Linear Algebra Examples

$5 + (- 1) + 2 (- 8) = - 12$ , $- 2 \cdot 5 + 3 (- 1) + 3 (- 8) = 4$ , $2 (5) + 5 (- 1) + (- 8) = 0$

Step 1

Remove parentheses.

$5 - 1 + 2 (- 8) = - 12, - 2 \cdot 5 + 3 (- 1) + 3 (- 8) = 4, 2 (5) + 5 (- 1) - 8 = 0$

Step 2

Simplify.

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Step 2.1

Multiply $2$ by $- 8$ .

$5 - 1 - 16 = - 12, - 2 \cdot 5 + 3 (- 1) + 3 (- 8) = 4, 2 (5) + 5 (- 1) - 8 = 0$

Step 2.2

Subtract $1$ from $5$ .

$4 - 16 = - 12, - 2 \cdot 5 + 3 (- 1) + 3 (- 8) = 4, 2 (5) + 5 (- 1) - 8 = 0$

Step 2.3

Subtract $16$ from $4$ .

$- 12 = - 12, - 2 \cdot 5 + 3 (- 1) + 3 (- 8) = 4, 2 (5) + 5 (- 1) - 8 = 0$

Step 3

Add $12$ to both sides of the equation.

$0 = - 12 + 12, - 2 \cdot 5 + 3 (- 1) + 3 (- 8) = 4, 2 (5) + 5 (- 1) - 8 = 0$

Step 4

Add $- 12$ and $12$ .

$0 = 0, - 2 \cdot 5 + 3 (- 1) + 3 (- 8) = 4, 2 (5) + 5 (- 1) - 8 = 0$

Step 5

Simplify.

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Step 5.1

Simplify each term.

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Step 5.1.1

Multiply $- 2$ by $5$ .

$0 = 0, - 10 + 3 (- 1) + 3 (- 8) = 4, 2 (5) + 5 (- 1) - 8 = 0$

Step 5.1.2

Multiply $3$ by $- 1$ .

$0 = 0, - 10 - 3 + 3 (- 8) = 4, 2 (5) + 5 (- 1) - 8 = 0$

Step 5.1.3

Multiply $3$ by $- 8$ .

$0 = 0, - 10 - 3 - 24 = 4, 2 (5) + 5 (- 1) - 8 = 0$

Step 5.2

Subtract $3$ from $- 10$ .

$0 = 0, - 13 - 24 = 4, 2 (5) + 5 (- 1) - 8 = 0$

Step 5.3

Subtract $24$ from $- 13$ .

$0 = 0, - 37 = 4, 2 (5) + 5 (- 1) - 8 = 0$

Step 6

Add $37$ to both sides of the equation.

$0 = 0, 0 = 4 + 37, 2 (5) + 5 (- 1) - 8 = 0$

Step 7

Add $4$ and $37$ .

$0 = 0, 0 = 41, 2 (5) + 5 (- 1) - 8 = 0$

Step 8

Simplify.

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Step 8.1

Simplify each term.

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Step 8.1.1

Multiply $2$ by $5$ .

$0 = 0, 0 = 41, 10 + 5 (- 1) - 8 = 0$

Step 8.1.2

Multiply $5$ by $- 1$ .

$0 = 0, 0 = 41, 10 - 5 - 8 = 0$

Step 8.2

Subtract $5$ from $10$ .

$0 = 0, 0 = 41, 5 - 8 = 0$

Step 8.3

Subtract $8$ from $5$ .

$0 = 0, 0 = 41, - 3 = 0$

Step 9

Add $3$ to both sides of the equation.

$0 = 0, 0 = 41, 0 = 3$

Step 10

Write the system of equations in matrix form.

$[\begin{array}{c} 0 \\ 41 \\ 3 \end{array}]$

Step 11

Find the reduced row echelon form.

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Step 11.1

Swap $R_{2}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

$[\begin{array}{c} 41 \\ 0 \\ 3 \end{array}]$

Step 11.2

Multiply each element of $R_{1}$ by $\frac{1}{41}$ to make the entry at $1, 1$ a $1$ .

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Step 11.2.1

Multiply each element of $R_{1}$ by $\frac{1}{41}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{41}{41} \\ 0 \\ 3 \end{array}]$

Step 11.2.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 \\ 0 \\ 3 \end{array}]$

Step 11.3

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

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Step 11.3.1

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

$[\begin{array}{c} 1 \\ 0 \\ 3 - 3 \cdot 1 \end{array}]$

Step 11.3.2

Simplify $R_{3}$ .

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

Step 12

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

$0 = 1$

Step 13

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

No solution