For how many values of \( x \) is the matrix \( A \) symmetric?
Show 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 $x=0$.
Step 1 : Understanding the Question:
The question asks to find the number of real values of $x$ for which the given $3 \times 3$ matrix $A$ is symmetric. Step 2 : Key Formulas and Approach:
A matrix $A$ is symmetric if and only if it is equal to its transpose ($A = A^T$), which implies $a_{ij} = a_{ji}$ for all indices $i$ and $j$.
We will set up the equality condition for the off-diagonal entries and solve the resulting transcendental equation. Step 3 : Detailed Explanation:
Let us compare the symmetric entries $a_{ij}$ and $a_{ji}$ of the matrix $A$:
• $a_{12} = -1$ and $a_{21} = -1$ (already equal)
• $a_{23} = 25$ and $a_{32} = 25$ (already equal)
• $a_{13} = \cos x$ and $a_{31} = x^2 + 1$
For $A$ to be symmetric, we must have:
\[ \cos x = x^2 + 1 \]
Let us analyze the range of the functions on both sides of this equation:
• For any real number $x$, the range of the cosine function is:
\[ -1 \le \cos x \le 1 \]
• For any real number $x$, we know that $x^2 \ge 0$. Adding 1 to both sides gives:
\[ x^2 + 1 \ge 1 \]
The only way for these two expressions to be equal is if both sides are simultaneously equal to $1$:
\[ \cos x = 1 \quad \text{and} \quad x^2 + 1 = 1 \]
Solving the quadratic equation:
\[ x^2 = 0 \implies x = 0 \]
Let us verify if $x = 0$ satisfies the first condition:
\[ \cos(0) = 1 \]
This is true. Thus, $x = 0$ is the unique real solution to the equation. Step 4 : Final Answer:
There is exactly 1 real value of $x$ ($x = 0$) for which the matrix $A$ is symmetric.
This corresponds to Option (A).
