Linear Algebra Examples

$x - 5 y + z = 30$ , $5 x - y - z = 18$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & - 5 & 1 & 30 \\ 5 & - 1 & - 1 & 18 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} 1 & - 5 & 1 & 30 \\ 5 - 5 \cdot 1 & - 1 - 5 \cdot - 5 & - 1 - 5 \cdot 1 & 18 - 5 \cdot 30 \end{array}]$

Step 2.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 5 & 1 & 30 \\ 0 & 24 & - 6 & - 132 \end{array}]$

Step 2.2

Multiply each element of $R_{2}$ by $\frac{1}{24}$ to make the entry at $2, 2$ a $1$ .

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Step 2.2.1

Multiply each element of $R_{2}$ by $\frac{1}{24}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 5 & 1 & 30 \\ \frac{0}{24} & \frac{24}{24} & \frac{- 6}{24} & \frac{- 132}{24} \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 5 & 1 & 30 \\ 0 & 1 & - \frac{1}{4} & - \frac{11}{2} \end{array}]$

Step 2.3

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

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Step 2.3.1

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

$[\begin{array}{c} 1 + 5 \cdot 0 & - 5 + 5 \cdot 1 & 1 + 5 (- \frac{1}{4}) & 30 + 5 (- \frac{11}{2}) \\ 0 & 1 & - \frac{1}{4} & - \frac{11}{2} \end{array}]$

Step 2.3.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - \frac{1}{4} & \frac{5}{2} \\ 0 & 1 & - \frac{1}{4} & - \frac{11}{2} \end{array}]$

Step 3

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

$x - \frac{z}{4} = \frac{5}{2}$

$y - \frac{z}{4} = - \frac{11}{2}$

Step 4

Add $\frac{z}{4}$ to both sides of the equation.

$x = \frac{5}{2} + \frac{z}{4}$

$y - \frac{z}{4} = - \frac{11}{2}$

Step 5

Add $\frac{z}{4}$ to both sides of the equation.

$y = - \frac{11}{2} + \frac{z}{4}$

$x = \frac{5}{2} + \frac{z}{4}$

Step 6

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

$(\frac{5}{2} + \frac{z}{4}, - \frac{11}{2} + \frac{z}{4}, z)$

Step 7

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.

$X = [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{5}{2} + \frac{z}{4} \\ - \frac{11}{2} + \frac{z}{4} \\ z \end{array}]$