Linear Algebra Examples

$x - 1 = 5$ , $x + 1 = 6$

Step 1

Add $1$ to both sides of the equation.

$x = 5 + 1, x + 1 = 6$

Step 2

Add $5$ and $1$ .

$x = 6, x + 1 = 6$

Step 3

Subtract $1$ from both sides of the equation.

$x = 6, x = 6 - 1$

Step 4

Subtract $1$ from $6$ .

$x = 6, x = 5$

Step 5

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 6 \\ 1 & 5 \end{array}]$

Step 6

Find the reduced row echelon form.

Tap for more steps...

Step 6.1

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

Tap for more steps...

Step 6.1.1

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

$[\begin{array}{c} 1 & 6 \\ 1 - 1 & 5 - 6 \end{array}]$

Step 6.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 6 \\ 0 & - 1 \end{array}]$

Step 6.2

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

Tap for more steps...

Step 6.2.1

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

$[\begin{array}{c} 1 & 6 \\ - 0 & - - 1 \end{array}]$

Step 6.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 6 \\ 0 & 1 \end{array}]$

Step 6.3

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

Tap for more steps...

Step 6.3.1

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

$[\begin{array}{c} 1 - 6 \cdot 0 & 6 - 6 \cdot 1 \\ 0 & 1 \end{array}]$

Step 6.3.2

Simplify $R_{1}$ .

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

Step 7

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

$x = 0$

$0 = 1$

Step 8

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

No solution