Linear Algebra Examples

$x - 4 y = - 4$ , $4 y = x + 4$

Step 1

Subtract $x$ from both sides of the equation.

$x - 4 y = - 4, 4 y - x = 4$

Step 2

Write the system of equations in matrix form.

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

Step 3

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

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

$[\begin{array}{c} 1 & - 4 & - 4 \\ - 1 + 1 \cdot 1 & 4 - 4 & 4 - 4 \end{array}]$

Step 3.2

Simplify $R_{2}$ .

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

Step 4

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

$x - 4 y = - 4$

Step 5

Solve the equation for $y$ .

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Step 5.1

Subtract $x$ from both sides of the equation.

$- 4 y = - 4 - x$

Step 5.2

Divide each term in $- 4 y = - 4 - x$ by $- 4$ and simplify.

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Step 5.2.1

Divide each term in $- 4 y = - 4 - x$ by $- 4$ .

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

Step 5.2.2

Simplify the left side.

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Step 5.2.2.1

Cancel the common factor of $- 4$ .

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Step 5.2.2.1.1

Cancel the common factor.

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

Step 5.2.2.1.2

Divide $y$ by $1$ .

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

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

Simplify each term.

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Step 5.2.3.1.1

Divide $- 4$ by $- 4$ .

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

Step 5.2.3.1.2

Dividing two negative values results in a positive value.

$y = 1 + \frac{x}{4}$

Step 6

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

$(x, 1 + \frac{x}{4})$

Step 7

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} x \\ 1 + \frac{x}{4} \end{array}]$