Linear Algebra Examples

$8 x - 3 y = 5 x y$ , $6 x - 5 y = - 2 x y$

Step 1

Subtract $5 x y$ from both sides of the equation.

$8 x - 3 y - 5 x y = 0, 6 x - 5 y = - 2 x y$

Step 2

Add $2 x y$ to both sides of the equation.

$8 x - 3 y - 5 x y = 0, 6 x - 5 y + 2 x y = 0$

Step 3

Write the system of equations in matrix form.

$[\begin{array}{c} - 5 y & - 3 & 0 \\ 2 y & - 5 & 0 \end{array}]$

Step 4

Find the reduced row echelon form.

Tap for more steps...

Step 4.1

Multiply each element of $R_{1}$ by $\frac{1}{- 5 y}$ to make the entry at $1, 1$ a $1$ .

Tap for more steps...

Step 4.1.1

Multiply each element of $R_{1}$ by $\frac{1}{- 5 y}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{- 5 y}{- 5 y} & \frac{- 3}{- 5 y} & \frac{0}{- 5 y} \\ 2 y & - 5 & 0 \end{array}]$

Step 4.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3}{5 y} & 0 \\ 2 y & - 5 & 0 \end{array}]$

Step 4.2

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

Tap for more steps...

Step 4.2.1

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

$[\begin{array}{c} 1 & \frac{3}{5 y} & 0 \\ 2 y - 2 y \cdot 1 & - 5 - 2 y \frac{3}{5 y} & 0 - 2 y \cdot 0 \end{array}]$

Step 4.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{5 y} & 0 \\ 0 & - \frac{31}{5} & 0 \end{array}]$

Step 4.3

Multiply each element of $R_{2}$ by $- \frac{5}{31}$ to make the entry at $2, 2$ a $1$ .

Tap for more steps...

Step 4.3.1

Multiply each element of $R_{2}$ by $- \frac{5}{31}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{3}{5 y} & 0 \\ - \frac{5}{31} \cdot 0 & - \frac{5}{31} (- \frac{31}{5}) & - \frac{5}{31} \cdot 0 \end{array}]$

Step 4.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{5 y} & 0 \\ 0 & 1 & 0 \end{array}]$

Step 4.4

Perform the row operation $R_{1} = R_{1} - \frac{3}{5 y} R_{2}$ to make the entry at $1, 2$ a $0$ .

Tap for more steps...

Step 4.4.1

Perform the row operation $R_{1} = R_{1} - \frac{3}{5 y} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{3}{5 y} \cdot 0 & \frac{3}{5 y} - \frac{3}{5 y} \cdot 1 & 0 - \frac{3}{5 y} \cdot 0 \\ 0 & 1 & 0 \end{array}]$

Step 4.4.2

Simplify $R_{1}$ .

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

Step 5

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

$x = 0$

$y = 0$

Step 6

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

$(0, 0)$

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 0 \\ 0 \end{array}]$