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Let \alpha be a root of the equation x^{2}+x+1=0 and the matrix A=\frac{1}{\sqrt{3}}\begin{bmatrix} 1 & 1 & 1\\ 1 & \alpha & \alpha ^{2}\\ 1 & \alpha ^{2}& \alpha ^{4} \end{bmatrix}, Then the matrix A^{31} is equal to :
Option: 1 A
Option: 2A^{2}
Option: 3 A^{3}
Option: 4 I_{3}
 

Answers (1)

best_answer

 

 

Cube roots of unity -

z is a complex number

Let z3 = 1  

⇒ z3 - 1 = 0

⇒ (z - 1)(z2 + z + 1) = 0

⇒ z - 1 = 0  or z2 + z + 1 = 0

 

\\\mathrm{\therefore \;z=1\;\;or\;\;z=\frac{-1\pm\sqrt{(1-4)}}{2}=\frac{-1\pm i\sqrt{3}}{2}} \\\mathrm{Therefore, \;\mathbf{z=1},\;\mathbf{z=\frac{-1+ i\sqrt{3}}{2}}\;\;and\;\;\mathbf{z=\frac{-1- i\sqrt{3}}{2}}}

 

If the second root is represented by ?, then the third root will be represented by ?2.

 

 

\\\mathrm{\mathbf{\omega=\frac{-1+ i\sqrt{3}}{2}},\;\;\omega^2=\mathbf{\frac{-1- i\sqrt{3}}{2}}}

 

Properties of cube roots:

i) 1 + ? + ?2 = 0 and ?3 = 1

ii) to find ?n , first we write ? in multiple of 3 with remainder belonging to 0,1,2 like n=3q + r

Where r is from 0,1,2. Now ?n = ?3q + r = (?3)q·?r  = ?r.

-

 

 

 

 

Multiplication of two matrices -

Matrix multiplication: 

Two matrices  A and B are conformable for the product AB if the number of columns in A and the number of rows in B is equal. Otherwise, these two matrices will be non-conformable for matrix multiplication. So on that basis,

i) AB is defined only if col(A) = row(B)

ii) BA is defined only if col(B) = row(A)

If 

    \\\mathrm{A = \left [ a_{ij} \right ]_{m\times n}} \\\mathrm{\\B=\left [ b_{ij} \right ]_{n\times p}}

    \\\mathrm{C = AB = \left [ c_{ij} \right ]_{m\times p}} \\\mathrm{Where\;\; c_{ij} = \sum_{j=1}^{n}a_{ij}b_{jk}, 1\leq i\leq m,1\leq k\leq p} \\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=a_{i1}b_{1k} + a_{i2}b_{2k} + a_{i3}b_{3k}+ ... + a_{in}b_{nk}}

For examples

\\\mathrm{Suppose,\;two\;matrices\;are\;given}\\\mathrm{A=\begin{bmatrix} a_{11} &a_{12} &a_{13} \\ a_{21} &a_{22} & a_{33} \end{bmatrix}_{2\times3}\;\;\;and\;\;\;B=\begin{bmatrix} b_{11}& b_{12} &b_{13} \\b_{21} &b_{22} &b_{23} \\b_{31} &b_{32} &b_{33} \end{bmatrix}_{3\times3}}\\\\\mathrm{To\:obtain\:the\:entries\:in\:row\:\mathit{i}\:of\:AB,\:we\:multiply\:the\:entries\:in\:row\:\mathit{i}\:of\:A\:by\:}\\\mathrm{column\:\mathit{j}\:in\:B\:and\:add.}\\\mathrm{given\:matrices\:A\:and\:B,\:where\:the\:order\:of\:A\:are\:2\times3\:and\:the\:order\:of\:B\:are\:3\times3,}\\\mathrm{the\:product\:of\:AB\:will\:be\:a\:2\times3\:matrix.}\\\\\mathrm{To\:obtain\:the\:entry\:in\:row\:1,\:column\:1\:of\:AB,\:multiply\:the\:}\\\mathrm{first\:row\:in\:A\:by\:the\:first\:column\:in\:B,and\:add.}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\begin{bmatrix} a_{11} &a_{12} &a_{13} \end{bmatrix}\begin{bmatrix} b_{11}\\b_{21} \\b_{31} \end{bmatrix}=a_{11}\cdot b_{11}+a_{12}\cdot b_{21}+a_{13}\cdot b_{31}}

