Linear Algebra Examples

$x + 3 y + 6 = 0$ , $- x - 3 y = - 6$

Step 1

Subtract $6$ from both sides of the equation.

$x + 3 y = - 6, - x - 3 y = - 6$

Step 2

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 3 & - 6 \\ - 1 & - 3 & - 6 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

$[\begin{array}{c} 1 & 3 & - 6 \\ - 1 + 1 \cdot 1 & - 3 + 1 \cdot 3 & - 6 - 6 \end{array}]$

Step 3.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 3 & - 6 \\ 0 & 0 & - 12 \end{array}]$

Step 3.2

Multiply each element of $R_{2}$ by $- \frac{1}{12}$ to make the entry at $2, 3$ a $1$ .

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Step 3.2.1

Multiply each element of $R_{2}$ by $- \frac{1}{12}$ to make the entry at $2, 3$ a $1$ .

$[\begin{array}{c} 1 & 3 & - 6 \\ - \frac{1}{12} \cdot 0 & - \frac{1}{12} \cdot 0 & - \frac{1}{12} \cdot - 12 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

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

Step 3.3

Perform the row operation $R_{1} = R_{1} + 6 R_{2}$ to make the entry at $1, 3$ a $0$ .

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Step 3.3.1

Perform the row operation $R_{1} = R_{1} + 6 R_{2}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 + 6 \cdot 0 & 3 + 6 \cdot 0 & - 6 + 6 \cdot 1 \\ 0 & 0 & 1 \end{array}]$

Step 3.3.2

Simplify $R_{1}$ .

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

Step 4

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

$x + 3 y = 0$

$0 = 1$

Step 5

Subtract $3 y$ from both sides of the equation.

$x = - 3 y$

$0 = 1$

Step 6

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

No solution