Linear Algebra Examples

$2 x - y + 3 z = - 9$ , $x + 3 y - 2 = 10$ , $3 x + y - 2 = 8$

Step 1

Add $2$ to both sides of the equation.

$2 x - y + 3 z = - 9, x + 3 y = 10 + 2, 3 x + y - 2 = 8$

Step 2

Add $10$ and $2$ .

$2 x - y + 3 z = - 9, x + 3 y = 12, 3 x + y - 2 = 8$

Step 3

Add $2$ to both sides of the equation.

$2 x - y + 3 z = - 9, x + 3 y = 12, 3 x + y = 8 + 2$

Step 4

Add $8$ and $2$ .

$2 x - y + 3 z = - 9, x + 3 y = 12, 3 x + y = 10$

Step 5

Write the system of equations in matrix form.

$[\begin{array}{c} 2 & - 1 & 3 & - 9 \\ 1 & 3 & 0 & 12 \\ 3 & 1 & 0 & 10 \end{array}]$

Step 6

Find the reduced row echelon form.

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Step 6.1

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

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Step 6.1.1

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

$[\begin{array}{c} \frac{2}{2} & \frac{- 1}{2} & \frac{3}{2} & \frac{- 9}{2} \\ 1 & 3 & 0 & 12 \\ 3 & 1 & 0 & 10 \end{array}]$

Step 6.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{3}{2} & - \frac{9}{2} \\ 1 & 3 & 0 & 12 \\ 3 & 1 & 0 & 10 \end{array}]$

Step 6.2

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

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Step 6.2.1

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

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{3}{2} & - \frac{9}{2} \\ 1 - 1 & 3 + \frac{1}{2} & 0 - \frac{3}{2} & 12 + \frac{9}{2} \\ 3 & 1 & 0 & 10 \end{array}]$

Step 6.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{3}{2} & - \frac{9}{2} \\ 0 & \frac{7}{2} & - \frac{3}{2} & \frac{33}{2} \\ 3 & 1 & 0 & 10 \end{array}]$

Step 6.3

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 6.3.1

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

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{3}{2} & - \frac{9}{2} \\ 0 & \frac{7}{2} & - \frac{3}{2} & \frac{33}{2} \\ 3 - 3 \cdot 1 & 1 - 3 (- \frac{1}{2}) & 0 - 3 (\frac{3}{2}) & 10 - 3 (- \frac{9}{2}) \end{array}]$

Step 6.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{3}{2} & - \frac{9}{2} \\ 0 & \frac{7}{2} & - \frac{3}{2} & \frac{33}{2} \\ 0 & \frac{5}{2} & - \frac{9}{2} & \frac{47}{2} \end{array}]$

Step 6.4

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

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Step 6.4.1

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

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{3}{2} & - \frac{9}{2} \\ \frac{2}{7} \cdot 0 & \frac{2}{7} \cdot \frac{7}{2} & \frac{2}{7} (- \frac{3}{2}) & \frac{2}{7} \cdot \frac{33}{2} \\ 0 & \frac{5}{2} & - \frac{9}{2} & \frac{47}{2} \end{array}]$

Step 6.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{3}{2} & - \frac{9}{2} \\ 0 & 1 & - \frac{3}{7} & \frac{33}{7} \\ 0 & \frac{5}{2} & - \frac{9}{2} & \frac{47}{2} \end{array}]$

Step 6.5

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

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Step 6.5.1

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

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{3}{2} & - \frac{9}{2} \\ 0 & 1 & - \frac{3}{7} & \frac{33}{7} \\ 0 - \frac{5}{2} \cdot 0 & \frac{5}{2} - \frac{5}{2} \cdot 1 & - \frac{9}{2} - \frac{5}{2} (- \frac{3}{7}) & \frac{47}{2} - \frac{5}{2} \cdot \frac{33}{7} \end{array}]$

Step 6.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{3}{2} & - \frac{9}{2} \\ 0 & 1 & - \frac{3}{7} & \frac{33}{7} \\ 0 & 0 & - \frac{24}{7} & \frac{82}{7} \end{array}]$

Step 6.6

Multiply each element of $R_{3}$ by $- \frac{7}{24}$ to make the entry at $3, 3$ a $1$ .

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Step 6.6.1

Multiply each element of $R_{3}$ by $- \frac{7}{24}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{3}{2} & - \frac{9}{2} \\ 0 & 1 & - \frac{3}{7} & \frac{33}{7} \\ - \frac{7}{24} \cdot 0 & - \frac{7}{24} \cdot 0 & - \frac{7}{24} (- \frac{24}{7}) & - \frac{7}{24} \cdot \frac{82}{7} \end{array}]$

Step 6.6.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{3}{2} & - \frac{9}{2} \\ 0 & 1 & - \frac{3}{7} & \frac{33}{7} \\ 0 & 0 & 1 & - \frac{41}{12} \end{array}]$

Step 6.7

Perform the row operation $R_{2} = R_{2} + \frac{3}{7} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 6.7.1

Perform the row operation $R_{2} = R_{2} + \frac{3}{7} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{3}{2} & - \frac{9}{2} \\ 0 + \frac{3}{7} \cdot 0 & 1 + \frac{3}{7} \cdot 0 & - \frac{3}{7} + \frac{3}{7} \cdot 1 & \frac{33}{7} + \frac{3}{7} (- \frac{41}{12}) \\ 0 & 0 & 1 & - \frac{41}{12} \end{array}]$

Step 6.7.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{3}{2} & - \frac{9}{2} \\ 0 & 1 & 0 & \frac{13}{4} \\ 0 & 0 & 1 & - \frac{41}{12} \end{array}]$

Step 6.8

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 6.8.1

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

$[\begin{array}{c} 1 - \frac{3}{2} \cdot 0 & - \frac{1}{2} - \frac{3}{2} \cdot 0 & \frac{3}{2} - \frac{3}{2} \cdot 1 & - \frac{9}{2} - \frac{3}{2} (- \frac{41}{12}) \\ 0 & 1 & 0 & \frac{13}{4} \\ 0 & 0 & 1 & - \frac{41}{12} \end{array}]$

Step 6.8.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 0 & \frac{5}{8} \\ 0 & 1 & 0 & \frac{13}{4} \\ 0 & 0 & 1 & - \frac{41}{12} \end{array}]$

Step 6.9

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 6.9.1

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

$[\begin{array}{c} 1 + \frac{1}{2} \cdot 0 & - \frac{1}{2} + \frac{1}{2} \cdot 1 & 0 + \frac{1}{2} \cdot 0 & \frac{5}{8} + \frac{1}{2} \cdot \frac{13}{4} \\ 0 & 1 & 0 & \frac{13}{4} \\ 0 & 0 & 1 & - \frac{41}{12} \end{array}]$

Step 6.9.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{9}{4} \\ 0 & 1 & 0 & \frac{13}{4} \\ 0 & 0 & 1 & - \frac{41}{12} \end{array}]$

Step 7

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

$x = \frac{9}{4}$

$y = \frac{13}{4}$

$z = - \frac{41}{12}$

Step 8

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

$(\frac{9}{4}, \frac{13}{4}, - \frac{41}{12})$

Step 9

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.

$X = [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{9}{4} \\ \frac{13}{4} \\ - \frac{41}{12} \end{array}]$