\\\mathrm{To\:obtain\:the\:entry\:in\:row\:1,\:column\:2\:of\:AB,\:multiply\:the\:}\\\mathrm{first\:row\:in\:A\:by\:the\:second\:column\:in\:B,and\:add.}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\begin{bmatrix} a_{11} &a_{12} &a_{13} \end{bmatrix}\begin{bmatrix} b_{12}\\b_{22} \\b_{32} \end{bmatrix}=a_{11}\cdot b_{12}+a_{12}\cdot b_{22}+a_{13}\cdot b_{32}}\\\\\mathrm{To\:obtain\:the\:entry\:in\:row\:1,\:column\:3\:of\:AB,\:multiply\:the\:}\\\mathrm{first\:row\:in\:A\:by\:the\:thired\:column\:in\:B,and\:add.}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\begin{bmatrix} a_{11} &a_{12} &a_{13} \end{bmatrix}\begin{bmatrix} b_{13}\\b_{23} \\b_{33} \end{bmatrix}=a_{11}\cdot b_{13}+a_{12}\cdot b_{23}+a_{13}\cdot b_{33}}\\\\\mathrm{We\:proceed\:the\:same\:way\:to\:obtain\:the\:second\:row\:of\:AB.\:In\:other\:words,\:}\\\mathrm{row\:2\:of\:A\:times\:column\:1\:of\:B;}\\\mathrm{row\:2\:of\:A\:times\:column\:2\:of\:B;}\\\mathrm{row\:2\:of\:A\;times\:column\:3\:of\:B.}

\\\mathrm{When\:complete,\:the\:product\:matrix\:will\:be}\\\\\mathrm{AB=\begin{bmatrix} a_{11}\cdot b_{11}+a_{12}\cdot b_{21}+a_{13}\cdot b_{31}\;\;& a_{11}\cdot b_{12}+a_{12}\cdot b_{22}+a_{13}\cdot b_{32}\;\; &a_{11}\cdot b_{13}+a_{12}\cdot b_{23}+a_{13}\cdot b_{33} \\ a_{21}\cdot b_{11}+a_{22}\cdot b_{21}+a_{23}\cdot b_{31} \;\;& a_{21}\cdot b_{12}+a_{22}\cdot b_{22}+a_{23}\cdot b_{32} \;\;& a_{21}\cdot b_{13}+a_{22}\cdot b_{23}+a_{23}\cdot b_{33} \end{bmatrix}}

 

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Given

\\x^2+x+1=0\\\Rightarrow \alpha\;is\;the\;root\;of\;the\;eq\\\alpha=\omega,\;\;\omega^2\\\text{A}=\frac{1}{\sqrt3}\begin{bmatrix} 1 &1 &1 \\ 1& \omega &\omega^2 \\ 1 &\omega^2 &\omega \end{bmatrix}\\\text{A}^2=\frac{1}{3}\begin{bmatrix} 3 &0 &0 \\ 0& 0 &3 \\ 0 &3 &0 \end{bmatrix}

\\\text{A}^4=\frac{1}{3}\begin{bmatrix} 3 &0 &0 \\ 0& 0 &3 \\ 0 &3 &0 \end{bmatrix}\times \frac{1}{3}\begin{bmatrix} 3 &0 &0 \\ 0& 0 &3 \\ 0 &3 &0 \end{bmatrix}\\\text{A}^4=I\\\therefore A^{31}=A^{28}\cdot A^3=A^3

Correct option (3)

Posted by

Ritika Jonwal

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