Linear Algebra Examples

$x = 9 y + 8$ , $x + 8 = 9 y$

Step 1

Subtract $9 y$ from both sides of the equation.

$x - 9 y = 8, x + 8 = 9 y$

Step 2

Subtract $8$ from both sides of the equation.

$x - 9 y = 8, x = 9 y - 8$

Step 3

Subtract $9 y$ from both sides of the equation.

$x - 9 y = 8, x - 9 y = - 8$

Step 4

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & - 9 & 8 \\ 1 & - 9 & - 8 \end{array}]$

Step 5

Find the reduced row echelon form.

Tap for more steps...

Step 5.1

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

Tap for more steps...

Step 5.1.1

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

$[\begin{array}{c} 1 & - 9 & 8 \\ 1 - 1 & - 9 + 9 & - 8 - 8 \end{array}]$

Step 5.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 9 & 8 \\ 0 & 0 & - 16 \end{array}]$

Step 5.2

Multiply each element of $R_{2}$ by $- \frac{1}{16}$ to make the entry at $2, 3$ a $1$ .

Tap for more steps...

Step 5.2.1

Multiply each element of $R_{2}$ by $- \frac{1}{16}$ to make the entry at $2, 3$ a $1$ .

$[\begin{array}{c} 1 & - 9 & 8 \\ - \frac{1}{16} \cdot 0 & - \frac{1}{16} \cdot 0 & - \frac{1}{16} \cdot - 16 \end{array}]$

Step 5.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 9 & 8 \\ 0 & 0 & 1 \end{array}]$

Step 5.3

Perform the row operation $R_{1} = R_{1} - 8 R_{2}$ to make the entry at $1, 3$ a $0$ .

Tap for more steps...

Step 5.3.1

Perform the row operation $R_{1} = R_{1} - 8 R_{2}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - 8 \cdot 0 & - 9 - 8 \cdot 0 & 8 - 8 \cdot 1 \\ 0 & 0 & 1 \end{array}]$

Step 5.3.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 9 & 0 \\ 0 & 0 & 1 \end{array}]$

Step 6

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

$x - 9 y = 0$

$0 = 1$

Step 7

Add $9 y$ to both sides of the equation.

$x = 9 y$

$0 = 1$

Step 8

Since $0 \neq 1$ , there are no solutions.

No solution