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Question: If n skew-symmetric matrices of same order are \({A_1},{A_2},.......................{A_{2n - 1}}\), ...

If n skew-symmetric matrices of same order are A1,A2,.......................A2n−1{A_1},{A_2},.......................{A_{2n - 1}}, then B=∑r=1n(2r−1)(A2r−1)2r−1B = \sum\limits_{r = 1}^n {\left( {2r - 1} \right){{\left( {{A_{2r - 1}}} \right)}^{2r - 1}}} will be
(a) symmetric (b) skew - symmetric (c) neither symmetric now - symmetric (d) data not adequate   (a){\text{ symmetric}} \\\ (b){\text{ skew - symmetric}} \\\ (c){\text{ neither symmetric now - symmetric}} \\\ (d){\text{ data not adequate }} \\\

Explanation

Solution

Hint – In this question it is given as A1,A2,.......................A2n−1{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 isBT=−B{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 A1,A2,.......................A2n−1{A_1},{A_2},.......................{A_{2n - 1}} are n skew-symmetric matrices of the same order.
So, we have to find out B=∑r=1n(2r−1)(A2r−1)2r−1B = \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
⇒A1T=−A1,A3T=−A3,.........................A2n−1T=−A2n−1\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,
⇒B=∑r=1n(2r−1)(A2r−1)2r−1\Rightarrow B = \sum\limits_{r = 1}^n {\left( {2r - 1} \right){{\left( {{A_{2r - 1}}} \right)}^{2r - 1}}}
⇒B=A1+3(A3)3+5(A5)5+.................+(2n−1)(A2n−1)2n−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,
⇒BT=A1T+3(A3T)3+5(A5T)5+.................+(2n−1)(A2n−1T)2n−1\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,
⇒BT=−A1+3(−A3)3+5(−A5)5+.................+(2n−1)(−A2n−1)2n−1\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,
⇒BT=−[A1+3(A3)3+5(A5)5+.................+(2n−1)(A2n−1)2n−1]\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,
⇒BT=−B\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.