Linear Algebra Examples

$x + y = 90$ , $0.5 + 0.3 = (0.6) (90)$

Step 1

Simplify.

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Step 1.1

Add $0.5$ and $0.3$ .

$x + y = 90, 0.8 = (0.6) (90)$

Step 1.2

Multiply $0.6$ by $90$ .

$x + y = 90, 0.8 = 54$

Step 2

Subtract $0.8$ from both sides of the equation.

$x + y = 90, 0 = 54 - 0.8$

Step 3

Subtract $0.8$ from $54$ .

$x + y = 90, 0 = 53.2$

Step 4

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 1 & 90 \\ 0 & 0 & 53.2 \end{array}]$

Step 5

Find the reduced row echelon form.

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Step 5.1

Multiply each element of $R_{2}$ by $\frac{1}{53.2}$ to make the entry at $2, 3$ a $1$ .

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Step 5.1.1

Multiply each element of $R_{2}$ by $\frac{1}{53.2}$ to make the entry at $2, 3$ a $1$ .

$[\begin{array}{c} 1 & 1 & 90 \\ \frac{0}{53.2} & \frac{0}{53.2} & \frac{53.2}{53.2} \end{array}]$

Step 5.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 90 \\ 0 & 0 & 1 \end{array}]$

Step 5.2

Perform the row operation $R_{1} = R_{1} - 90 R_{2}$ to make the entry at $1, 3$ a $0$ .

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Step 5.2.1

Perform the row operation $R_{1} = R_{1} - 90 R_{2}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - 90 \cdot 0 & 1 - 90 \cdot 0 & 90 - 90 \cdot 1 \\ 0 & 0 & 1 \end{array}]$

Step 5.2.2

Simplify $R_{1}$ .

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

Step 6

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

$x + y = 0$

$0 = 1$

Step 7

Subtract $y$ from both sides of the equation.

$x = - y$

$0 = 1$

Step 8

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

No solution