Linear Algebra Examples

$\frac{2}{x} - \frac{3}{y} = 2$ , $- \frac{10}{x} + \frac{9}{y} = - 8$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} \frac{2}{1} & - \frac{3}{1} & 2 \\ - \frac{10}{1} & \frac{9}{1} & - 8 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

Divide $2$ by $1$ .

$[\begin{array}{c} 2 & - \frac{3}{1} & 2 \\ - \frac{10}{1} & \frac{9}{1} & - 8 \end{array}]$

Step 2.2

Divide $3$ by $1$ .

$[\begin{array}{c} 2 & - 1 \cdot 3 & 2 \\ - \frac{10}{1} & \frac{9}{1} & - 8 \end{array}]$

Step 2.3

Multiply $- 1$ by $3$ .

$[\begin{array}{c} 2 & - 3 & 2 \\ - \frac{10}{1} & \frac{9}{1} & - 8 \end{array}]$

Step 2.4

Divide $10$ by $1$ .

$[\begin{array}{c} 2 & - 3 & 2 \\ - 1 \cdot 10 & \frac{9}{1} & - 8 \end{array}]$

Step 2.5

Multiply $- 1$ by $10$ .

$[\begin{array}{c} 2 & - 3 & 2 \\ - 10 & \frac{9}{1} & - 8 \end{array}]$

Step 2.6

Divide $9$ by $1$ .

$[\begin{array}{c} 2 & - 3 & 2 \\ - 10 & 9 & - 8 \end{array}]$

Step 2.7

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

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Step 2.7.1

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

$[\begin{array}{c} \frac{2}{2} & \frac{- 3}{2} & \frac{2}{2} \\ - 10 & 9 & - 8 \end{array}]$

Step 2.7.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{3}{2} & 1 \\ - 10 & 9 & - 8 \end{array}]$

Step 2.8

Perform the row operation $R_{2} = R_{2} + 10 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 2.8.1

Perform the row operation $R_{2} = R_{2} + 10 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{3}{2} & 1 \\ - 10 + 10 \cdot 1 & 9 + 10 (- \frac{3}{2}) & - 8 + 10 \cdot 1 \end{array}]$

Step 2.8.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{3}{2} & 1 \\ 0 & - 6 & 2 \end{array}]$

Step 2.9

Multiply each element of $R_{2}$ by $- \frac{1}{6}$ to make the entry at $2, 2$ a $1$ .

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Step 2.9.1

Multiply each element of $R_{2}$ by $- \frac{1}{6}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{3}{2} & 1 \\ - \frac{1}{6} \cdot 0 & - \frac{1}{6} \cdot - 6 & - \frac{1}{6} \cdot 2 \end{array}]$

Step 2.9.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{3}{2} & 1 \\ 0 & 1 & - \frac{1}{3} \end{array}]$

Step 2.10

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

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Step 2.10.1

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

$[\begin{array}{c} 1 + \frac{3}{2} \cdot 0 & - \frac{3}{2} + \frac{3}{2} \cdot 1 & 1 + \frac{3}{2} (- \frac{1}{3}) \\ 0 & 1 & - \frac{1}{3} \end{array}]$

Step 2.10.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{1}{2} \\ 0 & 1 & - \frac{1}{3} \end{array}]$

Step 3

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

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

$y = - \frac{1}{3}$

Step 4

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

$(\frac{1}{2}, - \frac{1}{3})$

Step 5

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{1}{2} \\ - \frac{1}{3} \end{array}]$