Question

If

\[ A = \begin{pmatrix} x & 2 \\ 2 & x \end{pmatrix} \quad \text{and} \quad \det(A^2) = 25, \] then \( x \) is equal to:

• A. \( \pm 3 \)
• B. \( \pm 1 \)
• C. \( \pm 2 \)
• D. \( \pm 4 \)
• E. \( \pm 5 \)

Hint:

When dealing with determinants, remember that \( \det(A^2) = (\det(A))^2 \). This simplifies the problem of finding values for \( x \) by equating the determinant expression to the given value.

Answer 1

The Correct Option is A

We are given that:

\[ A = \begin{pmatrix} x & 2 \\ 2 & x \end{pmatrix} \] and \[ \det(A^2) = 25. \] We are asked to find the value of \( x \).

Step 1: Recall that the determinant of a matrix product is the product of the determinants:

\[ \det(A^2) = \det(A) \cdot \det(A). \] Thus, we need to first calculate \( \det(A) \).

Step 2: The determinant of \( A \) is given by:

\[ \det(A) = \det \begin{pmatrix} x & 2 \\ 2 & x \end{pmatrix} = (x)(x) - (2)(2) = x^2 - 4. \]
Step 3: Now, we can calculate \( \det(A^2) \):

\[ \det(A^2) = (x^2 - 4)^2. \] We are given that \( \det(A^2) = 25 \), so:

\[ (x^2 - 4)^2 = 25. \]
Step 4: Taking the square root of both sides:

\[ x^2 - 4 = \pm 5. \]
Step 5: Solving for \( x^2 \):

- If \( x^2 - 4 = 5 \), then \( x^2 = 9 \), so \( x = \pm 3 \).

- If \( x^2 - 4 = -5 \), then \( x^2 = -1 \), which has no real solutions.

Thus, the only possible solution is \( x = \pm 3 \).

Therefore, the correct answer is option (A).