Question

Let
\[ A= \begin{bmatrix} 5 & \sin^2\theta & \cos^2\theta \\ -\sin^2\theta & -5 & 1 \\ \cos^2\theta & 1 & 5 \end{bmatrix} \] Then, the maximum value of \(\det(A)\) is

• A. \(-125\)
• B. \(200\)
• C. \(-\dfrac{255}{2}\)
• D. \(145\)

Hint:

When a determinant contains \(\sin^2\theta\) and \(\cos^2\theta\), put \(\sin^2\theta=x\) and \(\cos^2\theta=1-x\), where \(0\leq x\leq 1\).

Answer 1

The Correct Option is A

Step 1: Substitute simple variables.
Let
\[ \sin^2\theta=x \] and
\[ \cos^2\theta=1-x \]
Since \(\sin^2\theta+\cos^2\theta=1\), we have
\[ 0\leq x\leq 1 \]

Step 2: Rewrite the matrix.
The matrix becomes
\[ A= \begin{bmatrix} 5 & x & 1-x \\ -x & -5 & 1 \\ 1-x & 1 & 5 \end{bmatrix} \]

Step 3: Find the determinant.
On expanding the determinant, we get
\[ \det(A)=10x^2-10x-125 \]
This can be written as
\[ \det(A)=10(x^2-x)-125 \]
Completing the square,
\[ x^2-x=\left(x-\frac12\right)^2-\frac14 \]
Therefore,
\[ \det(A)=10\left[\left(x-\frac12\right)^2-\frac14\right]-125 \]
\[ \det(A)=10\left(x-\frac12\right)^2-\frac{10}{4}-125 \]
\[ \det(A)=10\left(x-\frac12\right)^2-\frac{255}{2} \]

Step 4: Find the maximum value.
Since
\[ 0\leq x\leq 1 \] the term \(\left(x-\frac12\right)^2\) is maximum at \(x=0\) or \(x=1\).
At \(x=0\) or \(x=1\),
\[ \left(x-\frac12\right)^2=\frac14 \]
Hence,
\[ \det(A)=10\cdot \frac14-\frac{255}{2} \]
\[ \det(A)=\frac{5}{2}-\frac{255}{2} \]
\[ \det(A)=-125 \]

Step 5: Final conclusion.
Therefore, the maximum value of \(\det(A)\) is
\[ \boxed{-125} \]