Question

Let $d\in R$, and $A=\left[ \begin{matrix}
   -2 & 4+d & \sin \theta -2 \\
   1 & \sin \theta +2 & d \\
   5 & 2\sin \theta -d & \left( -\sin \theta \right)+2+2d \\
\end{matrix} \right]$, $\theta \in \left[ 0,2\pi \right]$. If the minimum value of det(A) is 8, then a value of d is?
(a) -7
(b) $2\left( \sqrt{2}+2 \right)$
(c) -5
(d) $2\left( \sqrt{2}+1 \right)$

Answer 1

Hint: First, by using the row transformation as ${{R}_{1}}\to {{R}_{1}}+{{R}_{3}}-2{{R}_{2}}$ to the determinant of the matrix A, we get the simple form. Then, by expanding the determinant along the first row, we get a solution of the determinant. Then, we can clearly see that determinant gives minimum value when $\theta =\dfrac{3\pi }{2}$ as $\theta \in \left[ 0,2\pi \right]$,then by substituting, we get the value of d.

Complete step-by-step answer:
In this question, we are supposed to find the value of d from the given condition that det(A) is 8 which is minimum for the given matrix A as:
$A=\left[ \begin{matrix}
   -2 & 4+d & \sin \theta -2 \\
   1 & \sin \theta +2 & d \\
   5 & 2\sin \theta -d & \left( -\sin \theta \right)+2+2d \\
\end{matrix} \right]$
So, before proceeding for this, we must write the matrix A in determinant form as:
$\det A=\left| \begin{matrix}
   -2 & 4+d & \sin \theta -2 \\
   1 & \sin \theta +2 & d \\
   5 & 2\sin \theta -d & \left( -\sin \theta \right)+2+2d \\
\end{matrix} \right|$
Now, by using the row transformation as ${{R}_{1}}\to {{R}_{1}}+{{R}_{3}}-2{{R}_{2}}$ to the determinant of the matrix A as:
$\det A=\left| \begin{matrix}
   -2+5-2 & 4+d+2\sin \theta -d-2\sin \theta -4 & \sin \theta -2-\sin \theta +2+2d-2d \\
   1 & \sin \theta +2 & d \\
   5 & 2\sin \theta -d & \left( -\sin \theta \right)+2+2d \\
\end{matrix} \right|$
Then, after solving row first, we get:
$\det A=\left| \begin{matrix}
   1 & 0 & 0 \\
   1 & \sin \theta +2 & d \\
   5 & 2\sin \theta -d & \left( -\sin \theta \right)+2+2d \\
\end{matrix} \right|$
Then, by expanding the determinant along first row, we get:
$1\left[ \left( \sin \theta +2 \right)\left( 2+2d-\sin \theta \right)-d\left( 2\sin \theta -d \right) \right]$
Then, by solving the above expression, we get:
$\begin{align}
  & 4+4d-2\sin \theta +2\sin \theta +2d\sin \theta -{{\sin }^{2}}\theta -2d\sin \theta +{{d}^{2}} \\
 & \Rightarrow {{d}^{2}}+4d+4-{{\sin }^{2}}\theta \\
\end{align}$
Now, we are given the condition in the question that we get the minimum value of the determinant of A when the determinant is minimum.
So, we can clearly see that determinant gives minimum value when $\theta =\dfrac{3\pi }{2}$ as $\theta \in \left[ 0,2\pi \right]$, which in turn gives the value as:
$\begin{align}
  & {{\sin }^{2}}\left( \dfrac{3\pi }{2} \right)={{\left( -1 \right)}^{2}} \\
 & \Rightarrow 1 \\
\end{align}$
Now, by equating the value of determinant to 8 and also substituting the value of ${{\sin }^{2}}\theta $ for minimum condition, we get:
$\begin{align}
  & {{d}^{2}}+4d+4-1=8 \\
 & \Rightarrow {{\left( d+2 \right)}^{2}}=9 \\
 & \Rightarrow \left( d+2 \right)=\pm 3 \\
\end{align}$
Then, by using the above condition, we can get two values of d as:
$\begin{align}
  & d=3-2 \\
 & \Rightarrow d=1 \\
\end{align}$ or $\begin{align}
  & d=-3-2 \\
 & \Rightarrow d=-5 \\
\end{align}$
So, we get the values of d as 1 or -5.

So, the correct answer is “Option c”.

Note: Now, to solve these types of questions we need to know some of the basic formulas for the perfect square and condition of determinants. So, there is no compulsion that we expand along the first row, we can expand along any row or column to get the same result. Moreover, the formula used for perfect square is ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$.