Linear Algebra Examples

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

Step 1

Write the system of equations in matrix form.

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

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

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

Step 2.1.2

Simplify $R_{2}$ .

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

Simplify $R_{2}$ .

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

Step 2.3

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

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Step 2.3.1

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

$[\begin{array}{c} 1 - 0 & 1 - 1 & 6 - (- 2 y + 6) \\ 0 & 1 & - 2 y + 6 \end{array}]$

Step 2.3.2

Simplify $R_{1}$ .

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

Step 3

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

$x = 2 y$

$y = - 2 y + 6$

Step 4

Solve the equation for $y$ .

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Step 4.1

Move all terms containing $y$ to the left side of the equation.

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Step 4.1.1

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

$y + 2 y = 6$

$x = 2 y$

Step 4.1.2

Add $y$ and $2 y$ .

$3 y = 6$

$x = 2 y$

$3 y = 6$

$x = 2 y$

Step 4.2

Divide each term in $3 y = 6$ by $3$ and simplify.

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Step 4.2.1

Divide each term in $3 y = 6$ by $3$ .

$\frac{3 y}{3} = \frac{6}{3}$

$x = 2 y$

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.

$\frac{3 y}{3} = \frac{6}{3}$

$x = 2 y$

Step 4.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{6}{3}$

$x = 2 y$

$y = \frac{6}{3}$

$x = 2 y$

$y = \frac{6}{3}$

$x = 2 y$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Divide $6$ by $3$ .

$y = 2$

$x = 2 y$

$y = 2$

$x = 2 y$

$y = 2$

$x = 2 y$

$y = 2$

$x = 2 y$

Step 5

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

$(2 y, 2)$

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.

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