Linear Algebra Examples

$3 x_{1} - 2 x_{2} + x_{3} = - 1$ $- x_{1} - x_{2} + 6 x_{3} = 29$ $5 x_{1} - 3 x_{2} - 4 x_{3} = - 25$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 3 & - 2 & 1 & - 1 \\ - 1 & - 1 & 6 & 29 \\ 5 & - 3 & - 4 & - 25 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} \frac{3}{3} & \frac{- 2}{3} & \frac{1}{3} & \frac{- 1}{3} \\ - 1 & - 1 & 6 & 29 \\ 5 & - 3 & - 4 & - 25 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{1}{3} & - \frac{1}{3} \\ - 1 & - 1 & 6 & 29 \\ 5 & - 3 & - 4 & - 25 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{1}{3} & - \frac{1}{3} \\ - 1 + 1 \cdot 1 & - 1 - \frac{2}{3} & 6 + \frac{1}{3} & 29 - \frac{1}{3} \\ 5 & - 3 & - 4 & - 25 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{1}{3} & - \frac{1}{3} \\ 0 & - \frac{5}{3} & \frac{19}{3} & \frac{86}{3} \\ 5 & - 3 & - 4 & - 25 \end{array}]$

Step 2.3

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 2.3.1

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

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{1}{3} & - \frac{1}{3} \\ 0 & - \frac{5}{3} & \frac{19}{3} & \frac{86}{3} \\ 5 - 5 \cdot 1 & - 3 - 5 (- \frac{2}{3}) & - 4 - 5 (\frac{1}{3}) & - 25 - 5 (- \frac{1}{3}) \end{array}]$

Step 2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{1}{3} & - \frac{1}{3} \\ 0 & - \frac{5}{3} & \frac{19}{3} & \frac{86}{3} \\ 0 & \frac{1}{3} & - \frac{17}{3} & - \frac{70}{3} \end{array}]$

Step 2.4

Multiply each element of $R_{2}$ by $- \frac{3}{5}$ to make the entry at $2, 2$ a $1$ .

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Step 2.4.1

Multiply each element of $R_{2}$ by $- \frac{3}{5}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{1}{3} & - \frac{1}{3} \\ - \frac{3}{5} \cdot 0 & - \frac{3}{5} (- \frac{5}{3}) & - \frac{3}{5} \cdot \frac{19}{3} & - \frac{3}{5} \cdot \frac{86}{3} \\ 0 & \frac{1}{3} & - \frac{17}{3} & - \frac{70}{3} \end{array}]$

Step 2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{1}{3} & - \frac{1}{3} \\ 0 & 1 & - \frac{19}{5} & - \frac{86}{5} \\ 0 & \frac{1}{3} & - \frac{17}{3} & - \frac{70}{3} \end{array}]$

Step 2.5

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

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Step 2.5.1

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

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{1}{3} & - \frac{1}{3} \\ 0 & 1 & - \frac{19}{5} & - \frac{86}{5} \\ 0 - \frac{1}{3} \cdot 0 & \frac{1}{3} - \frac{1}{3} \cdot 1 & - \frac{17}{3} - \frac{1}{3} (- \frac{19}{5}) & - \frac{70}{3} - \frac{1}{3} (- \frac{86}{5}) \end{array}]$

Step 2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{1}{3} & - \frac{1}{3} \\ 0 & 1 & - \frac{19}{5} & - \frac{86}{5} \\ 0 & 0 & - \frac{22}{5} & - \frac{88}{5} \end{array}]$

Step 2.6

Multiply each element of $R_{3}$ by $- \frac{5}{22}$ to make the entry at $3, 3$ a $1$ .

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Step 2.6.1

Multiply each element of $R_{3}$ by $- \frac{5}{22}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{1}{3} & - \frac{1}{3} \\ 0 & 1 & - \frac{19}{5} & - \frac{86}{5} \\ - \frac{5}{22} \cdot 0 & - \frac{5}{22} \cdot 0 & - \frac{5}{22} (- \frac{22}{5}) & - \frac{5}{22} (- \frac{88}{5}) \end{array}]$

Step 2.6.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{1}{3} & - \frac{1}{3} \\ 0 & 1 & - \frac{19}{5} & - \frac{86}{5} \\ 0 & 0 & 1 & 4 \end{array}]$

Step 2.7

Perform the row operation $R_{2} = R_{2} + \frac{19}{5} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 2.7.1

Perform the row operation $R_{2} = R_{2} + \frac{19}{5} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{1}{3} & - \frac{1}{3} \\ 0 + \frac{19}{5} \cdot 0 & 1 + \frac{19}{5} \cdot 0 & - \frac{19}{5} + \frac{19}{5} \cdot 1 & - \frac{86}{5} + \frac{19}{5} \cdot 4 \\ 0 & 0 & 1 & 4 \end{array}]$

Step 2.7.2

Simplify $R_{2}$ .

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

Step 2.8

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

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Step 2.8.1

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

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

Step 2.8.2

Simplify $R_{1}$ .

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

Step 2.9

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

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Step 2.9.1

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

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

Step 2.9.2

Simplify $R_{1}$ .

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

Step 3

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

$x_{1} = - 3$

$x_{2} = - 2$

$x_{3} = 4$

Step 4

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

$(- 3, - 2, 4)$