# 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 }} \\

$

Last updated date: 25th Mar 2023

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Answer

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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.

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