Linear Algebra Examples

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

Step 1

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

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

Step 2

Subtract $\frac{x}{2}$ from both sides of the equation.

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

Step 3

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

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

Step 4

Subtract $\frac{x}{2}$ from 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 $1$ .

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

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

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

$x = 2 y$

$0 = 1$

Step 9

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

No solution