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42. Express the following matrices as the sum of a symmetric and a skew symmetric

matrix:

(i) [351−1] (ii) [6−22−23−12−13]

(iii) [33−1−2−21−4−52] (iv) [15−12]

0 37 Views | Posted 4 months ago
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    Answered by

    Vishal Baghel | Contributor-Level 10

    4 months ago

    (i) Let A = [351−1].

    Then, A’ = [315−1].

    Let P = 12 (A + A’) = 12 {[351−1]+[315−1]}=12[3+35+11+5−1+]

    = 12[666−2]=[333−1].

    Then, P’ = [333−1] = P.

    ∴ P = 12 (A + A’) is symmetric matrix

    Let Q = 12 (A + A’) = 12{[351−1]−[315−1]}=12[3−35−11−5−1−(−1)]

    = 12[04−40]=[02−20].

    Then Q.’ = [0−220] = (-1) [02−20] = (-1) Q.

    ⇒ Q.’ = Q,

    ∴ Q = 12 (A - A’) is a symmetric matrix

    Now, P + Q = 12 (A + A’) + 12 (A - A’)

    ⇒ P + Q = [333−1]+[02−20]=[3+03+23−2−1+0]=[351−1] = A.

    This A is represented as a sun of symmetric and skew symmetric matrix

    Let A = [6−22−23−12−13].

    Then A’ = [6−22−23−12−13].

    Now, A + A’ = [6−22−23−12−13]+[6−22−23−12−13]

    = [6+6−2+(−2)2+2−2+(−2)3+3−1+(−1)2+2−1+(−1)3+3] = [12−44−46−24−26]

    Let P = 12 (A + A’) = 12[12−44−46−24−26]=[6−22−23−12−13]

    Then, P’ =&n

    ...more

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