Linear Algebra Examples

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

Step 1

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

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

Step 2

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

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

Step 3

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

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

Step 4

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

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

Step 5

Write the system of equations in matrix form.

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

Step 6

Row reduce to eliminate one of the variables.

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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 $1$ .

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

Step 6.1.2

Simplify $R_{1}$ .

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

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 $0$ .

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

Step 6.2.2

Simplify $R_{2}$ .

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

Step 7

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

$x - 2 y = - 2$

Step 8

Solve the equation for $y$ .

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Step 8.1

Subtract $x$ from both sides of the equation.

$- 2 y = - 2 - x$

Step 8.2

Divide each term in $- 2 y = - 2 - x$ by $- 2$ and simplify.

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Step 8.2.1

Divide each term in $- 2 y = - 2 - x$ by $- 2$ .

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

Step 8.2.2

Simplify the left side.

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Step 8.2.2.1

Cancel the common factor of $- 2$ .

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Step 8.2.2.1.1

Cancel the common factor.

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

Step 8.2.2.1.2

Divide $y$ by $1$ .

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

Step 8.2.3

Simplify the right side.

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Step 8.2.3.1

Simplify each term.

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Step 8.2.3.1.1

Divide $- 2$ by $- 2$ .

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

Step 8.2.3.1.2

Dividing two negative values results in a positive value.

$y = 1 + \frac{x}{2}$

Step 9

The solution is the set of ordered pairs that makes the system true.

$(x, 1 + \frac{x}{2})$

Step 10

Decompose a solution vector by re-arranging each equation represented in the row-reduced form of the augmented matrix by solving for the dependent variable in each row yields the vector equality.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} x \\ 1 + \frac{x}{2} \end{array}]$