Linear Algebra Examples

$x_{1} - 3 x_{2} + 2 x_{3} = 5$ $2 x_{1} + x_{2} + 4 x_{3} = 24$ $5 x_{1} - 3 x_{2} + x_{3} = - 14$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 1 & - 3 & 2 & 5 \\ 2 & 1 & 4 & 24 \\ 5 & - 3 & 1 & - 14 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & - 3 & 2 & 5 \\ 2 - 2 \cdot 1 & 1 - 2 \cdot - 3 & 4 - 2 \cdot 2 & 24 - 2 \cdot 5 \\ 5 & - 3 & 1 & - 14 \end{array}]$

Step 2.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 3 & 2 & 5 \\ 0 & 7 & 0 & 14 \\ 5 & - 3 & 1 & - 14 \end{array}]$

Step 2.2

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 2.2.1

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

$[\begin{array}{c} 1 & - 3 & 2 & 5 \\ 0 & 7 & 0 & 14 \\ 5 - 5 \cdot 1 & - 3 - 5 \cdot - 3 & 1 - 5 \cdot 2 & - 14 - 5 \cdot 5 \end{array}]$

Step 2.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 3 & 2 & 5 \\ 0 & 7 & 0 & 14 \\ 0 & 12 & - 9 & - 39 \end{array}]$

Step 2.3

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

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Step 2.3.1

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

$[\begin{array}{c} 1 & - 3 & 2 & 5 \\ \frac{0}{7} & \frac{7}{7} & \frac{0}{7} & \frac{14}{7} \\ 0 & 12 & - 9 & - 39 \end{array}]$

Step 2.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 3 & 2 & 5 \\ 0 & 1 & 0 & 2 \\ 0 & 12 & - 9 & - 39 \end{array}]$

Step 2.4

Perform the row operation $R_{3} = R_{3} - 12 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 2.4.1

Perform the row operation $R_{3} = R_{3} - 12 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - 3 & 2 & 5 \\ 0 & 1 & 0 & 2 \\ 0 - 12 \cdot 0 & 12 - 12 \cdot 1 & - 9 - 12 \cdot 0 & - 39 - 12 \cdot 2 \end{array}]$

Step 2.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 3 & 2 & 5 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & - 9 & - 63 \end{array}]$

Step 2.5

Multiply each element of $R_{3}$ by $- \frac{1}{9}$ to make the entry at $3, 3$ a $1$ .

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Step 2.5.1

Multiply each element of $R_{3}$ by $- \frac{1}{9}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - 3 & 2 & 5 \\ 0 & 1 & 0 & 2 \\ - \frac{1}{9} \cdot 0 & - \frac{1}{9} \cdot 0 & - \frac{1}{9} \cdot - 9 & - \frac{1}{9} \cdot - 63 \end{array}]$

Step 2.5.2

Simplify $R_{3}$ .

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

Step 2.6

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

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Step 2.6.1

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

$[\begin{array}{c} 1 - 2 \cdot 0 & - 3 - 2 \cdot 0 & 2 - 2 \cdot 1 & 5 - 2 \cdot 7 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 7 \end{array}]$

Step 2.6.2

Simplify $R_{1}$ .

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

Step 2.7

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 2.7.1

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

$[\begin{array}{c} 1 + 3 \cdot 0 & - 3 + 3 \cdot 1 & 0 + 3 \cdot 0 & - 9 + 3 \cdot 2 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 7 \end{array}]$

Step 2.7.2

Simplify $R_{1}$ .

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

Step 3

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

$x_{1} = - 3$

$x_{2} = 2$

$x_{3} = 7$

Step 4

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

$(- 3, 2, 7)$