Linear Algebra Examples

$x + 2 y = 6$ , $- x - 2 y = - 6$

Step 1

Write the system of equations in matrix form.

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

Step 2

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

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Step 2.1

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

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

Step 2.2

Simplify $R_{2}$ .

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

Step 3

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

$x + 2 y = 6$

Step 4

Solve the equation for $y$ .

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Step 4.1

Subtract $x$ from both sides of the equation.

$2 y = 6 - x$

Step 4.2

Divide each term in $2 y = 6 - x$ by $2$ and simplify.

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Step 4.2.1

Divide each term in $2 y = 6 - x$ by $2$ .

$\frac{2 y}{2} = \frac{6}{2} + \frac{- x}{2}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Cancel the common factor of $2$ .

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Step 4.2.2.1.1

Cancel the common factor.

$\frac{2 y}{2} = \frac{6}{2} + \frac{- x}{2}$

Step 4.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{6}{2} + \frac{- x}{2}$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Simplify each term.

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Step 4.2.3.1.1

Divide $6$ by $2$ .

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

Step 4.2.3.1.2

Move the negative in front of the fraction.

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

Step 5

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

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

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.

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