# Thread:

1. ## Find the solution set of the equation

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

2. Originally Posted by zorro
Question : Find the solution set of the equation $<br /> \begin{bmatrix}<br /> x & 3 & 7 \\<br /> 2 & x & 2\\<br /> 7 & 6 & x<br /> \end{bmatrix} = 0 <br />$
It is given that x= -9 is one of the roots
Your equation does not make any sense.

The zero matrix is $<br /> \begin{bmatrix}<br /> 0 & 0 & 0 \\<br /> 0 & 0 & 0\\<br /> 0 & 0 & 0<br /> \end{bmatrix}$

which is CLEARLY not the same as

$<br /> \begin{bmatrix}<br /> x & 3 & 7 \\<br /> 2 & x & 2\\<br /> 7 & 6 & x<br /> \end{bmatrix}$.

3. ## Might be this would make some sence

Originally Posted by Prove It
Your equation does not make any sense.

The zero matrix is $<br /> \begin{bmatrix}<br /> 0 & 0 & 0 \\<br /> 0 & 0 & 0\\<br /> 0 & 0 & 0<br /> \end{bmatrix}$

which is CLEARLY not the same as

$<br /> \begin{bmatrix}<br /> x & 3 & 7 \\<br /> 2 & x & 2\\<br /> 7 & 6 & x<br /> \end{bmatrix}$.

Might be this might make some think clear ......The matrix is not this $<br /> \begin{bmatrix}<br /> x & 3 & 7 \\<br /> 2 & x & 2\\<br /> 7 & 6 & x<br /> \end{bmatrix}<br />$ = 0 , but this $<br /> \begin{vmatrix}<br /> x & 3 & 7 \\<br /> 2 & x & 2\\<br /> 7 & 6 & x<br /> \end{vmatrix}<br />$ = 0

4. Originally Posted by zorro
Might be this might make some think clear ......The matrix is not this $<br /> \begin{bmatrix}<br /> x & 3 & 7 \\<br /> 2 & x & 2\\<br /> 7 & 6 & x<br /> \end{bmatrix}<br />$ = 0 , but this $<br /> \begin{vmatrix}<br /> x & 3 & 7 \\<br /> 2 & x & 2\\<br /> 7 & 6 & x<br /> \end{vmatrix}<br />$ = 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.

5. 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.

6. ## Is this correct

Originally Posted by mr fantastic
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 ???????

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

You've done this...

8. 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.)

9. ## Thank u for helping me

Originally Posted by HallsofIvy
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
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