Answer:
The set of solutions is ![\{\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}12\\-7-r\\r\end{array}\right]: \text{r is a real number} \}](https://tex.z-dn.net/?f=%5C%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D12%5C%5C-7-r%5C%5Cr%5Cend%7Barray%7D%5Cright%5D%3A%20%5Ctext%7Br%20is%20a%20real%20number%7D%20%20%5C%7D)
Step-by-step explanation:
The augmented matrix of the system is ![\left[\begin{array}{ccccc}3&6&6&-9\\-2&-3&-3&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D3%266%266%26-9%5C%5C-2%26-3%26-3%263%5Cend%7Barray%7D%5Cright%5D)
.
We will use rows operations for find the echelon form of the matrix.
- In row 2 we subtract
from row 1. (R2- 2/3R1) and we obtain the matrix ![\left[\begin{array}{cccc}3&6&6&-9\\0&1&1&-7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D3%266%266%26-9%5C%5C0%261%261%26-7%5Cend%7Barray%7D%5Cright%5D)
- We multiply the row 1 by
.
Now we solve for the unknown variables:
The system has a free variable, the the system has infinite solutions and the set of solutions is
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X = 4 ; x = 3 + i ; x = 3 - i
(If you get a zero that is adding or subtracting, you always need to write it twice but change the sign do they cancel out)
f(x) = (x-4)(x-3-i)(x-3+i)
Distributing the last two parenthesis first is always the best way to start off
(x-3-i)(x-3+i) has (x-3) in common so it can be separated to
(x-3)^2 + (-i)(+i)
(x^2 - 6x + 9) ; (-i)(+i) is always +1
(x^2 - 6x + 9) + 1
(x^2 - 6x + 10)
Now multiply this with (x-4)
x^3 - 6x^2 + 10x
- 4x^2 + 24x - 40
x^3 - 10x^2 + 34x - 40 = f(x)
,
,
then