entry at $1, 3$ a $0$ .

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

Step 2.6.2

Simplify $R_{1}$ .

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

Step 3

Use the result matrix to declare the final solution to the system of equations.

$x_{1} - 2 x_{2} - x_{4} + 3 x_{5} = 0$

$x_{3} + 2 x_{4} - 2 x_{5} = 0$

$0 = 0$

Step 4

Write a solution vector by solving in terms of the free variables in each row.

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{array}] = [\begin{array}{c} 2 x_{2} + x_{4} - 3 x_{5} \\ x_{2} \\ - 2 x_{4} + 2 x_{5} \\ x_{4} \\ x_{5} \end{array}]$

Step 5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{array}] = x_{2} [\begin{array}{c} 2 \\ 1 \\ 0 \\ 0 \\ 0 \end{array}] + x_{4} [\begin{array}{c} 1 \\ 0 \\ - 2 \\ 1 \\ 0 \end{array}] + x_{5} [\begin{array}{c} - 3 \\ 0 \\ 2 \\ 0 \\ 1 \end{array}]$

Step 6

Write as a solution set.

${x_{2} [\begin{array}{c} 2 \\ 1 \\ 0 \\ 0 \\ 0 \end{array}] + x_{4} [\begin{array}{c} 1 \\ 0 \\ - 2 \\ 1 \\ 0 \end{array}] + x_{5} [\begin{array}{c} - 3 \\ 0 \\ 2 \\ 0 \\ 1 \end{array}] | x_{2}, x_{4}, x_{5} \in R}$

Step 7

The solution is the set of vectors created from the free variables of the system.

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

Step 8

Check if the vectors are linearly independent.

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Step 8.1

List the vectors.

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

Step 8.2

Write the vectors as a matrix.

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

Step 8.3

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

Step 8.4

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

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

Step 8.5

Find the reduced row echelon form.

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Step 8.5.1

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

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Step 8.5.1.1

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

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

Step 8.5.1.2

Simplify $R_{1}$ .

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

Step 8.5.2

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

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Step 8.5.2.1

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

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

Step 8.5.2.2

Simplify $R_{2}$ .

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

Step 8.5.3

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

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Step 8.5.3.1

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

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

Step 8.5.3.2

Simplify $R_{2}$ .

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

Step 8.5.4

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

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Step 8.5.4.1

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

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

Step 8.5.4.2

Simplify $R_{3}$ .

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

Step 8.5.5

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

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Step 8.5.5.1

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

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

Step 8.5.5.2

Simplify $R_{4}$ .

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

Step 8.5.6

Multiply each element of $R_{3}$ by $- \frac{1}{4}$ to make the entry at $3, 3$ a $1$ .

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Step 8.5.6.1

Multiply each element of $R_{3}$ by $- \frac{1}{4}$ to make the entry at $3, 3$ a $1$ .

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

Step 8.5.6.2

Simplify $R_{3}$ .

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

Step 8.5.7

Perform the row operation $R_{4} = R_{4} - 3 R_{3}$ to make the entry at $4, 3$ a $0$ .

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Step 8.5.7.1

Perform the row operation $R_{4} = R_{4} - 3 R_{3}$ to make the entry at $4, 3$ a $0$ .

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

Step 8.5.7.2

Simplify $R_{4}$ .

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

Step 8.5.8

Perform the row operation $R_{5} = R_{5} - R_{3}$ to make the entry at $5, 3$ a $0$ .

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Step 8.5.8.1

Perform the row operation $R_{5} = R_{5} - R_{3}$ to make the entry at $5, 3$ a $0$ .

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

Step 8.5.8.2

Simplify $R_{5}$ .

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

Step 8.5.9

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

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Step 8.5.9.1

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

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

Step 8.5.9.2

Simplify $R_{2}$ .

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

Step 8.5.10

Perform the row operation $R_{1} = R_{1} + \frac{3}{2} R_{3}$ to make the entry at $1, 3$ a $0$ .

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Step 8.5.10.1

Perform the row operation $R_{1} = R_{1} + \frac{3}{2} R_{3}$ to make the entry at $1, 3$ a $0$ .

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

Step 8.5.10.2

Simplify $R_{1}$ .

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

Step 8.5.11

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

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Step 8.5.11.1

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

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

Step 8.5.11.2

Simplify $R_{1}$ .

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

Step 8.6

Remove rows that are all zeros.

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

Step 8.7

Write the matrix as a system of linear equations.

$x = 0$

$y = 0$

$z = 0$

Step 8.8

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

Linearly Independent

Step 9

Since the vectors are linearly independent, they form a basis for the null space of the matrix.

Basis of $Nul (A)$ : ${[\begin{array}{c} 2 \\ 1 \\ 0 \\ 0 \\ 0 \end{array}], [\begin{array}{c} 1 \\ 0 \\ - 2 \\ 1 \\ 0 \end{array}], [\begin{array}{c} - 3 \\ 0 \\ 2 \\ 0 \\ 1 \end{array}]}$

Dimension of $Nul (A)$ : $3$

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