Linear Algebra Examples

$y = 3 x + 2$ , $- 3 x + y = 2$

Step 1

Subtract $3 x$ from both sides of the equation.

$y - 3 x = 2, - 3 x + y = 2$

Step 2

Write the system of equations in matrix form.

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

Step 3

Row reduce to eliminate one of the variables.

Tap for more steps...

Step 3.1

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

Tap for more steps...

Step 3.1.1

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

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

Step 3.1.2

Simplify $R_{1}$ .

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

Step 3.2

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

Tap for more steps...

Step 3.2.1

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

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

Step 3.2.2

Simplify $R_{2}$ .

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

Step 4

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

$x - \frac{y}{3} = - \frac{2}{3}$

Step 5

Solve the equation for $y$ .

Tap for more steps...

Step 5.1

Subtract $x$ from both sides of the equation.

$- \frac{y}{3} = - \frac{2}{3} - x$

Step 5.2

Multiply both sides of the equation by $- 3$ .

$- 3 (- \frac{y}{3}) = - 3 (- \frac{2}{3} - x)$

Step 5.3

Simplify both sides of the equation.

Tap for more steps...

Step 5.3.1

Simplify the left side.

Tap for more steps...

Step 5.3.1.1

Simplify $- 3 (- \frac{y}{3})$ .

Tap for more steps...

Step 5.3.1.1.1

Cancel the common factor of $3$ .

Tap for more steps...

Step 5.3.1.1.1.1

Move the leading negative in $- \frac{y}{3}$ into the numerator.

$- 3 \frac{- y}{3} = - 3 (- \frac{2}{3} - x)$

Step 5.3.1.1.1.2

Factor $3$ out of $- 3$ .

$3 (- 1) \frac{- y}{3} = - 3 (- \frac{2}{3} - x)$

Step 5.3.1.1.1.3

Cancel the common factor.

$3 \cdot - 1 \frac{- y}{3} = - 3 (- \frac{2}{3} - x)$

Step 5.3.1.1.1.4

Rewrite the expression.

$- - y = - 3 (- \frac{2}{3} - x)$

Step 5.3.1.1.2

Multiply.

Tap for more steps...

Step 5.3.1.1.2.1

Multiply $- 1$ by $- 1$ .

$1 y = - 3 (- \frac{2}{3} - x)$

Step 5.3.1.1.2.2

Multiply $y$ by $1$ .

$y = - 3 (- \frac{2}{3} - x)$

Step 5.3.2

Simplify the right side.

Tap for more steps...

Step 5.3.2.1

Simplify $- 3 (- \frac{2}{3} - x)$ .

Tap for more steps...

Step 5.3.2.1.1

Apply the distributive property.

$y = - 3 (- \frac{2}{3}) - 3 (- x)$

Step 5.3.2.1.2

Cancel the common factor of $3$ .

Tap for more steps...

Step 5.3.2.1.2.1

Move the leading negative in $- \frac{2}{3}$ into the numerator.

$y = - 3 (\frac{- 2}{3}) - 3 (- x)$

Step 5.3.2.1.2.2

Factor $3$ out of $- 3$ .

$y = 3 (- 1) \frac{- 2}{3} - 3 (- x)$

Step 5.3.2.1.2.3

Cancel the common factor.

$y = 3 \cdot - 1 \frac{- 2}{3} - 3 (- x)$

Step 5.3.2.1.2.4

Rewrite the expression.

$y = - - 2 - 3 (- x)$

Step 5.3.2.1.3

Multiply.

Tap for more steps...

Step 5.3.2.1.3.1

Multiply $- 1$ by $- 2$ .

$y = 2 - 3 (- x)$

Step 5.3.2.1.3.2

Multiply $- 1$ by $- 3$ .

$y = 2 + 3 x$

Step 5.4

Reorder $2$ and $3 x$ .

$y = 3 x + 2$

Step 6

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

$(x, 3 x + 2)$

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.

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