If the trace of the matrix `A=[(x-2,e^(x), -sinx),("cos"x^(2),x^(2)-x+3,"In"|x|),(cot x,"tan"^(-1)x,x-7)]` is zero, then x is equal to :

Options

A. `-2 or 3`
B. `-3 or -2`
C. `-3 or 2`
D. 2 or 3

Answer

To solve the problem, we need to find the value of \( x \) such that the trace of the matrix \( A \) is zero. The trace of a matrix is defined as the sum of the elements on its main diagonal.

Given the matrix:
\[ A = \begin{pmatrix} x - 2 & e^x & -\sin x \\ \cos x^2 & x^2 - x + 3 & \ln |x| \\ \cot x & \tan^{-1} x & x - 7 \end{pmatrix} \]

### Step 1: Identify the elements on the main diagonal
The elements on the main diagonal of matrix \( A \) are:
- First element: \( x - 2 \)
- Second element: \( x^2 - x + 3 \)
- Third element: \( x - 7 \)

### Step 2: Write the expression for the trace
The trace of matrix \( A \) can be expressed as:
\[ \text{Trace}(A) = (x - 2) + (x^2 - x + 3) + (x - 7) \]

### Step 3: Simplify the expression
Now, let's simplify the expression for the trace:
\[ \text{Trace}(A) = (x - 2) + (x^2 - x + 3) + (x - 7) \]
Combine like terms:
\[ = x - 2 + x^2 - x + 3 + x - 7 \]
\[ = x^2 + (x - x + x) + (-2 + 3 - 7) \]
\[ = x^2 + x - 6 \]

### Step 4: Set the trace equal to zero
We are given that the trace of the matrix is zero:
\[ x^2 + x - 6 = 0 \]

### Step 5: Factor the quadratic equation
To solve the quadratic equation \( x^2 + x - 6 = 0 \), we can factor it:
\[ x^2 + 3x - 2x - 6 = 0 \]
\[ (x + 3)(x - 2) = 0 \]

### Step 6: Solve for \( x \)
Setting each factor equal to zero gives us:
\[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \]
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]

Thus, the values of \( x \) that satisfy the equation are \( x = -3 \) or \( x = 2 \).

### Conclusion
The possible values of \( x \) are \( -3 \) and \( 2 \).