Linear Algebra Examples

$x - 7 + 7 z = - 27$ , $- 6 x + y + 8 z = - 38$ , $2 x + 3 y + 7 z = - 26$

Step 1

Add $7$ to both sides of the equation.

$x + 7 z = - 27 + 7, - 6 x + y + 8 z = - 38, 2 x + 3 y + 7 z = - 26$

Step 2

Add $- 27$ and $7$ .

$x + 7 z = - 20, - 6 x + y + 8 z = - 38, 2 x + 3 y + 7 z = - 26$

Step 3

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 0 & 7 & - 20 \\ - 6 & 1 & 8 & - 38 \\ 2 & 3 & 7 & - 26 \end{array}]$

Step 4

Find the reduced row echelon form.

Tap for more steps...

Step 4.1

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 4.1.1

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

$[\begin{array}{c} 1 & 0 & 7 & - 20 \\ - 6 + 6 \cdot 1 & 1 + 6 \cdot 0 & 8 + 6 \cdot 7 & - 38 + 6 \cdot - 20 \\ 2 & 3 & 7 & - 26 \end{array}]$

Step 4.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & 7 & - 20 \\ 0 & 1 & 50 & - 158 \\ 2 & 3 & 7 & - 26 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & 0 & 7 & - 20 \\ 0 & 1 & 50 & - 158 \\ 2 - 2 \cdot 1 & 3 - 2 \cdot 0 & 7 - 2 \cdot 7 & - 26 - 2 \cdot - 20 \end{array}]$

Step 4.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 0 & 7 & - 20 \\ 0 & 1 & 50 & - 158 \\ 0 & 3 & - 7 & 14 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & 0 & 7 & - 20 \\ 0 & 1 & 50 & - 158 \\ 0 - 3 \cdot 0 & 3 - 3 \cdot 1 & - 7 - 3 \cdot 50 & 14 - 3 \cdot - 158 \end{array}]$

Step 4.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 0 & 7 & - 20 \\ 0 & 1 & 50 & - 158 \\ 0 & 0 & - 157 & 488 \end{array}]$

Step 4.4

Multiply each element of $R_{3}$ by $- \frac{1}{157}$ to make the entry at $3, 3$ a $1$ .

Tap for more steps...

Step 4.4.1

Multiply each element of $R_{3}$ by $- \frac{1}{157}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 0 & 7 & - 20 \\ 0 & 1 & 50 & - 158 \\ - \frac{1}{157} \cdot 0 & - \frac{1}{157} \cdot 0 & - \frac{1}{157} \cdot - 157 & - \frac{1}{157} \cdot 488 \end{array}]$

Step 4.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 0 & 7 & - 20 \\ 0 & 1 & 50 & - 158 \\ 0 & 0 & 1 & - \frac{488}{157} \end{array}]$

Step 4.5

Perform the row operation $R_{2} = R_{2} - 50 R_{3}$ to make the entry at $2, 3$ a $0$ .

Tap for more steps...

Step 4.5.1

Perform the row operation $R_{2} = R_{2} - 50 R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & 0 & 7 & - 20 \\ 0 - 50 \cdot 0 & 1 - 50 \cdot 0 & 50 - 50 \cdot 1 & - 158 - 50 (- \frac{488}{157}) \\ 0 & 0 & 1 & - \frac{488}{157} \end{array}]$

Step 4.5.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & 7 & - 20 \\ 0 & 1 & 0 & - \frac{406}{157} \\ 0 & 0 & 1 & - \frac{488}{157} \end{array}]$

Step 4.6

Perform the row operation $R_{1} = R_{1} - 7 R_{3}$ to make the entry at $1, 3$ a $0$ .

Tap for more steps...

Step 4.6.1

Perform the row operation $R_{1} = R_{1} - 7 R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - 7 \cdot 0 & 0 - 7 \cdot 0 & 7 - 7 \cdot 1 & - 20 - 7 (- \frac{488}{157}) \\ 0 & 1 & 0 & - \frac{406}{157} \\ 0 & 0 & 1 & - \frac{488}{157} \end{array}]$

Step 4.6.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{276}{157} \\ 0 & 1 & 0 & - \frac{406}{157} \\ 0 & 0 & 1 & - \frac{488}{157} \end{array}]$

Step 5

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

$x = \frac{276}{157}$

$y = - \frac{406}{157}$

$z = - \frac{488}{157}$

Step 6

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

$(\frac{276}{157}, - \frac{406}{157}, - \frac{488}{157})$

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.

$X = [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{276}{157} \\ - \frac{406}{157} \\ - \frac{488}{157} \end{array}]$