Find the solution set of the equation

zorro · #1 $Old$ October 31st, 2009, 10:59 PM

Question : Find the solution set of the equation $\begin{bmatrix}x & 3 & 7 \\2 & x & 2\\7 & 6 & x\end{bmatrix} = 0$
It is given that x= -9 is one of the roots

Prove It · #2 $Old$ October 31st, 2009, 11:30 PM

Quote:

Originally Posted by zorro $View Post$

Question : Find the solution set of the equation $\begin{bmatrix}x & 3 & 7 \\2 & x & 2\\7 & 6 & x\end{bmatrix} = 0$
It is given that x= -9 is one of the roots

Your equation does not make any sense.

The zero matrix is $\begin{bmatrix}0 & 0 & 0 \\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}$

which is CLEARLY not the same as

$\begin{bmatrix}x & 3 & 7 \\2 & x & 2\\7 & 6 & x\end{bmatrix}$ .

zorro · #3 $Old$ November 1st, 2009, 12:02 AM

Quote:

Originally Posted by Prove It $View Post$

Your equation does not make any sense.

The zero matrix is $\begin{bmatrix}0 & 0 & 0 \\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}$

which is CLEARLY not the same as

$\begin{bmatrix}x & 3 & 7 \\2 & x & 2\\7 & 6 & x\end{bmatrix}$ .

Thanks fr ur reply
Might be this might make some think clear ......The matrix is not this $\begin{bmatrix}x & 3 & 7 \\2 & x & 2\\7 & 6 & x\end{bmatrix}$ = 0 , but this $\begin{vmatrix}x & 3 & 7 \\2 & x & 2\\7 & 6 & x\end{vmatrix}$ = 0

mr fantastic · #4 $Old$ November 1st, 2009, 01:10 AM

Quote:

Originally Posted by zorro $View Post$

Thanks fr ur reply
Might be this might make some think clear ......The matrix is not this $\begin{bmatrix}x & 3 & 7 \\2 & x & 2\\7 & 6 & x\end{bmatrix}$ = 0 , but this $\begin{vmatrix}x & 3 & 7 \\2 & x & 2\\7 & 6 & x\end{vmatrix}$ = 0

The first step is to get an expression for the determinant. Then equate this expression to zero and solve for x. Please show what you've done and where you get stuck.

HallsofIvy · #5 $Old$ November 1st, 2009, 04:57 AM

The determinant will clearly be a cubic in x. Since you are already given that x= 9 is a root, you can factor it as (x-9) times some quadratic and then, if necessary, use the quadratic formula.

zorro · #6 $Old$ December 8th, 2009, 03:46 PM

Quote:

Originally Posted by mr fantastic $View Post$

The first step is to get an expression for the determinant. Then equate this expression to zero and solve for x. Please show what you've done and where you get stuck.

After taking the deteminant this is what i have got

$x(x^2 - 12) - 3(2x - 14) + 7(12 - 7x) = 0$

$x^3 - 12x - 6x + 42 + 84 - 49x = 0$

$x^3 - 67x + 126 = 0$

$(x + 9) (x^2 - 9x + 14)$

$(x + 9) (x - 7) (x - 2)$

So what is the answer
as $x = -9 , \ 7 , \ 2$

i didnt actually understand the question therefore i dont know what to do after this ???????

Prove It · #7 $Old$ December 8th, 2009, 04:17 PM

Wasn't it just asking you to solve for $x$ ?

You've done this...

HallsofIvy · #8 $Old$ December 9th, 2009, 09:30 AM

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.)

zorro · #9 $Old$ December 18th, 2009, 11:50 PM

Quote:

Originally Posted by HallsofIvy $View Post$

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