Linear Algebra Examples

$0.6 x + 3.6 y = - 12$ , $- 0.9 + 0.6 = 6$

Step 1

Add $- 0.9$ and $0.6$ .

$0.6 x + 3.6 y = - 12, - 0.3 = 6$

Step 2

Add $0.3$ to both sides of the equation.

$0.6 x + 3.6 y = - 12, 0 = 6 + 0.3$

Step 3

Add $6$ and $0.3$ .

$0.6 x + 3.6 y = - 12, 0 = 6.3$

Step 4

Write the system of equations in matrix form.

$[\begin{array}{c} 0.6 & 3.6 & - 12 \\ 0 & 0 & 6.3 \end{array}]$

Step 5

Find the reduced row echelon form.

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Step 5.1

Multiply each element of $R_{1}$ by $\frac{1}{0.6}$ to make the entry at $1, 1$ a $1$ .

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Step 5.1.1

Multiply each element of $R_{1}$ by $\frac{1}{0.6}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{0.6}{0.6} & \frac{3.6}{0.6} & \frac{- 12}{0.6} \\ 0 & 0 & 6.3 \end{array}]$

Step 5.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 6 & - 20 \\ 0 & 0 & 6.3 \end{array}]$

Step 5.2

Multiply each element of $R_{2}$ by $\frac{1}{6.3}$ to make the entry at $2, 3$ a $1$ .

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Step 5.2.1

Multiply each element of $R_{2}$ by $\frac{1}{6.3}$ to make the entry at $2, 3$ a $1$ .

$[\begin{array}{c} 1 & 6 & - 20 \\ \frac{0}{6.3} & \frac{0}{6.3} & \frac{6.3}{6.3} \end{array}]$

Step 5.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 6 & - 20 \\ 0 & 0 & 1 \end{array}]$

Step 5.3

Perform the row operation $R_{1} = R_{1} + 20 R_{2}$ to make the entry at $1, 3$ a $0$ .

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Step 5.3.1

Perform the row operation $R_{1} = R_{1} + 20 R_{2}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 + 20 \cdot 0 & 6 + 20 \cdot 0 & - 20 + 20 \cdot 1 \\ 0 & 0 & 1 \end{array}]$

Step 5.3.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 6 & 0 \\ 0 & 0 & 1 \end{array}]$

Step 6

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

$x + 6 y = 0$

$0 = 1$

Step 7

Subtract $6 y$ from both sides of the equation.

$x = - 6 y$

$0 = 1$

Step 8

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

No solution