Question

Which of the following values of α satisfy the equation \[\left| \begin{matrix}
   {{(1+\alpha )}^{2}} & {{(1+2\alpha )}^{2}} & {{(1+3\alpha )}^{2}} \\
   {{(2+\alpha )}^{2}} & {{(2+2\alpha )}^{2}} & {{(2+3\alpha )}^{2}} \\
   {{(3+\alpha )}^{2}} & {{(3+2\alpha )}^{2}} & {{(3+3\alpha )}^{2}} \\
\end{matrix} \right|=-648\alpha \]
$\begin{align}
  & \text{a) }\text{-4} \\
 & \text{b) 9} \\
 & \text{c) -9} \\
 & \text{d) 4} \\
\end{align}$

Answer 1

Hint: Now in the question we are given with the determinant whose entries are quadratic equations. We will use row and column transformations successively as to bring the determinant is a simpler form and then equate it to the given value of determinant. Hence we will find all the possible solutions of α.

Complete step by step answer:
Now we are given that \[\left| \begin{matrix}
   {{(1+\alpha )}^{2}} & {{(1+2\alpha )}^{2}} & {{(1+3\alpha )}^{2}} \\
   {{(2+\alpha )}^{2}} & {{(2+2\alpha )}^{2}} & {{(2+3\alpha )}^{2}} \\
   {{(3+\alpha )}^{2}} & {{(3+2\alpha )}^{2}} & {{(3+3\alpha )}^{2}} \\
\end{matrix} \right|=-648\alpha \]
Consider \[\left| \begin{matrix}
   {{(1+\alpha )}^{2}} & {{(1+2\alpha )}^{2}} & {{(1+3\alpha )}^{2}} \\
   {{(2+\alpha )}^{2}} & {{(2+2\alpha )}^{2}} & {{(2+3\alpha )}^{2}} \\
   {{(3+\alpha )}^{2}} & {{(3+2\alpha )}^{2}} & {{(3+3\alpha )}^{2}} \\
\end{matrix} \right|\]
To solve this we will use row transformations
Now let us first use the transformation ${{R}_{3}}={{R}_{3}}-{{R}_{2}}$
\[\left| \begin{matrix}
   {{(1+\alpha )}^{2}} & {{(1+2\alpha )}^{2}} & {{(1+3\alpha )}^{2}} \\
   {{(2+\alpha )}^{2}} & {{(2+2\alpha )}^{2}} & {{(2+3\alpha )}^{2}} \\
   {{(3+\alpha )}^{2}}-{{(2+\alpha )}^{2}} & {{(3+2\alpha )}^{2}}-{{(2+2\alpha )}^{2}} & {{(3+3\alpha )}^{2}}-{{(2+3\alpha )}^{2}} \\
\end{matrix} \right|\]
Now we know the formula for ${{(a+b)}^{2}}$ is $({{a}^{2}}+2ab+{{b}^{2}})$. Using this we get
\[\begin{align}
  & \left| \begin{matrix}
   {{(1+\alpha )}^{2}} & {{(1+2\alpha )}^{2}} & {{(1+3\alpha )}^{2}} \\
   {{(2+\alpha )}^{2}} & {{(2+2\alpha )}^{2}} & {{(2+3\alpha )}^{2}} \\
   9+{{\alpha }^{2}}+6\alpha -(4+{{\alpha }^{2}}+4\alpha ) & (9+4{{\alpha }^{2}}+12\alpha )-(4+4{{\alpha }^{2}}+8\alpha ) & (9+9{{\alpha }^{2}}+18\alpha )-(4+9{{\alpha }^{2}}+12\alpha ) \\
\end{matrix} \right| \\
 & =\left| \begin{matrix}
   {{(1+\alpha )}^{2}} & {{(1+2\alpha )}^{2}} & {{(1+3\alpha )}^{2}} \\
   {{(2+\alpha )}^{2}} & {{(2+2\alpha )}^{2}} & {{(2+3\alpha )}^{2}} \\
   (5+2\alpha ) & (5+4\alpha ) & (5+6\alpha ) \\
\end{matrix} \right| \\
\end{align}\]
Now let us first use the transformation ${{R}_{2}}={{R}_{2}}-{{R}_{1}}$ .
\[\left| \begin{matrix}
   {{(1+\alpha )}^{2}} & {{(1+2\alpha )}^{2}} & {{(1+3\alpha )}^{2}} \\
   {{(2+\alpha )}^{2}}-{{(1+\alpha )}^{2}} & {{(2+2\alpha )}^{2}}-{{(1+2\alpha )}^{2}} & {{(2+3\alpha )}^{2}}-{{(1+3\alpha )}^{2}} \\
   (5+2\alpha ) & (5+4\alpha ) & (5+6\alpha ) \\
\end{matrix} \right|\]
\[\left| \begin{matrix}
   {{(1+\alpha )}^{2}} & {{(1+2\alpha )}^{2}} & {{(1+3\alpha )}^{2}} \\
   {{(4+4\alpha +\alpha )}^{2}}-(1+{{\alpha }^{2}}+2\alpha ) & (4+4{{\alpha }^{2}}+8\alpha )-(1+4{{\alpha }^{2}}+4\alpha ) & {{(4+9{{\alpha }^{2}}+12\alpha )}}-{{(1+9{{\alpha }^{2}}+6\alpha )}} \\
   (5+2\alpha ) & (5+4\alpha ) & (5+6\alpha ) \\
\end{matrix} \right|\]
\[\left| \begin{matrix}
   {{(1+\alpha )}^{2}} & {{(1+2\alpha )}^{2}} & {{(1+3\alpha )}^{2}} \\
   (3+2\alpha ) & (3+4\alpha ) & (3+6\alpha ) \\
