Linear Algebra Examples

$- \frac{1}{8 x} + 5 y = \frac{1}{2}$ , $x - 40 y = - 4$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} - \frac{1}{8} & 5 & \frac{1}{2} \\ 1 & - 40 & - 4 \end{array}]$

Step 2

Row reduce to eliminate one of the variables.

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Step 2.1

Multiply each element of $R_{1}$ by $- 8$ to make the entry at $1, 1$ a $1$ .

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Step 2.1.1

Multiply each element of $R_{1}$ by $- 8$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - 8 (- \frac{1}{8}) & - 8 \cdot 5 & - 8 (\frac{1}{2}) \\ 1 & - 40 & - 4 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 40 & - 4 \\ 1 & - 40 & - 4 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} 1 & - 40 & - 4 \\ 1 - 1 & - 40 + 40 & - 4 + 4 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 40 & - 4 \\ 0 & 0 & 0 \end{array}]$

Step 3

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

$x - 40 y = - 4$

Step 4

Solve the equation for $y$ .

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Step 4.1

Subtract $x$ from both sides of the equation.

$- 40 y = - 4 - x$

Step 4.2

Divide each term in $- 40 y = - 4 - x$ by $- 40$ and simplify.

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Step 4.2.1

Divide each term in $- 40 y = - 4 - x$ by $- 40$ .

$\frac{- 40 y}{- 40} = \frac{- 4}{- 40} + \frac{- x}{- 40}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Cancel the common factor of $- 40$ .

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Step 4.2.2.1.1

Cancel the common factor.

$\frac{- 40 y}{- 40} = \frac{- 4}{- 40} + \frac{- x}{- 40}$

Step 4.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{- 4}{- 40} + \frac{- x}{- 40}$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Simplify each term.

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Step 4.2.3.1.1

Cancel the common factor of $- 4$ and $- 40$ .

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Step 4.2.3.1.1.1

Factor $- 4$ out of $- 4$ .

$y = \frac{- 4 (1)}{- 40} + \frac{- x}{- 40}$

Step 4.2.3.1.1.2

Cancel the common factors.

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Step 4.2.3.1.1.2.1

Factor $- 4$ out of $- 40$ .

$y = \frac{- 4 \cdot 1}{- 4 \cdot 10} + \frac{- x}{- 40}$

Step 4.2.3.1.1.2.2

Cancel the common factor.

$y = \frac{- 4 \cdot 1}{- 4 \cdot 10} + \frac{- x}{- 40}$

Step 4.2.3.1.1.2.3

Rewrite the expression.

$y = \frac{1}{10} + \frac{- x}{- 40}$

Step 4.2.3.1.2

Dividing two negative values results in a positive value.

$y = \frac{1}{10} + \frac{x}{40}$

Step 5

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

$(x, \frac{1}{10} + \frac{x}{40})$

Step 6

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} x \\ \frac{1}{10} + \frac{x}{40} \end{array}]$