Linear Algebra Examples

$288 = 75 x + 85 x$ , $288 = 100 x$

Step 1

Add $75 x$ and $85 x$ .

$288 = 160 x, 288 = 100 x$

Step 2

Subtract $288$ from both sides of the equation.

$0 = 160 x - 288, 288 = 100 x$

Step 3

Subtract $160 x$ from both sides of the equation.

$- 160 x = - 288, 288 = 100 x$

Step 4

Subtract $288$ from both sides of the equation.

$- 160 x = - 288, 0 = 100 x - 288$

Step 5

Subtract $100 x$ from both sides of the equation.

$- 160 x = - 288, - 100 x = - 288$

Step 6

Write the system of equations in matrix form.

$[\begin{array}{c} - 160 & - 288 \\ - 100 & - 288 \end{array}]$

Step 7

Find the reduced row echelon form.

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Step 7.1

Multiply each element of $R_{1}$ by $- \frac{1}{160}$ to make the entry at $1, 1$ a $1$ .

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Step 7.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{160}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{160} \cdot - 160 & - \frac{1}{160} \cdot - 288 \\ - 100 & - 288 \end{array}]$

Step 7.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{9}{5} \\ - 100 & - 288 \end{array}]$

Step 7.2

Perform the row operation $R_{2} = R_{2} + 100 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 7.2.1

Perform the row operation $R_{2} = R_{2} + 100 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{9}{5} \\ - 100 + 100 \cdot 1 & - 288 + 100 (\frac{9}{5}) \end{array}]$

Step 7.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{9}{5} \\ 0 & - 108 \end{array}]$

Step 7.3

Multiply each element of $R_{2}$ by $- \frac{1}{108}$ to make the entry at $2, 2$ a $1$ .

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Step 7.3.1

Multiply each element of $R_{2}$ by $- \frac{1}{108}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{9}{5} \\ - \frac{1}{108} \cdot 0 & - \frac{1}{108} \cdot - 108 \end{array}]$

Step 7.3.2

Simplify $R_{2}$ .

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

Step 7.4

Perform the row operation $R_{1} = R_{1} - \frac{9}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 7.4.1

Perform the row operation $R_{1} = R_{1} - \frac{9}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

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

Step 7.4.2

Simplify $R_{1}$ .

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

Step 8

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

$x = 0$

$0 = 1$

Step 9

Since $0 \neq 1$ , there are no solutions.

No solution