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Byju's Answer
Standard XII
Mathematics
Skew Symmetric Matrix
Express the f...
Question

Express the following matrices as the sum of a symmetric and a skew-symmetric matrices
(i)[351−1]

(ii)⎡⎢⎣6−22−23−12−13⎤⎥⎦

(iii)⎡⎢⎣33−1−2−21−4−52⎤⎥⎦

(iv)[15−12]

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Solution

Let A=[351−1], then A=P+Q
where, P=12(A+A′)andQ=12(A−A′)
Now, P=12(A+A′)=12([351−1]+[315−1])=12[666−2]=[333−1]
∴P′=[333−1]=[333−1]=P
(∴P′=P, then P is a symmetric matrix.)
Thus, P=12(A+A′) is a symmetric matrix.
Now, Q=12(A−A′)=12([351−1]−[315−1])=12[04−40]=[02−20],
∴,Q′=[02−20]′=[0−220]=−Q
[∵Q′=−Q, then Q is a skew-symmetric matrix]
Thus, Q=12(A−A′) is a skew -symmetric matrix.
Representing A as the sum of P and Q.
P+Q=[333−1]+[02−20]=[351−1]=A.

Let A=⎡⎢⎣6−22−23−12−13⎤⎥⎦, then A′=⎡⎢⎣6−22−23−12−13⎤⎥⎦=⎡⎢⎣6−22−23−12−13⎤⎥⎦=A
Now, A+A′=⎡⎢⎣6−22−23−12−13⎤⎥⎦+⎡⎢⎣6−22−23−12−13⎤⎥⎦=⎡⎢⎣12−44−46−24−26⎤⎥⎦
Let P=12(A+A′)=12⎡⎢⎣12−44−46−24−26⎤⎥⎦=⎡⎢⎣6−22−23−12−13⎤⎥⎦
Now, P′=⎡⎢⎣6−22−23−12−13⎤⎥⎦=⎡⎢⎣6−22−23−12−13⎤⎥⎦=P
Thus, P=12(A+A′) is a symmetric matrix.
Now, A−A′=⎡⎢⎣6−22−23−12−13⎤⎥⎦−⎡⎢⎣6−22−23−12−13⎤⎥⎦=⎡⎢⎣000000000⎤⎥⎦
Let Q=12(A−A′)=12⎡⎢⎣000000000⎤⎥⎦Now,Q′=⎡⎢⎣000000000⎤⎥⎦=−Q
Thus, Q=12(A−A′) is a skew-symmetric matrix.
Representing A as the sum of P and Q.
P+Q=⎡⎢⎣6−22−23−12−13⎤⎥⎦+⎡⎢⎣000000000⎤⎥⎦=⎡⎢⎣6−22−23−12−13⎤⎥⎦=A

Let A=⎡⎢⎣33−1−2−21−4−52⎤⎥⎦,thenA′=⎡⎢⎣3−2−43−2−5−112⎤⎥⎦
Now, A+A′=⎡⎢⎣33−1−2−21−4−52⎤⎥⎦+⎡⎢⎣3−2−43−2−5−112⎤⎥⎦=⎡⎢⎣61−51−4−4−5−44⎤⎥⎦
P=12(A+A′)=12⎡⎢⎣61−51−4−4−5−44⎤⎥⎦=⎡⎢ ⎢ ⎢⎣312−−5212−2−2−52−22⎤⎥ ⎥ ⎥⎦
Now, P′=⎡⎢ ⎢ ⎢⎣312−5212−2−2−52−22⎤⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢⎣312−5212−2−2−52−22⎤⎥ ⎥ ⎥⎦=P
Thus, P=12(A+A′) is a symmetric matrix.
Now, A−A′=⎡⎢⎣33−1−2−21−4−52⎤⎥⎦−⎡⎢⎣3−2−43−2−5−112⎤⎥⎦=⎡⎢⎣053−506−3−60⎤⎥⎦
Let Q=12(A−A′)=12⎡⎢⎣053−506−3−60⎤⎥⎦=⎡⎢ ⎢ ⎢⎣05232−5203−32−30⎤⎥ ⎥ ⎥⎦

Now, Q′=⎡⎢ ⎢ ⎢⎣05232−5203−32−30⎤⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢⎣0−52−32520−33230⎤⎥ ⎥ ⎥⎦=−Q
Thus, Q=12(A−A′) is a skew -symmetric matrix.
Representing A as the sum of P and Q.
P+Q=⎡⎢ ⎢ ⎢⎣312−5212−2−2−52−22⎤⎥ ⎥ ⎥⎦+⎡⎢ ⎢ ⎢⎣05232−5203−32−30⎤⎥ ⎥ ⎥⎦=⎡⎢⎣33−1−2−21−4−52⎤⎥⎦=A

Let A=[15−12]. Then, A′=[15−12]′=[1−152]
Now, A+A′=[1544]=[1−152 ]=[2444]

Let P=12(A+A′)=12[2444]=[1222],Now,P′=[1222]=[1222]=P
Thus, P=12(A+A′) is a symmetric matrix.
Now, A−A′=[15−12]−[1−152]=[06−60]
Let Q=12(A−A′)=12[06−60]=[03−30]
Now, Q′=[03−30]′=[0−330]=−Q
Thus, Q=12(A−A′) is a skew -symmetric matrix.
Represending A as the sum of P and Q.
P+Q=[1222]+[03−30]=[15−12]=A


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