Linear Algebra Examples

$y = - \frac{1}{2} x + 3$ , $y = x + 3$

Step 1

Combine $x$ and $\frac{1}{2}$ .

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

Step 2

Add $\frac{x}{2}$ to both sides of the equation.

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

Step 3

Subtract $x$ from both sides of the equation.

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

Step 4

Write the system of equations in matrix form.

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

Step 5

Find the reduced row echelon form.

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Step 5.1

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

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Step 5.1.1

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

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

Step 5.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 2 & 6 \\ - 1 & 1 & 3 \end{array}]$

Step 5.2

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

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Step 5.2.1

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

$[\begin{array}{c} 1 & 2 & 6 \\ - 1 + 1 \cdot 1 & 1 + 1 \cdot 2 & 3 + 1 \cdot 6 \end{array}]$

Step 5.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 2 & 6 \\ 0 & 3 & 9 \end{array}]$

Step 5.3

Multiply each element of $R_{2}$ by $\frac{1}{3}$ to make the entry at $2, 2$ a $1$ .

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Step 5.3.1

Multiply each element of $R_{2}$ by $\frac{1}{3}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 2 & 6 \\ \frac{0}{3} & \frac{3}{3} & \frac{9}{3} \end{array}]$

Step 5.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 2 & 6 \\ 0 & 1 & 3 \end{array}]$

Step 5.4

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

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Step 5.4.1

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

$[\begin{array}{c} 1 - 2 \cdot 0 & 2 - 2 \cdot 1 & 6 - 2 \cdot 3 \\ 0 & 1 & 3 \end{array}]$

Step 5.4.2

Simplify $R_{1}$ .

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

Step 6

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

$x = 0$

$y = 3$

Step 7

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

$(0, 3)$

Step 8

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 0 \\ 3 \end{array}]$