Linear Algebra Examples

$x + y + z = 116000$ , $0.08 + 0.06 y + 0.09 z = 9000$ , $0.08 x = 4 (0.06 y)$

Step 1

Subtract $0.08$ from both sides of the equation.

$x + y + z = 116000, 0.06 y + 0.09 z = 9000 - 0.08, 0.08 x = 4 (0.06 y)$

Step 2

Subtract $0.08$ from $9000$ .

$x + y + z = 116000, 0.06 y + 0.09 z = 8999.92, 0.08 x = 4 (0.06 y)$

Step 3

Multiply $0.06$ by $4$ .

$x + y + z = 116000, 0.06 y + 0.09 z = 8999.92, 0.08 x = 0.24 y$

Step 4

Subtract $0.24 y$ from both sides of the equation.

$x + y + z = 116000, 0.06 y + 0.09 z = 8999.92, 0.08 x - 0.24 y = 0$

Step 5

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 1 & 1 & 116000 \\ 0 & 0.06 & 0.09 & 8999.92 \\ 0.08 & - 0.24 & 0 & 0 \end{array}]$

Step 6

Find the reduced row echelon form.

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Step 6.1

Perform the row operation $R_{3} = R_{3} - 0.08 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 6.1.1

Perform the row operation $R_{3} = R_{3} - 0.08 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & 1 & 1 & 116000 \\ 0 & 0.06 & 0.09 & 8999.92 \\ 0.08 - 0.08 \cdot 1 & - 0.24 - 0.08 \cdot 1 & 0 - 0.08 \cdot 1 & 0 - 0.08 \cdot 116000 \end{array}]$

Step 6.1.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 116000 \\ 0 & 0.06 & 0.09 & 8999.92 \\ 0 & - 0.32 & - 0.08 & - 9280 \end{array}]$

Step 6.2

Multiply each element of $R_{2}$ by $\frac{1}{0.06}$ to make the entry at $2, 2$ a $1$ .

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Step 6.2.1

Multiply each element of $R_{2}$ by $\frac{1}{0.06}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 1 & 1 & 116000 \\ \frac{0}{0.06} & \frac{0.06}{0.06} & \frac{0.09}{0.06} & \frac{8999.92}{0.06} \\ 0 & - 0.32 & - 0.08 & - 9280 \end{array}]$

Step 6.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 116000 \\ 0 & 1 & 1.5 & 149998.\overline{6} \\ 0 & - 0.32 & - 0.08 & - 9280 \end{array}]$

Step 6.3

Perform the row operation $R_{3} = R_{3} + 0.32 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 6.3.1

Perform the row operation $R_{3} = R_{3} + 0.32 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & 1 & 1 & 116000 \\ 0 & 1 & 1.5 & 149998.\overline{6} \\ 0 + 0.32 \cdot 0 & - 0.32 + 0.32 \cdot 1 & - 0.08 + 0.32 \cdot 1.5 & - 9280 + 0.32 \cdot 149998.\overline{6} \end{array}]$

Step 6.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 116000 \\ 0 & 1 & 1.5 & 149998.\overline{6} \\ 0 & 0 & 0.4 & 38719.57\overline{3} \end{array}]$

Step 6.4

Multiply each element of $R_{3}$ by $\frac{1}{0.4}$ to make the entry at $3, 3$ a $1$ .

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Step 6.4.1

Multiply each element of $R_{3}$ by $\frac{1}{0.4}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 1 & 1 & 116000 \\ 0 & 1 & 1.5 & 149998.\overline{6} \\ \frac{0}{0.4} & \frac{0}{0.4} & \frac{0.4}{0.4} & \frac{38719.57\overline{3}}{0.4} \end{array}]$

Step 6.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 116000 \\ 0 & 1 & 1.5 & 149998.\overline{6} \\ 0 & 0 & 1 & 96798.9\overline{3} \end{array}]$

Step 6.5

Perform the row operation $R_{2} = R_{2} - 1.5 R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 6.5.1

Perform the row operation $R_{2} = R_{2} - 1.5 R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & 1 & 1 & 116000 \\ 0 - 1.5 \cdot 0 & 1 - 1.5 \cdot 0 & 1.5 - 1.5 \cdot 1 & 149998.\overline{6} - 1.5 \cdot 96798.9\overline{3} \\ 0 & 0 & 1 & 96798.9\overline{3} \end{array}]$

Step 6.5.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 116000 \\ 0 & 1 & 0 & 4800.2\overline{6} \\ 0 & 0 & 1 & 96798.9\overline{3} \end{array}]$

Step 6.6

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

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Step 6.6.1

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

$[\begin{array}{c} 1 - 0 & 1 - 0 & 1 - 1 & 116000 - 96798.9\overline{3} \\ 0 & 1 & 0 & 4800.2\overline{6} \\ 0 & 0 & 1 & 96798.9\overline{3} \end{array}]$

Step 6.6.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 1 & 0 & 19201.0\overline{6} \\ 0 & 1 & 0 & 4800.2\overline{6} \\ 0 & 0 & 1 & 96798.9\overline{3} \end{array}]$

Step 6.7

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

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Step 6.7.1

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

$[\begin{array}{c} 1 - 0 & 1 - 1 & 0 + 0 & 19201.0\overline{6} - 4800.2\overline{6} \\ 0 & 1 & 0 & 4800.2\overline{6} \\ 0 & 0 & 1 & 96798.9\overline{3} \end{array}]$

Step 6.7.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 14400.8 \\ 0 & 1 & 0 & 4800.2\overline{6} \\ 0 & 0 & 1 & 96798.9\overline{3} \end{array}]$

Step 7

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

$x = 14400.8$

$y = 4800.2\overline{6}$

$z = 96798.9\overline{3}$

Step 8

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

$(14400.8, 4800.2\overline{6}, 96798.9\overline{3})$

Step 9

Decompose a solution vector by re-arranging each equation represented in the row-reduced form of the augmented matrix by solving for the dependent variable in each row yields the vector equality.

$X = [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 14400.8 \\ 4800.2\overline{6} \\ 96798.9\overline{3} \end{array}]$