Linear Algebra Examples

$x_{1} - 3 x_{2} + 4 x_{3} = 7$ $2 x_{1} + 5 x_{2} - x_{3} = 1$ $3 x_{1} - 4 x_{2} + 5 x_{3} = 18$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 1 & - 3 & 4 & 7 \\ 2 & 5 & - 1 & 1 \\ 3 & - 4 & 5 & 18 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} 1 & - 3 & 4 & 7 \\ 2 - 2 \cdot 1 & 5 - 2 \cdot - 3 & - 1 - 2 \cdot 4 & 1 - 2 \cdot 7 \\ 3 & - 4 & 5 & 18 \end{array}]$

Step 2.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 3 & 4 & 7 \\ 0 & 11 & - 9 & - 13 \\ 3 & - 4 & 5 & 18 \end{array}]$

Step 2.2

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 2.2.1

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

$[\begin{array}{c} 1 & - 3 & 4 & 7 \\ 0 & 11 & - 9 & - 13 \\ 3 - 3 \cdot 1 & - 4 - 3 \cdot - 3 & 5 - 3 \cdot 4 & 18 - 3 \cdot 7 \end{array}]$

Step 2.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 3 & 4 & 7 \\ 0 & 11 & - 9 & - 13 \\ 0 & 5 & - 7 & - 3 \end{array}]$

Step 2.3

Multiply each element of $R_{2}$ by $\frac{1}{11}$ to make the entry at $2, 2$ a $1$ .

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Step 2.3.1

Multiply each element of $R_{2}$ by $\frac{1}{11}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 3 & 4 & 7 \\ \frac{0}{11} & \frac{11}{11} & \frac{- 9}{11} & \frac{- 13}{11} \\ 0 & 5 & - 7 & - 3 \end{array}]$

Step 2.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 3 & 4 & 7 \\ 0 & 1 & - \frac{9}{11} & - \frac{13}{11} \\ 0 & 5 & - 7 & - 3 \end{array}]$

Step 2.4

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 2.4.1

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

$[\begin{array}{c} 1 & - 3 & 4 & 7 \\ 0 & 1 & - \frac{9}{11} & - \frac{13}{11} \\ 0 - 5 \cdot 0 & 5 - 5 \cdot 1 & - 7 - 5 (- \frac{9}{11}) & - 3 - 5 (- \frac{13}{11}) \end{array}]$

Step 2.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 3 & 4 & 7 \\ 0 & 1 & - \frac{9}{11} & - \frac{13}{11} \\ 0 & 0 & - \frac{32}{11} & \frac{32}{11} \end{array}]$

Step 2.5

Multiply each element of $R_{3}$ by $- \frac{11}{32}$ to make the entry at $3, 3$ a $1$ .

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Step 2.5.1

Multiply each element of $R_{3}$ by $- \frac{11}{32}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - 3 & 4 & 7 \\ 0 & 1 & - \frac{9}{11} & - \frac{13}{11} \\ - \frac{11}{32} \cdot 0 & - \frac{11}{32} \cdot 0 & - \frac{11}{32} (- \frac{32}{11}) & - \frac{11}{32} \cdot \frac{32}{11} \end{array}]$

Step 2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 3 & 4 & 7 \\ 0 & 1 & - \frac{9}{11} & - \frac{13}{11} \\ 0 & 0 & 1 & - 1 \end{array}]$

Step 2.6

Perform the row operation $R_{2} = R_{2} + \frac{9}{11} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 2.6.1

Perform the row operation $R_{2} = R_{2} + \frac{9}{11} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & - 3 & 4 & 7 \\ 0 + \frac{9}{11} \cdot 0 & 1 + \frac{9}{11} \cdot 0 & - \frac{9}{11} + \frac{9}{11} \cdot 1 & - \frac{13}{11} + \frac{9}{11} \cdot - 1 \\ 0 & 0 & 1 & - 1 \end{array}]$

Step 2.6.2

Simplify $R_{2}$ .

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

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 $0$ .

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

Step 2.7.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 3 & 0 & 11 \\ 0 & 1 & 0 & - 2 \\ 0 & 0 & 1 & - 1 \end{array}]$

Step 2.8

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

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Step 2.8.1

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

$[\begin{array}{c} 1 + 3 \cdot 0 & - 3 + 3 \cdot 1 & 0 + 3 \cdot 0 & 11 + 3 \cdot - 2 \\ 0 & 1 & 0 & - 2 \\ 0 & 0 & 1 & - 1 \end{array}]$

Step 2.8.2

Simplify $R_{1}$ .

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

Step 3

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

$x_{1} = 5$

$x_{2} = - 2$

$x_{3} = - 1$

Step 4

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

$(5, - 2, - 1)$