Linear Algebra Examples

$x + 3 y = 10$ , $4 x + 12 y = 40$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 3 & 10 \\ 4 & 12 & 40 \end{array}]$

Step 2

Perform the row operation $R_{2} = R_{2} - 4 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 2.1

Perform the row operation $R_{2} = R_{2} - 4 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 3 & 10 \\ 4 - 4 \cdot 1 & 12 - 4 \cdot 3 & 40 - 4 \cdot 10 \end{array}]$

Step 2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 3 & 10 \\ 0 & 0 & 0 \end{array}]$

Step 3

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

$x + 3 y = 10$

Step 4

Solve the equation for $y$ .

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Step 4.1

Subtract $x$ from both sides of the equation.

$3 y = 10 - x$

Step 4.2

Divide each term in $3 y = 10 - x$ by $3$ and simplify.

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Step 4.2.1

Divide each term in $3 y = 10 - x$ by $3$ .

$\frac{3 y}{3} = \frac{10}{3} + \frac{- x}{3}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Cancel the common factor of $3$ .

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Step 4.2.2.1.1

Cancel the common factor.

$\frac{3 y}{3} = \frac{10}{3} + \frac{- x}{3}$

Step 4.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{10}{3} + \frac{- x}{3}$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Move the negative in front of the fraction.

$y = \frac{10}{3} - \frac{x}{3}$

Step 5

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

$(x, \frac{10}{3} - \frac{x}{3})$

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} x \\ \frac{10}{3} - \frac{x}{3} \end{array}]$