Solving for x in a determinant

[Kartik.]

Solve for x
x -6 -1
2 -3x x-3 = 0
-3 2x x+2Attempt-
x-2 3(x-2) -(x-2)
2 -3x x-3
-3 2x x+2

1 3 -1
2 -3x x-3
-3 2x x+2

1 3 -1
0 -3x-6 x-1 = 0
0 -x+3 2(x-1)
x =-3

By doing this I'm getting just one value of x. How do i get the values of x like 2 and 1?

 

[jedishrfu]

Its very hard to read your determinant or even that it is a determinant.

[ (x)   (-6)   (-1)  ]
[ (2)   (-3x)  (x-3) ]
[ (-3)  (2x)   (x+2) ]

Is this right?

You need to evaluate the determinant to get a polynomial in x and then factor it.

 

[Kartik.]

Its very hard to read your determinate or even that it is a determinate.

[ (x)   (-6)   (-1)  ]
[ (2)   (-3x)  (x-3) ]
[ (-3)  (2)   (x+2)  ]

Is this right?

You need to evaluate the determinate to get a polynomial in x and then factor it.

Its 2x in (column 2,row 3).

 

[jedishrfu]

Its 2x in (column 2,row 3).

Noted and fixed in my prior post.

Did my answer help?

 

[Mark44]

Solve for x
x -6 -1
2 -3x x-3 = 0
-3 2x x+2


Attempt-
x-2 3(x-2) -(x-2)
2 -3x x-3
-3 2x x+2

1 3 -1
2 -3x x-3
-3 2x x+2

1 3 -1
0 -3x-6 x-1 = 0
0 -x+3 2(x-1)
x =-3

By doing this I'm getting just one value of x. How do i get the values of x like 2 and 1?

Why don't you just expand the determinant directly?

It would be helpful to us if you told us what you are doing in each step above. Apparently in the 2nd step of your attempt, you have divided the first row by x - 2. In doing this, you are ignoring the possibility that x - 2 = 0, thereby losing x = 2 as a solution. Instead of dividing by x - 2, bring x - 2 out as a factor.

The basic idea is this:
$$ \begin{vmatrix} ka & kb \\ c & d \end{vmatrix} = k\begin{vmatrix} a & b \\ c & d\end{vmatrix}$$