Linear Algebra Examples

$2 x + 3 y - z = - 1$ , $6 x + 9 y - 3 z = - 3$ , $- 2 x - 3 y + 2 z = 1$

Step 1

Write the system of equations in matrix form.

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

Step 2

Find the reduced row echelon form.

Tap for more steps...

Step 2.1

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

Tap for more steps...

Step 2.1.1

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

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

Step 2.1.2

Simplify $R_{1}$ .

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

Step 2.2

Perform the row operation $R_{2} = R_{2} - 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

Tap for more steps...

Step 2.2.1

Perform the row operation $R_{2} = R_{2} - 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & - \frac{1}{2} \\ 6 - 6 \cdot 1 & 9 - 6 (\frac{3}{2}) & - 3 - 6 (- \frac{1}{2}) & - 3 - 6 (- \frac{1}{2}) \\ - 2 & - 3 & 2 & 1 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

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

Step 2.3

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 2.3.1

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

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

Step 2.3.2

Simplify $R_{3}$ .

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

Step 2.4

Swap $R_{3}$ with $R_{2}$ to put a nonzero entry at $2, 3$ .

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

Step 2.5

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

Tap for more steps...

Step 2.5.1

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

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

Step 2.5.2

Simplify $R_{1}$ .

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

Step 3

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

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

$z = 0$

Step 4

Subtract $\frac{3 y}{2}$ from both sides of the equation.

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

$z = 0$

Step 5

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

$(- \frac{1}{2} - \frac{3 y}{2}, y, 0)$

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.

$X = [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} - \frac{1}{2} - \frac{3 y}{2} \\ y \\ 0 \end{array}]$