Linear Algebra Examples

$y = \frac{3}{4} x - 3$ , $y = - \frac{2}{5} x + 1$

Step 1

Combine $\frac{3}{4}$ and $x$ .

$y = \frac{3 x}{4} - 3, y = - \frac{2}{5} x + 1$

Step 2

Subtract $\frac{3 x}{4}$ from both sides of the equation.

$y - \frac{3 x}{4} = - 3, y = - \frac{2}{5} x + 1$

Step 3

Simplify each term.

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Step 3.1

Combine $x$ and $\frac{2}{5}$ .

$y - \frac{3 x}{4} = - 3, y = - \frac{x \cdot 2}{5} + 1$

Step 3.2

Move $2$ to the left of $x$ .

$y - \frac{3 x}{4} = - 3, y = - \frac{2 x}{5} + 1$

Step 4

Add $\frac{2 x}{5}$ to both sides of the equation.

$y - \frac{3 x}{4} = - 3, y + \frac{2 x}{5} = 1$

Step 5

Write the system of equations in matrix form.

$[\begin{array}{c} - \frac{3}{4} & 1 & - 3 \\ \frac{2}{5} & 1 & 1 \end{array}]$

Step 6

Find the reduced row echelon form.

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Step 6.1

Multiply each element of $R_{1}$ by $- \frac{4}{3}$ to make the entry at $1, 1$ a $1$ .

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Step 6.1.1

Multiply each element of $R_{1}$ by $- \frac{4}{3}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{4}{3} (- \frac{3}{4}) & - \frac{4}{3} \cdot 1 & - \frac{4}{3} \cdot - 3 \\ \frac{2}{5} & 1 & 1 \end{array}]$

Step 6.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & 4 \\ \frac{2}{5} & 1 & 1 \end{array}]$

Step 6.2

Perform the row operation $R_{2} = R_{2} - \frac{2}{5} R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 6.2.1

Perform the row operation $R_{2} = R_{2} - \frac{2}{5} R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & 4 \\ \frac{2}{5} - \frac{2}{5} \cdot 1 & 1 - \frac{2}{5} (- \frac{4}{3}) & 1 - \frac{2}{5} \cdot 4 \end{array}]$

Step 6.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & 4 \\ 0 & \frac{23}{15} & - \frac{3}{5} \end{array}]$

Step 6.3

Multiply each element of $R_{2}$ by $\frac{15}{23}$ to make the entry at $2, 2$ a $1$ .

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Step 6.3.1

Multiply each element of $R_{2}$ by $\frac{15}{23}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & 4 \\ \frac{15}{23} \cdot 0 & \frac{15}{23} \cdot \frac{23}{15} & \frac{15}{23} (- \frac{3}{5}) \end{array}]$

Step 6.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & 4 \\ 0 & 1 & - \frac{9}{23} \end{array}]$

Step 6.4

Perform the row operation $R_{1} = R_{1} + \frac{4}{3} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 6.4.1

Perform the row operation $R_{1} = R_{1} + \frac{4}{3} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{4}{3} \cdot 0 & - \frac{4}{3} + \frac{4}{3} \cdot 1 & 4 + \frac{4}{3} (- \frac{9}{23}) \\ 0 & 1 & - \frac{9}{23} \end{array}]$

Step 6.4.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{80}{23} \\ 0 & 1 & - \frac{9}{23} \end{array}]$

Step 7

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

$x = \frac{80}{23}$

$y = - \frac{9}{23}$

Step 8

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

$(\frac{80}{23}, - \frac{9}{23})$

Step 9

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{80}{23} \\ - \frac{9}{23} \end{array}]$