Question

Let \( A \) be a \( 3 \times 3 \) matrix with real entries such that:

\[ A = \begin{bmatrix} 4 & -1 & \cos x \\ -1 & 5x & 25x^2 + 1 \\ 25 & 7 & ? \end{bmatrix} \]

For how many values of \( x \) is the matrix \( A \) symmetric?

• A. 1
• B. 2
• C. 4
• D. infinitely many

Hint:

Inequalities are powerful tools for solving transcendental equations.
Since $\cos x \le 1$ and $x^2 + 1 \ge 1$, the graphs of $y = \cos x$ and $y = x^2 + 1$ can only touch at their boundary value of $1$, which immediately restricts the solution to $x=0$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A matrix $A$ is symmetric if $A = A^T$, which means the elements $a_{ij} = a_{ji}$ for all $i, j$.
Step 2: Identifying Symmetric Pairs:
From the given matrix:
$a_{12} = a_{21} = -1$ (Already symmetric)
$a_{23} = a_{32} = 25$ (Already symmetric)
$a_{13} = \cos x$ and $a_{31} = x^2 + 1$.
For $A$ to be symmetric, we must have: $\cos x = x^2 + 1$.
Step 3: Solving the Equation:
We know that for any real $x$: $-1 \le \cos x \le 1$ and $x^2 + 1 \ge 1$. The only way these two can be equal is if both sides equal 1. $x^2 + 1 = 1 \implies x^2 = 0 \implies x = 0$. If $x = 0$, then $\cos(0) = 1$, which satisfies the equation. Wait, let's re-examine the question's provided options or matrix values. If $x=0$ is the only solution, the answer is 1. However, let's check if the question implies other elements depend on $x$. The only constraint is $\cos x = x^2 + 1$. In the real domain, only $x=0$ works.
Step 4: Final Answer:
There is 1 such value of $x$. (Note: If the option (B) is intended, there might be a typo in the provided matrix values or a complex domain is implied, but for real entries, $x=0$ is unique). Given standard real analysis, the answer is 1.