Linear Algebra Examples

5 + (- 1) + 2 (- 8) = - 12

$5 + (- 1) + 2 (- 8) = - 12$ ,

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

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

2 (5) + 5 (- 1) + (- 8) = 0

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

Step 1

Remove parentheses.

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

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

Step 2

Simplify.

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Step 2.1

Multiply

2

$2$ by

- 8

$- 8$ .

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

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

Step 2.2

Subtract

1

$1$ from

5

$5$ .

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

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

Step 2.3

Subtract

16

$16$ from

4

$4$ .

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

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

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

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

Step 3

Add

12

$12$ to both sides of the equation.

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

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

Step 4

Add

- 12

$- 12$ and

12

$12$ .

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

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

Step 5

Simplify.

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Step 5.1

Simplify each term.

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Step 5.1.1

Multiply

- 2

$- 2$ by

5

$5$ .

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

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

Step 5.1.2

Multiply

3

$3$ by

- 1

$- 1$ .

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

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

Step 5.1.3

Multiply

3

$3$ by

- 8

$- 8$ .

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

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

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

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

Step 5.2

Subtract

3

$3$ from

- 10

$- 10$ .

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

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

Step 5.3

Subtract

24

$24$ from

- 13

$- 13$ .

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

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

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

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

Step 6

Add

37

$37$ to both sides of the equation.

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

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

Step 7

Add

4

$4$ and

37

$37$ .

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

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

Step 8

Simplify.

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Step 8.1

Simplify each term.

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Step 8.1.1

Multiply

2

$2$ by

5

$5$ .

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

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

Step 8.1.2

Multiply

5

$5$ by

- 1

$- 1$ .

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

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

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

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

Step 8.2

Subtract

5

$5$ from

10

$10$ .

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

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

Step 8.3

Subtract

8

$8$ from

5

$5$ .

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

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

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

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

Step 9

Add

3

$3$ to both sides of the equation.

0 = 0, 0 = 41, 0 = 3

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

Step 10

Write the system of equations in matrix form.

⎡ ⎢ ⎣ \begin{matrix} 0413 \end{matrix} ⎤ ⎥ ⎦

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

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