Question : Find the solution set of the equation $\displaystyle
\begin{bmatrix}
x & 3 & 7 \\
2 & x & 2\\
7 & 6 & x
\end{bmatrix} = 0
$
It is given that x= -9 is one of the roots
Thanks fr ur reply
Might be this might make some think clear ......The matrix is not this $\displaystyle
\begin{bmatrix}
x & 3 & 7 \\
2 & x & 2\\
7 & 6 & x
\end{bmatrix}
$ = 0 , but this $\displaystyle
\begin{vmatrix}
x & 3 & 7 \\
2 & x & 2\\
7 & 6 & x
\end{vmatrix}
$ = 0
After taking the deteminant this is what i have got
$\displaystyle x(x^2 - 12) - 3(2x - 14) + 7(12 - 7x) = 0$
$\displaystyle x^3 - 12x - 6x + 42 + 84 - 49x = 0$
$\displaystyle x^3 - 67x + 126 = 0$
$\displaystyle (x + 9) (x^2 - 9x + 14)$
$\displaystyle (x + 9) (x - 7) (x - 2)$
So what is the answer
as $\displaystyle x = -9 , \ 7 , \ 2$
i didnt actually understand the question therefore i dont know what to do after this ???????
If you want to get really, really "technical" (and mathematicians are notorious for that!), since the problem asked for "solution set", the answer is the set of those solutions: {-9, 7, 2}. (Warning: I am assuming those are the correct solutions; I didn't check them.)
Thanks HallsofIvy and Prove It for helping me
Cheers