Solving for x in a determinant

  • Thread starter Kartik.
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    Determinant
In summary: In the 3rd step, you are multiplying the last row of the determinant by -3. Doing so, you are creating a new matrix that is the inverse of the first matrix. This new matrix is created by multiplying the first matrix by -3 and then inverting the matrix.In summary, x=-6,-1.2.3x=-3,2x+2.
  • #1
Kartik.
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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?
 
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  • #2
Its very hard to read your determinant or even that it is a determinant.

Code:
[ (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.
 
Last edited:
  • #3
jedishrfu said:
Its very hard to read your determinate or even that it is a determinate.

Code:
[ (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).
 
Last edited:
  • #4
Kartik. said:
Its 2x in (column 2,row 3).

Noted and fixed in my prior post.

Did my answer help?
 
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  • #5
Kartik. said:
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:
$$ \begin{vmatrix} ka & kb \\ c & d \end{vmatrix} = k\begin{vmatrix} a & b \\ c & d\end{vmatrix}$$
 
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1. What is the purpose of solving for x in a determinant?

The purpose of solving for x in a determinant is to find the unknown variable x that satisfies the given system of equations. This allows us to find the specific values of the variables in the system and solve for the equations.

2. How do you solve for x in a determinant?

To solve for x in a determinant, we use the properties of determinants and perform operations such as row reduction, cofactor expansion, and substitution to simplify the equations and isolate the variable x. This allows us to find the unique solution for x.

3. What are the properties of determinants?

The properties of determinants include the fact that they are only defined for square matrices, they can be multiplied by a constant factor, their rows or columns can be interchanged, and they follow certain rules when adding or subtracting rows or columns. These properties allow us to manipulate the determinant to solve for x.

4. Can you solve for x in a determinant with non-linear equations?

Yes, it is possible to solve for x in a determinant with non-linear equations. However, the process may be more complex and involve methods such as substitution, elimination, or graphical methods. It may also result in multiple solutions or no solution at all.

5. Is solving for x in a determinant the same as solving for x in a system of equations?

No, solving for x in a determinant is not the same as solving for x in a system of equations. A determinant is a mathematical object that represents a system of equations, while solving for x in a system of equations involves finding the specific values of the variables in the equations. However, solving for x in a determinant is a step towards solving for x in a system of equations.

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