Linear Algebra Examples

$x - 15 = 27$ , $x = - 27 + 15$ , $x = - 12$

Step 1

Add $15$ to both sides of the equation.

$x = 27 + 15, x = - 27 + 15, x = - 12$

Step 2

Add $27$ and $15$ .

$x = 42, x = - 27 + 15, x = - 12$

Step 3

Add $- 27$ and $15$ .

$x = 42, x = - 12, x = - 12$

Step 4

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 42 \\ 1 & - 12 \\ 1 & - 12 \end{array}]$

Step 5

Find the reduced row echelon form.

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Step 5.1

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

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Step 5.1.1

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

$[\begin{array}{c} 1 & 42 \\ 1 - 1 & - 12 - 42 \\ 1 & - 12 \end{array}]$

Step 5.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 42 \\ 0 & - 54 \\ 1 & - 12 \end{array}]$

Step 5.2

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

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Step 5.2.1

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

$[\begin{array}{c} 1 & 42 \\ 0 & - 54 \\ 1 - 1 & - 12 - 42 \end{array}]$

Step 5.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 42 \\ 0 & - 54 \\ 0 & - 54 \end{array}]$

Step 5.3

Multiply each element of $R_{2}$ by $- \frac{1}{54}$ to make the entry at $2, 2$ a $1$ .

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Step 5.3.1

Multiply each element of $R_{2}$ by $- \frac{1}{54}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 42 \\ - \frac{1}{54} \cdot 0 & - \frac{1}{54} \cdot - 54 \\ 0 & - 54 \end{array}]$

Step 5.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 42 \\ 0 & 1 \\ 0 & - 54 \end{array}]$

Step 5.4

Perform the row operation $R_{3} = R_{3} + 54 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 5.4.1

Perform the row operation $R_{3} = R_{3} + 54 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & 42 \\ 0 & 1 \\ 0 + 54 \cdot 0 & - 54 + 54 \cdot 1 \end{array}]$

Step 5.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 42 \\ 0 & 1 \\ 0 & 0 \end{array}]$

Step 5.5

Perform the row operation $R_{1} = R_{1} - 42 R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 5.5.1

Perform the row operation $R_{1} = R_{1} - 42 R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - 42 \cdot 0 & 42 - 42 \cdot 1 \\ 0 & 1 \\ 0 & 0 \end{array}]$

Step 5.5.2

Simplify $R_{1}$ .

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

Step 6

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

$x = 0$

$0 = 1$

Step 7

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

No solution