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Matrices and Determinants question

2014 · Shift 0 · Q32
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  5. /2014 · Shift 0 · Q32

Matrices and Determinants question

2014 · Shift 0 · Q32

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
If α,βe0,\alpha ,\beta e 0,α,βe0, and f(n)=αn+βnf\left( n \right) = {\alpha ^n} + {\beta ^n}f(n)=αn+βn and ∣31+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4)∣\left| {\begin{matrix} 3 & {1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} \\ {1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} \\ {1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} & {1 + f\left( 4 \right)} \\ \end{matrix} } \right|​31+f(1)1+f(2)​1+f(1)1+f(2)1+f(3)​1+f(2)1+f(3)1+f(4)​​=K(1−α)2(1−β)2(α−β)2,= K{\left( {1 - \alpha } \right)^2}{\left( {1 - \beta } \right)^2}{\left( {\alpha - \beta } \right)^2},=K(1−α)2(1−β)2(α−β)2, then KKK is equal to :
  1. A
    111
  2. B
    −1-1−1
  3. C
    αβ\alpha \betaαβ
  4. D
    1αβ{1 \over {\alpha \beta }}αβ1​
View written solutionFree

Correct answer: A

  1. Write the determinant in a simpler form

Given f(n)=αn+βn,f(n)=\alpha^n+\beta^n,f(n)=αn+βn, so the matrix is

3 & 1+f(1) & 1+f(2)\\ 1+f(1) & 1+f(2) & 1+f(3)\\ 1+f(2) & 1+f(3) & 1+f(4) \end{pmatrix}.$$ Now, $$3=1+1+1,$$ $$1+f(1)=1+\alpha+\beta,$$ $$1+f(2)=1+\alpha^2+\beta^2,$$ and so on. Thus each entry is of the form $$1+\alpha^k+\beta^k.$$ This suggests writing the matrix as a sum of rank-one matrices. 2. **Observe the Gram-type structure** Define the vectors $$u=\begin{pmatrix}1\\1\\1\end{pmatrix},\qquad v=\begin{pmatrix}1\\\alpha\\\alpha^2\end{pmatrix},\qquad w=\begin{pmatrix}1\\\beta\\\beta^2\end{pmatrix}.$$ Then $$uu^T=\begin{pmatrix} 1&1&1\\1&1&1\\1&1&1 \end{pmatrix},$$ $$vv^T=\begin{pmatrix} 1&\alpha&\alpha^2\\ \alpha&\alpha^2&\alpha^3\\ \alpha^2&\alpha^3&\alpha^4 \end{pmatrix},$$ $$ww^T=\begin{pmatrix} 1&\beta&\beta^2\\ \beta&\beta^2&\beta^3\\ \beta^2&\beta^3&\beta^4 \end{pmatrix}.$$ Adding, $$uu^T+vv^T+ww^T= \begin{pmatrix} 3&1+\alpha+\beta&1+\alpha^2+\beta^2\\ 1+\alpha+\beta&1+\alpha^2+\beta^2&1+\alpha^3+\beta^3\\ 1+\alpha^2+\beta^2&1+\alpha^3+\beta^3&1+\alpha^4+\beta^4 \end{pmatrix}=M.$$ Hence $$M=AA^T,$$ where $$A=\begin{pmatrix} 1&1&1\\ 1&\alpha&\beta\\ 1&\alpha^2&\beta^2 \end{pmatrix}.$$ 3. **Use determinant property** Since $$M=AA^T,$$ we have $$\det(M)=\det(A)\det(A^T)=(\det A)^2.$$ So we only need to compute $$\det A=\begin{vmatrix} 1&1&1\\ 1&\alpha&\beta\\ 1&\alpha^2&\beta^2 \end{vmatrix}.$$ 4. **Compute $\det A$** This is a Vandermonde determinant with parameters $1,\alpha,\beta$: $$\det A=(\alpha-1)(\beta-1)(\beta-\alpha).$$ Equivalently, $$\det A=(1-\alpha)(1-\beta)(\alpha-\beta),$$ because changing signs twice gives the same value: $$ (\alpha-1)=-(1-\alpha),\quad (\beta-1)=-(1-\beta),\quad (\beta-\alpha)=-(\alpha-\beta),$$ so overall product remains $$ (1-\alpha)(1-\beta)(\alpha-\beta). $$ Therefore $$\det(M)=(\det A)^2=(1-\alpha)^2(1-\beta)^2(\alpha-\beta)^2.$$ 5. **Compare with the given form** Given $$\det(M)=K(1-\alpha)^2(1-\beta)^2(\alpha-\beta)^2,$$ so directly, $$K=1.$$ 6. **Check options** - **A: $1$** — correct - **B: $-1$** — incorrect - **C: $\alpha\beta$** — incorrect - **D: $\dfrac1{\alpha\beta}$** — incorrect Therefore the correct option is **A**.
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