Linear Algebra Examples

$10 x + 2 y = - 6$ , $y = - 5 x$

Step 1

Add $5 x$ to both sides of the equation.

$10 x + 2 y = - 6, y + 5 x = 0$

Step 2

Write the system of equations in matrix form.

$[\begin{array}{c} 10 & 2 & - 6 \\ 5 & 1 & 0 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Multiply each element of $R_{1}$ by $\frac{1}{10}$ to make the entry at $1, 1$ a $1$ .

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Step 3.1.1

Multiply each element of $R_{1}$ by $\frac{1}{10}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{10}{10} & \frac{2}{10} & \frac{- 6}{10} \\ 5 & 1 & 0 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{1}{5} & - \frac{3}{5} \\ 5 & 1 & 0 \end{array}]$

Step 3.2

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

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Step 3.2.1

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

$[\begin{array}{c} 1 & \frac{1}{5} & - \frac{3}{5} \\ 5 - 5 \cdot 1 & 1 - 5 (\frac{1}{5}) & 0 - 5 (- \frac{3}{5}) \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{5} & - \frac{3}{5} \\ 0 & 0 & 3 \end{array}]$

Step 3.3

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

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Step 3.3.1

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

$[\begin{array}{c} 1 & \frac{1}{5} & - \frac{3}{5} \\ \frac{0}{3} & \frac{0}{3} & \frac{3}{3} \end{array}]$

Step 3.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{5} & - \frac{3}{5} \\ 0 & 0 & 1 \end{array}]$

Step 3.4

Perform the row operation $R_{1} = R_{1} + \frac{3}{5} R_{2}$ to make the entry at $1, 3$ a $0$ .

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Step 3.4.1

Perform the row operation $R_{1} = R_{1} + \frac{3}{5} R_{2}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 + \frac{3}{5} \cdot 0 & \frac{1}{5} + \frac{3}{5} \cdot 0 & - \frac{3}{5} + \frac{3}{5} \cdot 1 \\ 0 & 0 & 1 \end{array}]$

Step 3.4.2

Simplify $R_{1}$ .

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

Step 4

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

$x + \frac{y}{5} = 0$

$0 = 1$

Step 5

Subtract $\frac{y}{5}$ from both sides of the equation.

$x = - \frac{y}{5}$

$0 = 1$

Step 6

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

No solution