If the trace of matrix `A=[[x-2,e^x,-sinx],[cos x^2, x^2-x+3,ln |x|],[0,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 values of \( x \) for which the trace of the matrix \( A \) is zero. The trace of a matrix is defined as the sum of its diagonal elements.

Given the matrix:
\[ A = \begin{bmatrix} x - 2 & e^x & -\sin x \\ \cos x^2 & x^2 - x + 3 & \ln |x| \\ 0 & \tan^{-1} x & x - 7 \end{bmatrix} \]

### Step 1: Identify the diagonal elements
The diagonal elements of the matrix \( A \) are:
1. \( x - 2 \) (from the first row)
2. \( x^2 - x + 3 \) (from the second row)
3. \( x - 7 \) (from the third row)

### Step 2: Write the equation for the trace
The trace of the matrix \( A \) is given by the sum of its diagonal elements:
\[ \text{Trace}(A) = (x - 2) + (x^2 - x + 3) + (x - 7) \]

### Step 3: Simplify the equation
Now, we simplify the expression:
\[ \text{Trace}(A) = (x - 2) + (x^2 - x + 3) + (x - 7) \]
Combining 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
Since we are given that the trace of the matrix is zero, we set the equation to zero:
\[ x^2 + x - 6 = 0 \]

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

### Step 6: Solve for \( x \)
Setting each factor equal to zero gives us:
1. \( x - 2 = 0 \) → \( x = 2 \)
2. \( x + 3 = 0 \) → \( x = -3 \)

### Conclusion
The values of \( x \) for which the trace of the matrix \( A \) is zero are:
\[ x = 2 \quad \text{and} \quad x = -3 \]