Linear Algebra Examples

$x + y + z = 2$ , $- x + 3 y + 2 z = 8$ , $4 x + y + 0 z = 4$

Step 1

Simplify.

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Step 1.1

Multiply $0$ by $z$ .

$x + y + z = 2, - x + 3 y + 2 z = 8, 4 x + y + 0 = 4$

Step 1.2

Add $4 x + y$ and $0$ .

$x + y + z = 2, - x + 3 y + 2 z = 8, 4 x + y = 4$

Step 2

Write the system of equations in matrix form.

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

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

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

Step 3.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 2 \\ 0 & 4 & 3 & 10 \\ 4 & 1 & 0 & 4 \end{array}]$

Step 3.2

Perform the row operation $R_{3} = R_{3} - 4 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 3.2.1

Perform the row operation $R_{3} = R_{3} - 4 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & 1 & 1 & 2 \\ 0 & 4 & 3 & 10 \\ 4 - 4 \cdot 1 & 1 - 4 \cdot 1 & 0 - 4 \cdot 1 & 4 - 4 \cdot 2 \end{array}]$

Step 3.2.2

Simplify $R_{3}$ .

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

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

Simplify $R_{2}$ .

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

Step 3.4

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

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Step 3.4.1

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

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

Step 3.4.2

Simplify $R_{3}$ .

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

Step 3.5

Multiply each element of $R_{3}$ by $- \frac{4}{7}$ to make the entry at $3, 3$ a $1$ .

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Step 3.5.1

Multiply each element of $R_{3}$ by $- \frac{4}{7}$ to make the entry at $3, 3$ a $1$ .

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

Step 3.5.2

Simplify $R_{3}$ .

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

Step 3.6

Perform the row operation $R_{2} = R_{2} - \frac{3}{4} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 3.6.1

Perform the row operation $R_{2} = R_{2} - \frac{3}{4} R_{3}$ to make the entry at $2, 3$ a $0$ .

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

Step 3.6.2

Simplify $R_{2}$ .

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

Step 3.7

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

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Step 3.7.1

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

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

Step 3.7.2

Simplify $R_{1}$ .

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

Step 3.8

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

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Step 3.8.1

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

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

Step 3.8.2

Simplify $R_{1}$ .

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

Step 4

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

$x = 0$

$y = 4$

$z = - 2$

Step 5

The solution is the set of ordered pairs that makes the system true.

$(0, 4, - 2)$

Step 6

Decompose a solution vector by re-arranging each equation represented in the row-reduced form of the augmented matrix by solving for the dependent variable in each row yields the vector equality.

$X = [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 0 \\ 4 \\ - 2 \end{array}]$