Linear Algebra Examples

$4 x + 6 y = 17$ , $y = 5 - 5 x$

Step 1

Add $5 x$ to both sides of the equation.

$4 x + 6 y = 17, y + 5 x = 5$

Step 2

Write the system of equations in matrix form.

$[\begin{array}{c} 4 & 6 & 17 \\ 5 & 1 & 5 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Multiply each element of $R_{1}$ by $\frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

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Step 3.1.1

Multiply each element of $R_{1}$ by $\frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{4}{4} & \frac{6}{4} & \frac{17}{4} \\ 5 & 1 & 5 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & \frac{17}{4} \\ 5 & 1 & 5 \end{array}]$

Step 3.2

Perform the row operation $R_{2} = R_{2} - 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.2.1

Perform the row operation $R_{2} = R_{2} - 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{2} & \frac{17}{4} \\ 5 - 5 \cdot 1 & 1 - 5 (\frac{3}{2}) & 5 - 5 (\frac{17}{4}) \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & \frac{17}{4} \\ 0 & - \frac{13}{2} & - \frac{65}{4} \end{array}]$

Step 3.3

Multiply each element of $R_{2}$ by $- \frac{2}{13}$ to make the entry at $2, 2$ a $1$ .

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Step 3.3.1

Multiply each element of $R_{2}$ by $- \frac{2}{13}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{3}{2} & \frac{17}{4} \\ - \frac{2}{13} \cdot 0 & - \frac{2}{13} (- \frac{13}{2}) & - \frac{2}{13} (- \frac{65}{4}) \end{array}]$

Step 3.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & \frac{17}{4} \\ 0 & 1 & \frac{5}{2} \end{array}]$

Step 3.4

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

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Step 3.4.1

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

$[\begin{array}{c} 1 - \frac{3}{2} \cdot 0 & \frac{3}{2} - \frac{3}{2} \cdot 1 & \frac{17}{4} - \frac{3}{2} \cdot \frac{5}{2} \\ 0 & 1 & \frac{5}{2} \end{array}]$

Step 3.4.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{1}{2} \\ 0 & 1 & \frac{5}{2} \end{array}]$

Step 4

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

$x = \frac{1}{2}$

$y = \frac{5}{2}$

Step 5

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

$(\frac{1}{2}, \frac{5}{2})$

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{1}{2} \\ \frac{5}{2} \end{array}]$