Linear Algebra Examples

$\frac{x}{7} + \frac{y}{8} = 0$ , $\frac{1}{7} x - \frac{3}{4} y = 7$

Step 1

Simplify each term.

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Step 1.1

Combine $\frac{1}{7}$ and $x$ .

$\frac{x}{7} + \frac{y}{8} = 0, \frac{x}{7} - \frac{3}{4} y = 7$

Step 1.2

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

$\frac{x}{7} + \frac{y}{8} = 0, \frac{x}{7} - \frac{y \cdot 3}{4} = 7$

Step 1.3

Move $3$ to the left of $y$ .

$\frac{x}{7} + \frac{y}{8} = 0, \frac{x}{7} - \frac{3 y}{4} = 7$

Step 2

Write the system of equations in matrix form.

$[\begin{array}{c} \frac{1}{7} & \frac{1}{8} & 0 \\ \frac{1}{7} & - \frac{3}{4} & 7 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Multiply each element of $R_{1}$ by $7$ to make the entry at $1, 1$ a $1$ .

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Step 3.1.1

Multiply each element of $R_{1}$ by $7$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} 7 (\frac{1}{7}) & 7 (\frac{1}{8}) & 7 \cdot 0 \\ \frac{1}{7} & - \frac{3}{4} & 7 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{7}{8} & 0 \\ \frac{1}{7} & - \frac{3}{4} & 7 \end{array}]$

Step 3.2

Perform the row operation $R_{2} = R_{2} - \frac{1}{7} R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.2.1

Perform the row operation $R_{2} = R_{2} - \frac{1}{7} R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{7}{8} & 0 \\ \frac{1}{7} - \frac{1}{7} \cdot 1 & - \frac{3}{4} - \frac{1}{7} \cdot \frac{7}{8} & 7 - \frac{1}{7} \cdot 0 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{7}{8} & 0 \\ 0 & - \frac{7}{8} & 7 \end{array}]$

Step 3.3

Multiply each element of $R_{2}$ by $- \frac{8}{7}$ to make the entry at $2, 2$ a $1$ .

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Step 3.3.1

Multiply each element of $R_{2}$ by $- \frac{8}{7}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{7}{8} & 0 \\ - \frac{8}{7} \cdot 0 & - \frac{8}{7} (- \frac{7}{8}) & - \frac{8}{7} \cdot 7 \end{array}]$

Step 3.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{7}{8} & 0 \\ 0 & 1 & - 8 \end{array}]$

Step 3.4

Perform the row operation $R_{1} = R_{1} - \frac{7}{8} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 3.4.1

Perform the row operation $R_{1} = R_{1} - \frac{7}{8} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{7}{8} \cdot 0 & \frac{7}{8} - \frac{7}{8} \cdot 1 & 0 - \frac{7}{8} \cdot - 8 \\ 0 & 1 & - 8 \end{array}]$

Step 3.4.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 7 \\ 0 & 1 & - 8 \end{array}]$

Step 4

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

$x = 7$

$y = - 8$

Step 5

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

$(7, - 8)$

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 7 \\ - 8 \end{array}]$