If the trace of the matrix `A= [{:( x-1 ,0,2,5),( 3, x^(2) - 2 ,4,1),( -1,-2,x-3,1),(2,0,4,x^(2)-6) :}]` is 0 then x is equal to

Options

A. ` (-2,3) `
B. ` (2,-3) `
C. ` ( -3,2) `
D. `( 3,-2) `

Answer

To solve the problem, we need to find the values of \( x \) for which the trace of the matrix \( A \) is equal to 0. The trace of a matrix is defined as the sum of its diagonal elements.

Given the matrix \( A \):

\[ A = \begin{pmatrix} x - 1 & 0 & 2 & 5 \\ 3 & x^2 - 2 & 4 & 1 \\ -1 & -2 & x - 3 & 1 \\ 2 & 0 & 4 & x^2 - 6 \end{pmatrix} \]

### Step 1: Identify the diagonal elements
The diagonal elements of the matrix \( A \) are:
- First row, first column: \( x - 1 \)
- Second row, second column: \( x^2 - 2 \)
- Third row, third column: \( x - 3 \)
- Fourth row, fourth column: \( x^2 - 6 \)

### Step 2: Write the expression for the trace
The trace \( \text{Tr}(A) \) is given by the sum of the diagonal elements:

\[ \text{Tr}(A) = (x - 1) + (x^2 - 2) + (x - 3) + (x^2 - 6) \]

### Step 3: Simplify the expression
Now, we simplify the expression:

\[ \text{Tr}(A) = x - 1 + x^2 - 2 + x - 3 + x^2 - 6 \]

Combining like terms:

\[ \text{Tr}(A) = 2x^2 + (x + x - 1 - 2 - 3 - 6) \]
\[ = 2x^2 + 2x - 12 \]

### Step 4: Set the trace equal to 0
We need to find \( x \) such that:

\[ 2x^2 + 2x - 12 = 0 \]

### Step 5: Divide the equation by 2
To simplify, divide the entire equation by 2:

\[ x^2 + x - 6 = 0 \]

### Step 6: Factor the quadratic equation
Now we factor the quadratic equation:

\[ (x + 3)(x - 2) = 0 \]

### Step 7: Solve for \( x \)
Setting each factor to zero gives us:

1. \( x + 3 = 0 \) → \( x = -3 \)
2. \( x - 2 = 0 \) → \( x = 2 \)

### Final Answer
Thus, the values of \( x \) for which the trace of the matrix \( A \) is 0 are:

\[ x = -3 \quad \text{and} \quad x = 2 \]