Question

Let $A\left(t\right)=\left[a_{ij}\right]$ be a matrix of order $3\times 3$ given by $a_{ij}=\{\begin{array}{cc}2\cos t,& \text{if}i=j\\ 1,& \text{if}|i-j|=1\\ 0,& \text{otherwise}\end{array}$ where $\left|A\right(t\left)\right|$ is determinant value of matrix $A\left(t\right).$ Then which of the following is/are CORRECT?

• A. $\underset{t\to 0}{lim}\left|A\right(t\left)\right|\left|A\right(4t\left)\right|=16$
• B. Maximum value of $\left|A\right(t\left)\right|\left|A\right(3t\left)\right|$ is $16$
• C. $\underset{0}{\overset{\pi }{\int }}\left|A\right(t\left)\right|\left|A\right(4t\left)\right|dt=0$
• D. $\left|A\left(\frac{\pi }{17}\right)\right|\left|A\left(\frac{4\pi }{17}\right)\right|=1$

Answer 1

Answer: (A), (B), (C), (D)

$\left|A\left(\frac{\pi }{17}\right)\right|\left|A\left(\frac{4\pi }{17}\right)\right|=1$

$A\left(t\right)=\left[a_{ij}\right]=\left[\begin{array}{ccc}2\cos t& 1& 0\\ 1& 2\cos t& 1\\ 0& 1& 2\cos t\end{array}\right]$

$\left|A\right(t\left)\right|=2\cos t\cdot (4{\cos }^{2}t-1)-1\cdot \left(2\cos t\right)$
$=8{\cos }^{3}t-4\cos t=4\cos t\cos 2t$

$\underset{t\to 0}{lim}\left|A\right(t\left)\right|\left|A\right(4t\left)\right|$ $=\underset{t\to 0}{lim}(16\cos t\cdot \cos 2t\cdot \cos 4t\cdot \cos 8t)=16$

$\left|A\right(t\left)\right|\left|A\right(3t\left)\right|=16\cos t\cdot \cos 2t\cdot \cos 3t\cdot \cos 6t$
We know, $\cos x\in [-1,1]$
Clearly, $\left|A\right(t\left)\right|\left|A\right(3t\left)\right|$ is maximum when $\cos 2t=\cos t=\cos 3t=\cos 6t=1$
That is possible at $t=n\pi$ where $n\in Z$
Hence, maximum value of $\left|A\right(t\left)\right|\left|A\right(3t\left)\right|$ is $16$

Let $I=\underset{0}{\overset{\pi }{\int }}\left|A\right(t\left)\right|\left|A\right(4t\left)\right|dt$
$\Rightarrow I=\underset{0}{\overset{\pi }{\int }}(16\cos t\cdot \cos 2t\cdot \cos 4t\cdot \cos 8t)dt\dots \left(1\right)$
Applying $\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$
$I=\underset{0}{\overset{\pi }{\int }}\left(16\cos \right(\pi -t)\cdot \cos (2\pi -2t)\cdot \cos (4\pi -4t)\cdot \cos (8\pi -8t\left)\right)dt\Rightarrow I=\underset{0}{\overset{\pi }{\int }}(-16\cos t\cdot \cos 2t\cdot \cos 4t\cdot \cos 8t)dt\dots \left(2\right)$
On adding $\left(1\right)$ and $\left(2\right),$
$2I=0\Rightarrow I=0$

$\left|A\right(t\left)\right|\left|A\right(4t\left)\right|$
$=16\cos t\cdot \cos 2t\cdot \cos 4t\cdot \cos 8t=16\cdot \frac{\sin 16t}{16\sin t}=\frac{\sin 16t}{\sin t}$
At $t=\frac{\pi }{17},$
$\left|A\right(t\left)\right|\left|A\right(4t\left)\right|=\frac{\sin \frac{16\pi }{17}}{\sin \frac{\pi }{17}}=\frac{\sin (\pi -\frac{\pi }{17})}{\sin \frac{\pi }{17}}=1$