If n skew-symmetric matrices of same order are ${A_1},{A_2},.......................{A_{2n - 1}}$, then $B = \sum\limits_{r = 1}^n {\left( {2r - 1} \right){{\left( {{A_{2r - 1}}} \right)}^{2r - 1}}} $will be
$
(a){\text{ symmetric}} \\
(b){\text{ skew - symmetric}} \\
(c){\text{ neither symmetric now - symmetric}} \\
(d){\text{ data not adequate }} \\
$
Answer
648.9k+ views
Hint – In this question it is given as ${A_1},{A_2},.......................{A_{2n - 1}}$ are n skew-symmetric matrices. A skew-symmetric matrix is one whose transpose is equal to a matrix multiplied with a negative sign that is${B^T} = - B$, use this condition while evaluating the submission to check whether it satisfies the options given in the question or not.
Complete step-by-step answer:
It is given that ${A_1},{A_2},.......................{A_{2n - 1}}$ are n skew-symmetric matrices of the same order.
So, we have to find out $B = \sum\limits_{r = 1}^n {\left( {2r - 1} \right){{\left( {{A_{2r - 1}}} \right)}^{2r - 1}}} $will be.
Now as we know the condition of skew-symmetric matrices of same order is
$ \Rightarrow {A_1}^T = - {A_1},{A_3}^T = - {A_3},.........................{A_{2n - 1}}^T = - {A_{2n - 1}}$ ………………….. (1)
[Where T is the transpose of the matrix]
Now expand the summation (from r = 1 to n) we have,
$ \Rightarrow B = \sum\limits_{r = 1}^n {\left( {2r - 1} \right){{\left( {{A_{2r - 1}}} \right)}^{2r - 1}}} $
$ \Rightarrow B = {A_1} + 3{\left( {{A_3}} \right)^3} + 5{\left( {{A_5}} \right)^5} + ................. + \left( {2n - 1} \right){\left( {{A_{2n - 1}}} \right)^{2n - 1}}$………………. (2)
Now take transpose of matrix B we have,
$ \Rightarrow {B^T} = {A_1}^T + 3{\left( {{A_3}^T} \right)^3} + 5{\left( {{A_5}^T} \right)^5} + ................. + \left( {2n - 1} \right){\left( {{A_{2n - 1}}^T} \right)^{2n - 1}}$
Now from equation (1) we have,
$ \Rightarrow {B^T} = - {A_1} + 3{\left( { - {A_3}} \right)^3} + 5{\left( { - {A_5}} \right)^5} + ................. + \left( {2n - 1} \right){\left( { - {A_{2n - 1}}} \right)^{2n - 1}}$
Now take (-) common we have,
$ \Rightarrow {B^T} = - \left[ {{A_1} + 3{{\left( {{A_3}} \right)}^3} + 5{{\left( {{A_5}} \right)}^5} + ................. + \left( {2n - 1} \right){{\left( {{A_{2n - 1}}} \right)}^{2n - 1}}} \right]$
Now from equation (2) we have,
$ \Rightarrow {B^T} = - B$
Which is the condition of skew-symmetric.
So, the matrix B is a skew-symmetric matrix.
Hence option (b) is correct.
Note – Whenever we face such types of problems the key concept is to use the gist of the basic definition of symmetric and skew-symmetric matrix. A symmetric matrix is one which even after transposed gives us the same matrix. Use these concepts of symmetric and skew-symmetric matrix to get the right option for the question.
Complete step-by-step answer:
It is given that ${A_1},{A_2},.......................{A_{2n - 1}}$ are n skew-symmetric matrices of the same order.
So, we have to find out $B = \sum\limits_{r = 1}^n {\left( {2r - 1} \right){{\left( {{A_{2r - 1}}} \right)}^{2r - 1}}} $will be.
Now as we know the condition of skew-symmetric matrices of same order is
$ \Rightarrow {A_1}^T = - {A_1},{A_3}^T = - {A_3},.........................{A_{2n - 1}}^T = - {A_{2n - 1}}$ ………………….. (1)
[Where T is the transpose of the matrix]
Now expand the summation (from r = 1 to n) we have,
$ \Rightarrow B = \sum\limits_{r = 1}^n {\left( {2r - 1} \right){{\left( {{A_{2r - 1}}} \right)}^{2r - 1}}} $
$ \Rightarrow B = {A_1} + 3{\left( {{A_3}} \right)^3} + 5{\left( {{A_5}} \right)^5} + ................. + \left( {2n - 1} \right){\left( {{A_{2n - 1}}} \right)^{2n - 1}}$………………. (2)
Now take transpose of matrix B we have,
$ \Rightarrow {B^T} = {A_1}^T + 3{\left( {{A_3}^T} \right)^3} + 5{\left( {{A_5}^T} \right)^5} + ................. + \left( {2n - 1} \right){\left( {{A_{2n - 1}}^T} \right)^{2n - 1}}$
Now from equation (1) we have,
$ \Rightarrow {B^T} = - {A_1} + 3{\left( { - {A_3}} \right)^3} + 5{\left( { - {A_5}} \right)^5} + ................. + \left( {2n - 1} \right){\left( { - {A_{2n - 1}}} \right)^{2n - 1}}$
Now take (-) common we have,
$ \Rightarrow {B^T} = - \left[ {{A_1} + 3{{\left( {{A_3}} \right)}^3} + 5{{\left( {{A_5}} \right)}^5} + ................. + \left( {2n - 1} \right){{\left( {{A_{2n - 1}}} \right)}^{2n - 1}}} \right]$
Now from equation (2) we have,
$ \Rightarrow {B^T} = - B$
Which is the condition of skew-symmetric.
So, the matrix B is a skew-symmetric matrix.
Hence option (b) is correct.
Note – Whenever we face such types of problems the key concept is to use the gist of the basic definition of symmetric and skew-symmetric matrix. A symmetric matrix is one which even after transposed gives us the same matrix. Use these concepts of symmetric and skew-symmetric matrix to get the right option for the question.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Master Class 9 Maths: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Name the crygenes that control cotton bollworm and class 12 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

In a transcription unit the promoter is said to be class 12 biology CBSE

Sulphuric acid is known as the king of acids State class 12 chemistry CBSE

