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Matrices and Determinants question

2015 · Shift 1 · Q38
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  5. /2015 · Shift 1 · Q38

Matrices and Determinants question

2015 · Shift 1 · Q38

JEE AdvancedMathematicsMatrices and DeterminantsMultiple correct+4 / −2
Which of the following values of α\alphaα 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​(1−α)2(2+α)2(3+α)2​(1+2α)2(2+2α)2(3+2α)2​(1+3α)2(2+3α)2(3+3α)2​​=−648α ?
  1. A
    −-− 4
  2. B
    9
  3. C
    −-− 9
  4. D
    4
View written solutionFree

Correct answer: CORRECT SOLUTION SET: $\ALPHA=0,\ \PM \DFRAC{3\SQRT6}{2}$; HENCE NONE OF THE OPTIONS IS CORRECT.

  1. Write the matrix in a structured form

Let Mij=(i+jα)2,i,j=1,2,3M_{ij}=(i+j\alpha)^2,\qquad i,j=1,2,3Mij​=(i+jα)2,i,j=1,2,3 with the given entries matching:

  • first column: (1−α)2,(2+α)2,(3+α)2(1-\alpha)^2,(2+\alpha)^2,(3+\alpha)^2(1−α)2,(2+α)2,(3+α)2
  • second column: (1+2α)2,(2+2α)2,(3+2α)2(1+2\alpha)^2,(2+2\alpha)^2,(3+2\alpha)^2(1+2α)2,(2+2α)2,(3+2α)2
  • third column: (1+3α)2,(2+3α)2,(3+3α)2(1+3\alpha)^2,(2+3\alpha)^2,(3+3\alpha)^2(1+3α)2,(2+3α)2,(3+3α)2

It is convenient to view each column as a quadratic in the row variable x=1,2,3x=1,2,3x=1,2,3.

For fixed column parameter ttt, we have (x+t)2=x2+2tx+t2.(x+t)^2=x^2+2tx+t^2.(x+t)2=x2+2tx+t2. So each column is a linear combination of the three vectors u=[111],v=[123],w=[149].u=\begin{bmatrix}1\\1\\1\end{bmatrix},\quad v=\begin{bmatrix}1\\2\\3\end{bmatrix},\quad w=\begin{bmatrix}1\\4\\9\end{bmatrix}. u=​111​​,v=​123​​,w=​149​​.

The three column shifts are t1=−α,t2=2α,t3=3α.t_1=-\alpha,\quad t_2=2\alpha,\quad t_3=3\alpha.t1​=−α,t2​=2α,t3​=3α. Hence Cj=w+2tjv+tj2u.C_j=w+2t_j v+t_j^2 u.Cj​=w+2tj​v+tj2​u. Therefore

t_1^2 & t_2^2 & t_3^2\\ 2t_1 & 2t_2 & 2t_3\\ 1 & 1 & 1 \end{bmatrix}.$$ So, $$\det M=\det[u\ v\ w]\cdot \det\begin{bmatrix} t_1^2 & t_2^2 & t_3^2\\ 2t_1 & 2t_2 & 2t_3\\ 1 & 1 & 1 \end{bmatrix}. $$ --- 2. **Compute $\det[u\ v\ w]$** $$[u\ v\ w]=\begin{bmatrix} 1&1&1\\ 1&2&4\\ 1&3&9 \end{bmatrix}.$$ Its determinant is $$\det\begin{bmatrix} 1&1&1\\ 1&2&4\\ 1&3&9 \end{bmatrix} =1\begin{vmatrix}2&4\\3&9\end{vmatrix}-1\begin{vmatrix}1&4\\1&9\end{vmatrix}+1\begin{vmatrix}1&2\\1&3\end{vmatrix}$$ $$=(18-12)-(9-4)+(3-2)=6-5+1=2.$$ Thus, $$\det[u\ v\ w]=2.$$ --- 3. **Compute the second determinant** Let $$N=\begin{bmatrix} t_1^2 & t_2^2 & t_3^2\\ 2t_1 & 2t_2 & 2t_3\\ 1 & 1 & 1 \end{bmatrix}.$$ Factor $2$ from the second row: $$\det N=2\det\begin{bmatrix} t_1^2 & t_2^2 & t_3^2\\ t_1 & t_2 & t_3\\ 1 & 1 & 1 \end{bmatrix}. $$ Now compare with the Vandermonde determinant: $$\det\begin{bmatrix} 1&1&1\\ t_1&t_2&t_3\\ t_1^2&t_2^2&t_3^2 \end{bmatrix}=(t_2-t_1)(t_3-t_1)(t_3-t_2).$$ Our matrix has rows reversed, which introduces a minus sign: $$\det\begin{bmatrix} t_1^2 & t_2^2 & t_3^2\\ t_1 & t_2 & t_3\\ 1 & 1 & 1 \end{bmatrix}=-(t_2-t_1)(t_3-t_1)(t_3-t_2).$$ Therefore, $$\det N=-2(t_2-t_1)(t_3-t_1)(t_3-t_2).$$ Now substitute $$t_1=-\alpha,\quad t_2=2\alpha,\quad t_3=3\alpha.$$ Then $$t_2-t_1=2\alpha-(-\alpha)=3\alpha,$$ $$t_3-t_1=3\alpha-(-\alpha)=4\alpha,$$ $$t_3-t_2=3\alpha-2\alpha=\alpha.$$ So $$\det N=-2(3\alpha)(4\alpha)(\alpha)=-24\alpha^3.$$ Hence $$\det M=2(-24\alpha^3)=-48\alpha^3.$$ --- 4. **Use the given equation** The equation is $$\det M=-648\alpha.$$ So, $$-48\alpha^3=-648\alpha.$$ Divide by $-24$: $$2\alpha^3=27\alpha.$$ Bring all terms to one side: $$2\alpha^3-27\alpha=0.$$ Factor: $$\alpha(2\alpha^2-27)=0.$$ Thus, $$\alpha=0\quad \text{or}\quad \alpha^2=\frac{27}{2}.$$ This gives $$\alpha=0,\quad \alpha=\pm \frac{3\sqrt6}{2}.$$ These are the actual solutions of the equation. --- 5. **Check the given options** Options are: - A: $-4$ - B: $9$ - C: $-9$ - D: $4$ Test quickly using $$\det M=-48\alpha^3.$$ The equation requires $$-48\alpha^3=-648\alpha.$$ - For $\alpha=9$: $$-48\cdot 9^3\ne -648\cdot 9$$ so **false**. - For $\alpha=-9$: $$-48\cdot (-9)^3\ne -648\cdot (-9)$$ so **false**. - For $\alpha=4$: $$-48\cdot 64=-3072,\quad -648\cdot 4=-2592$$ not equal, so **false**. - For $\alpha=-4$: $$-48\cdot (-64)=3072,\quad -648\cdot (-4)=2592$$ not equal, so **false**. So **none of the given options satisfy the equation**. --- 6. **Conclusion** The determinant simplifies to $$\det M=-48\alpha^3,$$ and solving $$-48\alpha^3=-648\alpha$$ gives $$\alpha=0,\ \pm \frac{3\sqrt6}{2}.$$ None of the listed options match these values.
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