Linear Algebra Examples

$[\begin{array}{c} 5 & - 10 & 25 & - 20 \\ 1 & - 2 & 5 & - 4 \\ 3 & - 6 & 15 & 12 \end{array}]$

Step 1

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

$[\begin{array}{c} 5 & - 10 & 25 & - 20 & 0 \\ 1 & - 2 & 5 & - 4 & 0 \\ 3 & - 6 & 15 & 12 & 0 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} \frac{5}{5} & \frac{- 10}{5} & \frac{25}{5} & \frac{- 20}{5} & \frac{0}{5} \\ 1 & - 2 & 5 & - 4 & 0 \\ 3 & - 6 & 15 & 12 & 0 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 2 & 5 & - 4 & 0 \\ 1 & - 2 & 5 & - 4 & 0 \\ 3 & - 6 & 15 & 12 & 0 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & - 2 & 5 & - 4 & 0 \\ 1 - 1 & - 2 + 2 & 5 - 5 & - 4 + 4 & 0 - 0 \\ 3 & - 6 & 15 & 12 & 0 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 2 & 5 & - 4 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 3 & - 6 & 15 & 12 & 0 \end{array}]$

Step 2.3

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

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Step 2.3.1

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

$[\begin{array}{c} 1 & - 2 & 5 & - 4 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 3 - 3 \cdot 1 & - 6 - 3 \cdot - 2 & 15 - 3 \cdot 5 & 12 - 3 \cdot - 4 & 0 - 3 \cdot 0 \end{array}]$

Step 2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 2 & 5 & - 4 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 24 & 0 \end{array}]$

Step 2.4

Swap $R_{3}$ with $R_{2}$ to put a nonzero entry at $2, 4$ .

$[\begin{array}{c} 1 & - 2 & 5 & - 4 & 0 \\ 0 & 0 & 0 & 24 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}]$

Step 2.5

Multiply each element of $R_{2}$ by $\frac{1}{24}$ to make the entry at $2, 4$ a $1$ .

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Step 2.5.1

Multiply each element of $R_{2}$ by $\frac{1}{24}$ to make the entry at $2, 4$ a $1$ .

$[\begin{array}{c} 1 & - 2 & 5 & - 4 & 0 \\ \frac{0}{24} & \frac{0}{24} & \frac{0}{24} & \frac{24}{24} & \frac{0}{24} \\ 0 & 0 & 0 & 0 & 0 \end{array}]$

Step 2.5.2

Simplify $R_{2}$ .

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

Step 2.6

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

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Step 2.6.1

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

$[\begin{array}{c} 1 + 4 \cdot 0 & - 2 + 4 \cdot 0 & 5 + 4 \cdot 0 & - 4 + 4 \cdot 1 & 0 + 4 \cdot 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}]$

Step 2.6.2

Simplify $R_{1}$ .

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

Step 3

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

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

$x_{4} = 0$

$0 = 0$

Step 4

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

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

Step 5

Write the solution as a linear combination of vectors.

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

Step 6

Write as a solution set.

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