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Question: Which of the following values of α satisfy the equation \[\left| \begin{matrix} {{(1+\alpha )}^...

Which of the following values of α satisfy the equation ∣(1+α)2(1+2α)2(1+3α)2 (2+α)2(2+2α)2(2+3α)2 (3+α)2(3+2α)2(3+3α)2 ∣=−648α\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
a) -4 b) 9 c) -9 d) 4 \begin{aligned} & \text{a) }\text{-4} \\\ & \text{b) 9} \\\ & \text{c) -9} \\\ & \text{d) 4} \\\ \end{aligned}

Explanation

Solution

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 ∣(1+α)2(1+2α)2(1+3α)2 (2+α)2(2+2α)2(2+3α)2 (3+α)2(3+2α)2(3+3α)2 ∣=−648α\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 ∣(1+α)2(1+2α)2(1+3α)2 (2+α)2(2+2α)2(2+3α)2 (3+α)2(3+2α)2(3+3α)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}} & {{(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 R3=R3−R2{{R}_{3}}={{R}_{3}}-{{R}_{2}}

{{(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{aligned} & \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{aligned}$$ Now let us first use the transformation ${{R}_{2}}={{R}_{2}}-{{R}_{1}}$ . $$\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{aligned} & \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{aligned}$$ Now we will do a Column transformation ${{C}_{3}}={{C}_{3}}-{{C}_{2}}$ , hence we get $$\begin{aligned} & \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{aligned}$$ Now we will do a Column transformation ${{C}_{2}}={{C}_{2}}-{{C}_{1}}$ , hence we get $$\begin{aligned} & \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{aligned}$$ 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 $\left| \begin{matrix} {{(1+\alpha )}^{2}} & 3{{\alpha }^{2}}+2\alpha & (5{{\alpha }^{2}}+2\alpha ) \\\ (3+2\alpha ) & 2\alpha & 2\alpha \\\ 2 & 0 & 0 \\\ \end{matrix} \right|=2\left[ \left( 3{{\alpha }^{2}}+2\alpha \right)2\alpha -2\alpha (5{{\alpha }^{2}}+2\alpha ) \right]-0+0$ $\begin{aligned} & 2(2\alpha )\left[ \left( 3{{\alpha }^{2}}+2\alpha \right)-(5{{\alpha }^{2}}+2\alpha ) \right] \\\ & =4\alpha [-2{{\alpha }^{2}}] \\\ & =-8{{\alpha }^{3}} \\\ \end{aligned}$ 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 $