   (5+2\alpha ) & (5+4\alpha ) & (5+6\alpha ) \\
\end{matrix} \right|\]
Now let us take ${{R}_{3}}={{R}_{3}}-{{R}_{2}}$
\[\begin{align}
  & \left| \begin{matrix}
   {{(1+\alpha )}^{2}} & {{(1+2\alpha )}^{2}} & {{(1+3\alpha )}^{2}} \\
   (3+2\alpha ) & (3+4\alpha ) & (3+6\alpha ) \\
   (5+2\alpha )-(3+2\alpha ) & (5+4\alpha )-(3+4\alpha ) & (5+6\alpha )-(3+6\alpha ) \\
\end{matrix} \right| \\
 & \Rightarrow \left| \begin{matrix}
   {{(1+\alpha )}^{2}} & {{(1+2\alpha )}^{2}} & {{(1+3\alpha )}^{2}} \\
   (3+2\alpha ) & (3+4\alpha ) & (3+6\alpha ) \\
   2 & 2 & 2 \\
\end{matrix} \right| \\
\end{align}\]
Now we will do a Column transformation ${{C}_{3}}={{C}_{3}}-{{C}_{2}}$ , hence we get
\[\begin{align}
  & \Rightarrow \left| \begin{matrix}
   {{(1+\alpha )}^{2}} & {{(1+2\alpha )}^{2}} & {{(1+3\alpha )}^{2}}-{{(1+2\alpha )}^{2}} \\
   (3+2\alpha ) & (3+4\alpha ) & (3+6\alpha )-(3+4\alpha ) \\
   2 & 2 & 2-2 \\
\end{matrix} \right| \\
 & \Rightarrow \left| \begin{matrix}
   {{(1+\alpha )}^{2}} & {{(1+2\alpha )}^{2}} & (1+9{{\alpha }^{2}}+6\alpha )-(1+4{{\alpha }^{2}}+4\alpha ) \\
   (3+2\alpha ) & (3+4\alpha ) & 2\alpha \\
   2 & 2 & 0 \\
\end{matrix} \right| \\
 & \Rightarrow \left| \begin{matrix}
   {{(1+\alpha )}^{2}} & {{(1+2\alpha )}^{2}} & (5{{\alpha }^{2}}+2\alpha ) \\
   (3+2\alpha ) & (3+4\alpha ) & (32{{\alpha }^{2}}+24\alpha ) \\
   2 & 2 & 0 \\
\end{matrix} \right| \\
\end{align}\]
Now we will do a Column transformation ${{C}_{2}}={{C}_{2}}-{{C}_{1}}$ , hence we get
\[\begin{align}
  & \left| \begin{matrix}
   {{(1+\alpha )}^{2}} & {{(1+2\alpha )}^{2}}-{{(1+\alpha )}^{2}} & (5{{\alpha }^{2}}+2\alpha ) \\
   (3+2\alpha ) & (3+4\alpha )-(3+2\alpha ) & 2\alpha \\
   2 & 2-2 & 0 \\
\end{matrix} \right| \\
 & \Rightarrow \left| \begin{matrix}
   {{(1+\alpha )}^{2}} & (1+4{{\alpha }^{2}}+4\alpha )-(1+{{\alpha }^{2}}+2\alpha ) & (5{{\alpha }^{2}}+2\alpha ) \\
   (3+2\alpha ) & 2\alpha & 2\alpha \\
   2 & 0 & 0 \\
\end{matrix} \right| \\
 & \Rightarrow \left| \begin{matrix}
   {{(1+\alpha )}^{2}} & 3{{\alpha }^{2}}+2\alpha & (5{{\alpha }^{2}}+2\alpha ) \\
   (3+2\alpha ) & 2\alpha & 2\alpha \\
   2 & 0 & 0 \\
\end{matrix} \right| \\
\end{align}\]
Now let us evaluate the determinant. Now since we have 2 elements = 0 in row 3 we will open the determinant with respect to third row
$\left| \begin{matrix}
   {{(1+\alpha )}^{2}} & 3{{\alpha }^{2}}+2\alpha & (5{{\alpha }^{2}}+2\alpha ) \\
   (3+2\alpha ) & 2\alpha & 2\alpha \\
   2 & 0 & 0 \\
\end{matrix} \right|=2\left[ \left( 3{{\alpha }^{2}}+2\alpha \right)2\alpha -2\alpha (5{{\alpha }^{2}}+2\alpha ) \right]-0+0$
$\begin{align}
  & 2(2\alpha )\left[ \left( 3{{\alpha }^{2}}+2\alpha \right)-(5{{\alpha }^{2}}+2\alpha ) \right] \\
 & =4\alpha [-2{{\alpha }^{2}}] \\
 & =-8{{\alpha }^{3}} \\
\end{align}$
Now we are given that the value of determinant is equal to $-648\alpha $
$-8{{\alpha }^{3}}=-648\alpha $
Multiplying – 1 to the equation we get
$8{{\alpha }^{3}}=648\alpha $
Now let us divide the equation by $8\alpha $, hence we get.
${{\alpha }^{2}}=81$
Hence the value of $\alpha $ will be $\alpha =\sqrt{81}=\pm 9$

So, the correct answer is “Option B and C”.

Note: Now when we Solve the determinant we get the $-8{{\alpha }^{3}}=-648\alpha $. After this step we have divided the equation by 8α. We can only do this if α is not equal to zero. Hence we have assumed that α is not equal to zero or else we have α = 0 also as solution of $-8{{\alpha }^{3}}=-648\alpha $