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Matrices and Determinants question

2020 · 7 Jan · Shift 1 · Q27
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Matrices and Determinants question

2020 · 7 Jan · Shift 1 · Q27

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let α\alphaα be a root of the equation x2 + x + 1 = 0 and the matrix A =13[1111αα21α2α4]{1 \over {\sqrt 3 }}\left[ {\begin{matrix} 1 & 1 & 1 \\ 1 & \alpha & {{\alpha ^2}} \\ 1 & {{\alpha ^2}} & {{\alpha ^4}} \\ \end{matrix} } \right]3​1​​111​1αα2​1α2α4​​ then the matrix A31 is equal to
  1. A
    A2
  2. B
    A
  3. C
    I3
  4. D
    A3
View written solutionFree

Correct answer: D

  1. Given data

We are given that α\alphaα is a root of x2+x+1=0.x^2+x+1=0.x2+x+1=0. Hence, α3=1,α≠1,\alpha^3=1, \quad \alpha\neq 1,α3=1,α=1, and also 1+α+α2=0.1+\alpha+\alpha^2=0.1+α+α2=0.

The matrix is A=13[1111αα21α2α4].A=\frac{1}{\sqrt{3}}\begin{bmatrix}1&1&1\\[4pt]1&\alpha&\alpha^2\\[4pt]1&\alpha^2&\alpha^4\end{bmatrix}.A=3​1​​111​1αα2​1α2α4​​.

Since α3=1\alpha^3=1α3=1, we have α4=α.\alpha^4=\alpha.α4=α. So

a\end{bmatrix}$$ which is the normalized Fourier-type matrix of order $3$. Let $$M=\begin{bmatrix}1&1&1\\1&\alpha&\alpha^2\\1&\alpha^2&\alpha\end{bmatrix},$$ so that $$A=\frac{1}{\sqrt{3}}M.$$ --- 2. **Compute $M^2$** We use $1+\alpha+\alpha^2=0$ and $\alpha^3=1$. ### First row of $M^2$ - Entry $(1,1)$: $$1+1+1=3$$ - Entry $(1,2)$: $$1+\alpha+\alpha^2=0$$ - Entry $(1,3)$: $$1+\alpha^2+\alpha=0$$ So first row is $$[3,0,0].$$ ### Second row of $M^2$ Second row of $M$ is $[1,\alpha,\alpha^2]$. - Entry $(2,1)$: $$1+\alpha+\alpha^2=0$$ - Entry $(2,2)$: $$1+\alpha^2+\alpha^4=1+\alpha^2+\alpha=0$$ since $\alpha^4=\alpha$. - Entry $(2,3)$: $$1+\alpha\cdot\alpha^2+\alpha^2\cdot\alpha =1+\alpha^3+\alpha^3=1+1+1=3.$$ So second row is $$[0,0,3].$$ ### Third row of $M^2$ Third row of $M$ is $[1,\alpha^2,\alpha]$. - Entry $(3,1)$: $$1+\alpha^2+\alpha=0$$ - Entry $(3,2)$: $$1+\alpha^2\cdot\alpha+\alpha\cdot\alpha^2 =1+\alpha^3+\alpha^3=3$$ - Entry $(3,3)$: $$1+\alpha^4+\alpha^2=1+\alpha+\alpha^2=0.$$ So third row is $$[0,3,0].$$ Therefore, $$M^2=\begin{bmatrix}3&0&0\\0&0&3\\0&3&0\end{bmatrix} =3\begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix}.$$ Hence, $$A^2=\frac{1}{3}M^2=\begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix}=:P.$$ --- 3. **Compute $A^3$** Now, $$A^3=A\cdot A^2=AP.$$ The matrix $$P=\begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix}$$ interchanges the 2nd and 3rd columns of $A$. So $AP$ is obtained from $A$ by swapping its 2nd and 3rd columns. But in $A$, - 2nd column is $\dfrac{1}{\sqrt{3}}\begin{bmatrix}1\\ \alpha\\ \alpha^2\end{bmatrix}$, - 3rd column is $\dfrac{1}{\sqrt{3}}\begin{bmatrix}1\\ \alpha^2\\ \alpha\end{bmatrix}$. Swapping them gives $$A^3=\frac{1}{\sqrt{3}}\begin{bmatrix}1&1&1\\1&\alpha^2&\alpha\\1&\alpha&\alpha^2\end{bmatrix}.$$ Now compare with $$A^{-1}=A^{\dagger}$$ since $A$ is unitary. Also, because $\bar\alpha=\alpha^2$ and $\overline{\alpha^2}=\alpha$, $$A^{\dagger}=\frac{1}{\sqrt{3}}\begin{bmatrix}1&1&1\\1&\alpha^2&\alpha\\1&\alpha&\alpha^2\end{bmatrix}.$$ Thus, $$A^3=A^{-1}.$$ Multiplying by $A$, $$A^4=I_3.$$ So $A$ satisfies $$A^4=I_3.$$ Hence, $$A^{31}=A^{28}A^3=(A^4)^7A^3=I_3\cdot A^3=A^3.$$ --- 4. **Check options** - **A:** $A^2$ Not correct, since $A^{31}=A^3$. - **B:** $A$ Not correct. - **C:** $I_3$ Not correct. - **D:** $A^3$ Correct. --- 5. **Final answer** $$\boxed{A^{31}=A^3}$$ So the correct option is **D**.
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