Linear Algebra Examples

$(\frac{2}{3}) x + z = 5$ , $y + (\frac{3}{2}) z = 6$ , $2 x + 3 y = 8$

Step 1

Combine $\frac{2}{3}$ and $x$ .

$\frac{2 x}{3} + z = 5, y + (\frac{3}{2}) z = 6, 2 x + 3 y = 8$

Step 2

Combine $\frac{3}{2}$ and $z$ .

$\frac{2 x}{3} + z = 5, y + \frac{3 z}{2} = 6, 2 x + 3 y = 8$

Step 3

Write the system of equations in matrix form.

$[\begin{array}{c} \frac{2}{3} & 0 & 1 & 5 \\ 0 & 1 & \frac{3}{2} & 6 \\ 2 & 3 & 0 & 8 \end{array}]$

Step 4

Find the reduced row echelon form.

Tap for more steps...

Step 4.1

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

Tap for more steps...

Step 4.1.1

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

$[\begin{array}{c} \frac{3}{2} \cdot \frac{2}{3} & \frac{3}{2} \cdot 0 & \frac{3}{2} \cdot 1 & \frac{3}{2} \cdot 5 \\ 0 & 1 & \frac{3}{2} & 6 \\ 2 & 3 & 0 & 8 \end{array}]$

Step 4.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{3}{2} & \frac{15}{2} \\ 0 & 1 & \frac{3}{2} & 6 \\ 2 & 3 & 0 & 8 \end{array}]$

Step 4.2

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

Tap for more steps...

Step 4.2.1

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

$[\begin{array}{c} 1 & 0 & \frac{3}{2} & \frac{15}{2} \\ 0 & 1 & \frac{3}{2} & 6 \\ 2 - 2 \cdot 1 & 3 - 2 \cdot 0 & 0 - 2 (\frac{3}{2}) & 8 - 2 (\frac{15}{2}) \end{array}]$

Step 4.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 0 & \frac{3}{2} & \frac{15}{2} \\ 0 & 1 & \frac{3}{2} & 6 \\ 0 & 3 & - 3 & - 7 \end{array}]$

Step 4.3

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

Tap for more steps...

Step 4.3.1

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

$[\begin{array}{c} 1 & 0 & \frac{3}{2} & \frac{15}{2} \\ 0 & 1 & \frac{3}{2} & 6 \\ 0 - 3 \cdot 0 & 3 - 3 \cdot 1 & - 3 - 3 (\frac{3}{2}) & - 7 - 3 \cdot 6 \end{array}]$

Step 4.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 0 & \frac{3}{2} & \frac{15}{2} \\ 0 & 1 & \frac{3}{2} & 6 \\ 0 & 0 & - \frac{15}{2} & - 25 \end{array}]$

Step 4.4

Multiply each element of $R_{3}$ by $- \frac{2}{15}$ to make the entry at $3, 3$ a $1$ .

Tap for more steps...

Step 4.4.1

Multiply each element of $R_{3}$ by $- \frac{2}{15}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 0 & \frac{3}{2} & \frac{15}{2} \\ 0 & 1 & \frac{3}{2} & 6 \\ - \frac{2}{15} \cdot 0 & - \frac{2}{15} \cdot 0 & - \frac{2}{15} (- \frac{15}{2}) & - \frac{2}{15} \cdot - 25 \end{array}]$

Step 4.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 0 & \frac{3}{2} & \frac{15}{2} \\ 0 & 1 & \frac{3}{2} & 6 \\ 0 & 0 & 1 & \frac{10}{3} \end{array}]$

Step 4.5

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

Tap for more steps...

Step 4.5.1

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

$[\begin{array}{c} 1 & 0 & \frac{3}{2} & \frac{15}{2} \\ 0 - \frac{3}{2} \cdot 0 & 1 - \frac{3}{2} \cdot 0 & \frac{3}{2} - \frac{3}{2} \cdot 1 & 6 - \frac{3}{2} \cdot \frac{10}{3} \\ 0 & 0 & 1 & \frac{10}{3} \end{array}]$

Step 4.5.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & \frac{3}{2} & \frac{15}{2} \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & \frac{10}{3} \end{array}]$

Step 4.6

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

Tap for more steps...

Step 4.6.1

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

$[\begin{array}{c} 1 - \frac{3}{2} \cdot 0 & 0 - \frac{3}{2} \cdot 0 & \frac{3}{2} - \frac{3}{2} \cdot 1 & \frac{15}{2} - \frac{3}{2} \cdot \frac{10}{3} \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & \frac{10}{3} \end{array}]$

Step 4.6.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{5}{2} \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & \frac{10}{3} \end{array}]$

Step 5

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

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

$y = 1$

$z = \frac{10}{3}$

Step 6

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

$(\frac{5}{2}, 1, \frac{10}{3})$

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.

$X = [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{5}{2} \\ 1 \\ \frac{10}{3} \end{array}]$