Linear Algebra Examples

$6 x + 3 y = 6$ , $y = 2 x - 4$ , $y = - 2 x + 6$

Step 1

Subtract $2 x$ from both sides of the equation.

$6 x + 3 y = 6, y - 2 x = - 4, y = - 2 x + 6$

Step 2

Add $2 x$ to both sides of the equation.

$6 x + 3 y = 6, y - 2 x = - 4, y + 2 x = 6$

Step 3

Write the system of equations in matrix form.

$[\begin{array}{c} 6 & 3 & 6 \\ - 2 & 1 & - 4 \\ 2 & 1 & 6 \end{array}]$

Step 4

Find the reduced row echelon form.

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Step 4.1

Multiply each element of $R_{1}$ by $\frac{1}{6}$ to make the entry at $1, 1$ a $1$ .

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Step 4.1.1

Multiply each element of $R_{1}$ by $\frac{1}{6}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{6}{6} & \frac{3}{6} & \frac{6}{6} \\ - 2 & 1 & - 4 \\ 2 & 1 & 6 \end{array}]$

Step 4.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{1}{2} & 1 \\ - 2 & 1 & - 4 \\ 2 & 1 & 6 \end{array}]$

Step 4.2

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

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Step 4.2.1

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

$[\begin{array}{c} 1 & \frac{1}{2} & 1 \\ - 2 + 2 \cdot 1 & 1 + 2 (\frac{1}{2}) & - 4 + 2 \cdot 1 \\ 2 & 1 & 6 \end{array}]$

Step 4.2.2

Simplify $R_{2}$ .

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

Step 4.3

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

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Step 4.3.1

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

$[\begin{array}{c} 1 & \frac{1}{2} & 1 \\ 0 & 2 & - 2 \\ 2 - 2 \cdot 1 & 1 - 2 (\frac{1}{2}) & 6 - 2 \cdot 1 \end{array}]$

Step 4.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{1}{2} & 1 \\ 0 & 2 & - 2 \\ 0 & 0 & 4 \end{array}]$

Step 4.4

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

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Step 4.4.1

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

$[\begin{array}{c} 1 & \frac{1}{2} & 1 \\ \frac{0}{2} & \frac{2}{2} & \frac{- 2}{2} \\ 0 & 0 & 4 \end{array}]$

Step 4.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{2} & 1 \\ 0 & 1 & - 1 \\ 0 & 0 & 4 \end{array}]$

Step 4.5

Multiply each element of $R_{3}$ by $\frac{1}{4}$ to make the entry at $3, 3$ a $1$ .

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Step 4.5.1

Multiply each element of $R_{3}$ by $\frac{1}{4}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{1}{2} & 1 \\ 0 & 1 & - 1 \\ \frac{0}{4} & \frac{0}{4} & \frac{4}{4} \end{array}]$

Step 4.5.2

Simplify $R_{3}$ .

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

Step 4.6

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

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Step 4.6.1

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

$[\begin{array}{c} 1 & \frac{1}{2} & 1 \\ 0 + 0 & 1 + 0 & - 1 + 1 \cdot 1 \\ 0 & 0 & 1 \end{array}]$

Step 4.6.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{2} & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$

Step 4.7

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

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Step 4.7.1

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

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

Step 4.7.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{1}{2} & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$

Step 4.8

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

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Step 4.8.1

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

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

Step 4.8.2

Simplify $R_{1}$ .

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

Step 5

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

$x = 0$

$y = 0$

$0 = 1$

Step 6

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

No solution