Question

If the value of matrix $A=\left[ \begin{matrix}
   1 & \sin \theta & 1 \\
   -\sin \theta & 1 & \sin \theta \\
   -1 & -\sin \theta & 1 \\
\end{matrix} \right]$ then for all $\theta \in \left( \dfrac{3\pi }{4},\dfrac{5\pi }{4} \right)$ then let (A) lies in the interval
\[\begin{align}
  & A.\left( \dfrac{5}{2},4 \right) \\
 & B.\left( \dfrac{3}{2},3 \right) \\
 & C.\left( 0,\dfrac{3}{2} \right) \\
 & D.\left( 1,\dfrac{5}{2} \right) \\
\end{align}\]

Answer 1

Hint: To solve this question, we will first use the formula of determinant of a matrix. If $M=\left[ \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right]$ is a matrix then its determinant $\left| M \right|=a\left( ei-fh \right)-b\left( di-gf \right)+c\left( dh-eg \right)$ so, we will calculate $\left| A \right|$ then as we are given that $\theta \in \left( \dfrac{3\pi }{4},\dfrac{5\pi }{4} \right)$ Using this we will try to arrange value obtained of $\left| A \right|$ in terms of an interval. Remember that, $\sin \left( \pi +\theta \right)=-\sin \theta \text{ and }\sin \left( \pi -\theta \right)=\sin \theta $

Complete step by step answer:
A determinant of a matrix is a scalar value that can be computed from the elements of a square matrix and which encodes certain properties of linear transformation described by a matrix. The determinant of a matrix A is denoted by $\left| A \right|$
We have \[A=\left[ \begin{matrix}
   1 & \sin \theta & 1 \\
   -\sin \theta & 1 & \sin \theta \\
   -1 & -\sin \theta & 1 \\
\end{matrix} \right]\]
When a matrix M is given as \[M=\left[ \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right]\]
Then the formula of its determinant opening from first row is:
\[\left| M \right|=a\left( ei-fh \right)-b\left( di-gf \right)+c\left( dh-eg \right)\]
Using this formula of determinant of opening from first row in matrix A we get:
\[\begin{align}
  & \left| A \right|=\left[ \begin{matrix}
   1 & \sin \theta & 1 \\
   -\sin \theta & 1 & \sin \theta \\
   -1 & -\sin \theta & 1 \\
\end{matrix} \right] \\
 & \left| A \right|=1\left( 1+{{\sin }^{2}}\theta \right)-\sin \theta \left( -\sin \theta +\sin \theta \right)+1\left( +{{\sin }^{2}}\theta +1 \right) \\
 & \left| A \right|=1+{{\sin }^{2}}\theta -0+{{\sin }^{2}}\theta +1 \\
 & \left| A \right|=2+2{{\sin }^{2}}\theta \\
\end{align}\]
Now, we have $\theta \in \left( \dfrac{3\pi }{4},\dfrac{5\pi }{4} \right)$ and value of \[\sin \dfrac{3\pi }{4}=\sin \left( \pi -\dfrac{\pi }{4} \right)\]
We have $\sin \left( \pi -\theta \right)=+\sin \theta $
Using this in above substituting $\theta =\dfrac{\pi }{4}$ we get:
\[\begin{align}
  & \sin \dfrac{3\pi }{4}=\sin \left( \pi -\dfrac{\pi }{4} \right)=+\sin \dfrac{\pi }{4} \\
 & \text{and }\sin \dfrac{\pi }{4}=+\dfrac{1}{\sqrt{2}}\Rightarrow +\sin \dfrac{\pi }{4}=+\dfrac{1}{\sqrt{2}} \\
 & \Rightarrow \sin \dfrac{3\pi }{4}=+\dfrac{1}{\sqrt{2}} \\
\end{align}\]
And value of \[\sin \dfrac{5\pi }{4}=\sin \left( \pi +\dfrac{\pi }{4} \right)\]
We have $\sin \left( \pi +\theta \right)=-\sin \theta $
Using this in above taking $\theta =\dfrac{\pi }{4}$
\[\sin \dfrac{5\pi }{4}=\sin \left( \pi +\dfrac{\pi }{4} \right)=-\sin \dfrac{\pi }{4}=-\dfrac{1}{\sqrt{2}}\]
So we have for \[\theta \in \left( \dfrac{3\pi }{4},\dfrac{5\pi }{4} \right)\Rightarrow \sin \in \left( \dfrac{-1}{\sqrt{2}},\dfrac{1}{\sqrt{2}} \right)\]
Now ${{\sin }^{2}}\theta $ always has all values positive as squaring makes all positive.
\[{{\sin }^{2}}\theta \in \left( 0,\dfrac{1}{\left( 2 \right)} \right)\]
Adding 1 both sides:
\[\begin{align}
  & 1+{{\sin }^{2}}\theta \in \left( 1,\dfrac{1+2}{\left( 2 \right)} \right) \\
 & \Rightarrow 1+{{\sin }^{2}}\theta \in \left( 1,\dfrac{3}{2} \right) \\
\end{align}\]
Multiplying 2 both sides:
\[2\left( 1+{{\sin }^{2}}\theta \right)\in \left( 2,3 \right)\]
So, $\left| A \right|\in \left( 2,3 \right)$
Because no option is given in the form of (2,3) so we will try to obtain the most suitable option or we can proceed to see which of the following given options is (2,3) a subset.
Option A \[\left( \dfrac{5}{2},4 \right)=\left( 2.5,4 \right)\]
Clearly $\left( 2,3 \right)\not\subset \left( 2.5,4 \right)$
So option A \[\left( \dfrac{5}{2},4 \right)\] is wrong.
Option B $\left( \dfrac{3}{2},3 \right)=\left( 1.5,3 \right)$
Clearly $\left( 2,3 \right)\subseteq \left( 1.5,3 \right)$
So option B is correct.
Option C $\left( 0,\dfrac{3}{2} \right)$
Clearly $\left( 2,3 \right)\not\subset \left( 0,\dfrac{3}{2} \right)$
So option C is wrong.
Option D $\left( 1,\dfrac{5}{2} \right)$
Clearly $\left( 2,3 \right)\not\subset \left( 1,\dfrac{5}{2} \right)$
So option D is wrong.

So, the correct answer is “Option B”.

Note: The key point to note in this question is at the step where we had \[\sin \in \left( \dfrac{-1}{\sqrt{2}},\dfrac{1}{\sqrt{2}} \right)\Rightarrow {{\sin }^{2}}\theta \in \left( 0,\dfrac{1}{2} \right)\]
Observe here that, $\sin \theta $ was belonging to an open set $\left( \dfrac{-1}{\sqrt{2}},\dfrac{1}{\sqrt{2}} \right)$ but ${{\sin }^{2}}\theta \in \left( 0,\dfrac{1}{2} \right)$ where point 0 is closed. This is so because ${{\sin }^{2}}\theta $ can be 0 it cannot be negative but it can have 0 value so 0 point is closed. Hence ${{\sin }^{2}}\theta \in \left( 0,\dfrac{1}{2} \right)$