Linear Algebra Examples

$y = \frac{- 1}{2} x + 4$ , $y = \frac{- 1}{2} x - 2$

Step 1

Simplify each term.

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Step 1.1

Move the negative in front of the fraction.

$y = - \frac{1}{2} x + 4, y = \frac{- 1}{2} x - 2$

Step 1.2

Combine $x$ and $\frac{1}{2}$ .

$y = - \frac{x}{2} + 4, y = \frac{- 1}{2} x - 2$

Step 2

Add $\frac{x}{2}$ to both sides of the equation.

$y + \frac{x}{2} = 4, y = \frac{- 1}{2} x - 2$

Step 3

Simplify each term.

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Step 3.1

Move the negative in front of the fraction.

$y + \frac{x}{2} = 4, y = - \frac{1}{2} x - 2$

Step 3.2

Combine $x$ and $\frac{1}{2}$ .

$y + \frac{x}{2} = 4, y = - \frac{x}{2} - 2$

Step 4

Add $\frac{x}{2}$ to both sides of the equation.

$y + \frac{x}{2} = 4, y + \frac{x}{2} = - 2$

Step 5

Write the system of equations in matrix form.

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

Step 6

Find the reduced row echelon form.

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Step 6.1

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

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Step 6.1.1

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

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

Step 6.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 2 & 8 \\ \frac{1}{2} & 1 & - 2 \end{array}]$

Step 6.2

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

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Step 6.2.1

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

$[\begin{array}{c} 1 & 2 & 8 \\ \frac{1}{2} - \frac{1}{2} \cdot 1 & 1 - \frac{1}{2} \cdot 2 & - 2 - \frac{1}{2} \cdot 8 \end{array}]$

Step 6.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 2 & 8 \\ 0 & 0 & - 6 \end{array}]$

Step 6.3

Multiply each element of $R_{2}$ by $- \frac{1}{6}$ to make the entry at $2, 3$ a $1$ .

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Step 6.3.1

Multiply each element of $R_{2}$ by $- \frac{1}{6}$ to make the entry at $2, 3$ a $1$ .

$[\begin{array}{c} 1 & 2 & 8 \\ - \frac{1}{6} \cdot 0 & - \frac{1}{6} \cdot 0 & - \frac{1}{6} \cdot - 6 \end{array}]$

Step 6.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 2 & 8 \\ 0 & 0 & 1 \end{array}]$

Step 6.4

Perform the row operation $R_{1} = R_{1} - 8 R_{2}$ to make the entry at $1, 3$ a $0$ .

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Step 6.4.1

Perform the row operation $R_{1} = R_{1} - 8 R_{2}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - 8 \cdot 0 & 2 - 8 \cdot 0 & 8 - 8 \cdot 1 \\ 0 & 0 & 1 \end{array}]$

Step 6.4.2

Simplify $R_{1}$ .

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

Step 7

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

$x + 2 y = 0$

$0 = 1$

Step 8

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

$x = - 2 y$

$0 = 1$

Step 9

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

No solution