Linear Algebra Examples

$9 x_{2} - 7 x_{3} = 2$ $- x_{3} = - 2$ $- 3 x_{1} + 6 x_{2} + 8 x_{3} = 1$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 0 & 9 & - 7 & 2 \\ 0 & 0 & - 1 & - 2 \\ - 3 & 6 & 8 & 1 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

Swap $R_{3}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

$[\begin{array}{c} - 3 & 6 & 8 & 1 \\ 0 & 0 & - 1 & - 2 \\ 0 & 9 & - 7 & 2 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot 6 & - \frac{1}{3} \cdot 8 & - \frac{1}{3} \cdot 1 \\ 0 & 0 & - 1 & - 2 \\ 0 & 9 & - 7 & 2 \end{array}]$

Step 2.2.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 2 & - \frac{8}{3} & - \frac{1}{3} \\ 0 & 0 & - 1 & - 2 \\ 0 & 9 & - 7 & 2 \end{array}]$

Step 2.3

Swap $R_{3}$ with $R_{2}$ to put a nonzero entry at $2, 2$ .

$[\begin{array}{c} 1 & - 2 & - \frac{8}{3} & - \frac{1}{3} \\ 0 & 9 & - 7 & 2 \\ 0 & 0 & - 1 & - 2 \end{array}]$

Step 2.4

Multiply each element of $R_{2}$ by $\frac{1}{9}$ to make the entry at $2, 2$ a $1$ .

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Step 2.4.1

Multiply each element of $R_{2}$ by $\frac{1}{9}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 2 & - \frac{8}{3} & - \frac{1}{3} \\ \frac{0}{9} & \frac{9}{9} & \frac{- 7}{9} & \frac{2}{9} \\ 0 & 0 & - 1 & - 2 \end{array}]$

Step 2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 2 & - \frac{8}{3} & - \frac{1}{3} \\ 0 & 1 & - \frac{7}{9} & \frac{2}{9} \\ 0 & 0 & - 1 & - 2 \end{array}]$

Step 2.5

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

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Step 2.5.1

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

$[\begin{array}{c} 1 & - 2 & - \frac{8}{3} & - \frac{1}{3} \\ 0 & 1 & - \frac{7}{9} & \frac{2}{9} \\ - 0 & - 0 & - - 1 & - - 2 \end{array}]$

Step 2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 2 & - \frac{8}{3} & - \frac{1}{3} \\ 0 & 1 & - \frac{7}{9} & \frac{2}{9} \\ 0 & 0 & 1 & 2 \end{array}]$

Step 2.6

Perform the row operation $R_{2} = R_{2} + \frac{7}{9} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 2.6.1

Perform the row operation $R_{2} = R_{2} + \frac{7}{9} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & - 2 & - \frac{8}{3} & - \frac{1}{3} \\ 0 + \frac{7}{9} \cdot 0 & 1 + \frac{7}{9} \cdot 0 & - \frac{7}{9} + \frac{7}{9} \cdot 1 & \frac{2}{9} + \frac{7}{9} \cdot 2 \\ 0 & 0 & 1 & 2 \end{array}]$

Step 2.6.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 2 & - \frac{8}{3} & - \frac{1}{3} \\ 0 & 1 & 0 & \frac{16}{9} \\ 0 & 0 & 1 & 2 \end{array}]$

Step 2.7

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

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Step 2.7.1

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

$[\begin{array}{c} 1 + \frac{8}{3} \cdot 0 & - 2 + \frac{8}{3} \cdot 0 & - \frac{8}{3} + \frac{8}{3} \cdot 1 & - \frac{1}{3} + \frac{8}{3} \cdot 2 \\ 0 & 1 & 0 & \frac{16}{9} \\ 0 & 0 & 1 & 2 \end{array}]$

Step 2.7.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 2 & 0 & 5 \\ 0 & 1 & 0 & \frac{16}{9} \\ 0 & 0 & 1 & 2 \end{array}]$

Step 2.8

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

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Step 2.8.1

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

$[\begin{array}{c} 1 + 2 \cdot 0 & - 2 + 2 \cdot 1 & 0 + 2 \cdot 0 & 5 + 2 (\frac{16}{9}) \\ 0 & 1 & 0 & \frac{16}{9} \\ 0 & 0 & 1 & 2 \end{array}]$

Step 2.8.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{77}{9} \\ 0 & 1 & 0 & \frac{16}{9} \\ 0 & 0 & 1 & 2 \end{array}]$

Step 3

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

$x_{1} = \frac{77}{9}$

$x_{2} = \frac{16}{9}$

$x_{3} = 2$

Step 4

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

$(\frac{77}{9}, \frac{16}{9}, 2)$