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

Question

If \[ A = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \] and $\theta = \frac{2\pi}{7}$, then $A^{100} = A \times A \times \ldots$ (100 times) is equal to:

Answer 1

Step 1: Replace the trig entries.
Let $x=\sin^2\theta$, so $\cos^2\theta=1-x$, and since both are squares of real numbers, $0\le x\le 1$.
Step 2: Rewrite the matrix.
The determinant becomes that of $\begin{bmatrix}5&x&1-x\\-x&-5&1\\1-x&1&5\end{bmatrix}$, a function of the single variable $x$.
Step 3: Expand to a polynomial in $x$.
Carrying out the expansion along the first row collapses the entries to $\det(A)=10x^2-10x-125$.
Step 4: Complete the square.
Write $10x^2-10x-125=10\left(x-\tfrac12\right)^2-\tfrac{255}{2}$, since $10\cdot\tfrac14=\tfrac52$ and $-125+\tfrac52$ is not used here; instead the constant collects as $-\tfrac{255}{2}$.
Step 5: Maximize on $[0,1]$.
The squared term $\left(x-\tfrac12\right)^2$ is largest at the ends $x=0$ or $x=1$, where it equals $\tfrac14$.
Step 6: Evaluate the maximum.
Then $\det(A)=10\cdot\tfrac14-\tfrac{255}{2}=\tfrac52-\tfrac{255}{2}=-125$. So the greatest value of the determinant is $-125$.
\[ \boxed{-125} \]

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

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The Correct Option is A

Solution and Explanation

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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