Linear Algebra Examples

$2 x_{1} + 3 x_{2} - 8 x_{3} = 105$ $- x_{1} + x_{2} - x_{3} = 10$ $8 x_{1} - 5 x_{2} + 3 x_{3} = - 10$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 2 & 3 & - 8 & 105 \\ - 1 & 1 & - 1 & 10 \\ 8 & - 5 & 3 & - 10 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} \frac{2}{2} & \frac{3}{2} & \frac{- 8}{2} & \frac{105}{2} \\ - 1 & 1 & - 1 & 10 \\ 8 & - 5 & 3 & - 10 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - 4 & \frac{105}{2} \\ - 1 & 1 & - 1 & 10 \\ 8 & - 5 & 3 & - 10 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - 4 & \frac{105}{2} \\ - 1 + 1 \cdot 1 & 1 + \frac{3}{2} & - 1 - 4 & 10 + \frac{105}{2} \\ 8 & - 5 & 3 & - 10 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - 4 & \frac{105}{2} \\ 0 & \frac{5}{2} & - 5 & \frac{125}{2} \\ 8 & - 5 & 3 & - 10 \end{array}]$

Step 2.3

Perform the row operation $R_{3} = R_{3} - 8 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 2.3.1

Perform the row operation $R_{3} = R_{3} - 8 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - 4 & \frac{105}{2} \\ 0 & \frac{5}{2} & - 5 & \frac{125}{2} \\ 8 - 8 \cdot 1 & - 5 - 8 (\frac{3}{2}) & 3 - 8 \cdot - 4 & - 10 - 8 (\frac{105}{2}) \end{array}]$

Step 2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - 4 & \frac{105}{2} \\ 0 & \frac{5}{2} & - 5 & \frac{125}{2} \\ 0 & - 17 & 35 & - 430 \end{array}]$

Step 2.4

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

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Step 2.4.1

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

$[\begin{array}{c} 1 & \frac{3}{2} & - 4 & \frac{105}{2} \\ \frac{2}{5} \cdot 0 & \frac{2}{5} \cdot \frac{5}{2} & \frac{2}{5} \cdot - 5 & \frac{2}{5} \cdot \frac{125}{2} \\ 0 & - 17 & 35 & - 430 \end{array}]$

Step 2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - 4 & \frac{105}{2} \\ 0 & 1 & - 2 & 25 \\ 0 & - 17 & 35 & - 430 \end{array}]$

Step 2.5

Perform the row operation $R_{3} = R_{3} + 17 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 2.5.1

Perform the row operation $R_{3} = R_{3} + 17 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - 4 & \frac{105}{2} \\ 0 & 1 & - 2 & 25 \\ 0 + 17 \cdot 0 & - 17 + 17 \cdot 1 & 35 + 17 \cdot - 2 & - 430 + 17 \cdot 25 \end{array}]$

Step 2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - 4 & \frac{105}{2} \\ 0 & 1 & - 2 & 25 \\ 0 & 0 & 1 & - 5 \end{array}]$

Step 2.6

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

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Step 2.6.1

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

$[\begin{array}{c} 1 & \frac{3}{2} & - 4 & \frac{105}{2} \\ 0 + 2 \cdot 0 & 1 + 2 \cdot 0 & - 2 + 2 \cdot 1 & 25 + 2 \cdot - 5 \\ 0 & 0 & 1 & - 5 \end{array}]$

Step 2.6.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - 4 & \frac{105}{2} \\ 0 & 1 & 0 & 15 \\ 0 & 0 & 1 & - 5 \end{array}]$

Step 2.7

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

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Step 2.7.1

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

$[\begin{array}{c} 1 + 4 \cdot 0 & \frac{3}{2} + 4 \cdot 0 & - 4 + 4 \cdot 1 & \frac{105}{2} + 4 \cdot - 5 \\ 0 & 1 & 0 & 15 \\ 0 & 0 & 1 & - 5 \end{array}]$

Step 2.7.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 0 & \frac{65}{2} \\ 0 & 1 & 0 & 15 \\ 0 & 0 & 1 & - 5 \end{array}]$

Step 2.8

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

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Step 2.8.1

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

$[\begin{array}{c} 1 - \frac{3}{2} \cdot 0 & \frac{3}{2} - \frac{3}{2} \cdot 1 & 0 - \frac{3}{2} \cdot 0 & \frac{65}{2} - \frac{3}{2} \cdot 15 \\ 0 & 1 & 0 & 15 \\ 0 & 0 & 1 & - 5 \end{array}]$

Step 2.8.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 10 \\ 0 & 1 & 0 & 15 \\ 0 & 0 & 1 & - 5 \end{array}]$

Step 3

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

$x_{1} = 10$

$x_{2} = 15$

$x_{3} = - 5$

Step 4

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

$(10, 15, - 5)$