Linear Algebra Examples

$x - y = 6$ , $2 y - z = - 1$ , $5 x + 5 + z = 7$

Step 1

Subtract $5$ from both sides of the equation.

$x - y = 6, 2 y - z = - 1, 5 x + z = 7 - 5$

Step 2

Subtract $5$ from $7$ .

$x - y = 6, 2 y - z = - 1, 5 x + z = 2$

Step 3

Write the system of equations in matrix form.

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

Step 4

Find the reduced row echelon form.

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Step 4.1

Perform the row operation $R_{3} = R_{3} - 5 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 4.1.1

Perform the row operation $R_{3} = R_{3} - 5 R_{1}$ to make the entry at $3, 1$ a $0$ .

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

Step 4.1.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & 0 & 6 \\ 0 & 2 & - 1 & - 1 \\ 0 & 5 & 1 & - 28 \end{array}]$

Step 4.2

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

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Step 4.2.1

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

$[\begin{array}{c} 1 & - 1 & 0 & 6 \\ \frac{0}{2} & \frac{2}{2} & \frac{- 1}{2} & \frac{- 1}{2} \\ 0 & 5 & 1 & - 28 \end{array}]$

Step 4.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & 0 & 6 \\ 0 & 1 & - \frac{1}{2} & - \frac{1}{2} \\ 0 & 5 & 1 & - 28 \end{array}]$

Step 4.3

Perform the row operation $R_{3} = R_{3} - 5 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 4.3.1

Perform the row operation $R_{3} = R_{3} - 5 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - 1 & 0 & 6 \\ 0 & 1 & - \frac{1}{2} & - \frac{1}{2} \\ 0 - 5 \cdot 0 & 5 - 5 \cdot 1 & 1 - 5 (- \frac{1}{2}) & - 28 - 5 (- \frac{1}{2}) \end{array}]$

Step 4.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & 0 & 6 \\ 0 & 1 & - \frac{1}{2} & - \frac{1}{2} \\ 0 & 0 & \frac{7}{2} & - \frac{51}{2} \end{array}]$

Step 4.4

Multiply each element of $R_{3}$ by $\frac{2}{7}$ to make the entry at $3, 3$ a $1$ .

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Step 4.4.1

Multiply each element of $R_{3}$ by $\frac{2}{7}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - 1 & 0 & 6 \\ 0 & 1 & - \frac{1}{2} & - \frac{1}{2} \\ \frac{2}{7} \cdot 0 & \frac{2}{7} \cdot 0 & \frac{2}{7} \cdot \frac{7}{2} & \frac{2}{7} (- \frac{51}{2}) \end{array}]$

Step 4.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & 0 & 6 \\ 0 & 1 & - \frac{1}{2} & - \frac{1}{2} \\ 0 & 0 & 1 & - \frac{51}{7} \end{array}]$

Step 4.5

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

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Step 4.5.1

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

$[\begin{array}{c} 1 & - 1 & 0 & 6 \\ 0 + \frac{1}{2} \cdot 0 & 1 + \frac{1}{2} \cdot 0 & - \frac{1}{2} + \frac{1}{2} \cdot 1 & - \frac{1}{2} + \frac{1}{2} (- \frac{51}{7}) \\ 0 & 0 & 1 & - \frac{51}{7} \end{array}]$

Step 4.5.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & 0 & 6 \\ 0 & 1 & 0 & - \frac{29}{7} \\ 0 & 0 & 1 & - \frac{51}{7} \end{array}]$

Step 4.6

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

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Step 4.6.1

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

$[\begin{array}{c} 1 + 0 & - 1 + 1 \cdot 1 & 0 + 0 & 6 - \frac{29}{7} \\ 0 & 1 & 0 & - \frac{29}{7} \\ 0 & 0 & 1 & - \frac{51}{7} \end{array}]$

Step 4.6.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{13}{7} \\ 0 & 1 & 0 & - \frac{29}{7} \\ 0 & 0 & 1 & - \frac{51}{7} \end{array}]$

Step 5

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

$x = \frac{13}{7}$

$y = - \frac{29}{7}$

$z = - \frac{51}{7}$

Step 6

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

$(\frac{13}{7}, - \frac{29}{7}, - \frac{51}{7})$

Step 7

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.

$X = [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{13}{7} \\ - \frac{29}{7} \\ - \frac{51}{7} \end{array}]$