1
Question
Find 12(A+A′) and 12(A−A′), when A=⎡⎢⎣0ab−a0c−b−c0⎤⎥⎦
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Solution
Given : A=⎡⎢⎣0ab−a0c−b−c0⎤⎥⎦
A′=⎡⎢⎣0−a−ba0−cbc0⎤⎥⎦
Now,
12(A+A′)=12⎛⎜⎝⎡⎢⎣0ab−a0c−b−c0⎤⎥⎦+⎡⎢⎣0−a−ba0−cbc0⎤⎥⎦⎞⎟⎠
⇒12(A+A′)=12⎡⎢⎣000000000⎤⎥⎦=⎡⎢⎣000000000⎤⎥⎦
12(A−A′)=12⎛⎜⎝⎡⎢⎣0ab−a0c−b−c0⎤⎥⎦−⎡⎢⎣0−a−ba0−cbc0⎤⎥⎦⎞⎟⎠
⇒12(A−A′)=12⎡⎢⎣02a2b−2a02c−2b−2c0⎤⎥⎦=⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣12×012×(2a)12×(2b)12×(−2a)12×012×(2c)12×(−2b)12×(−2c)12×0⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦
⇒12(A−A′)=⎡⎢⎣0ab−a0c−b−c0⎤⎥⎦
A′=⎡⎢⎣0−a−ba0−cbc0⎤⎥⎦
Now,
12(A+A′)=12⎛⎜⎝⎡⎢⎣0ab−a0c−b−c0⎤⎥⎦+⎡⎢⎣0−a−ba0−cbc0⎤⎥⎦⎞⎟⎠
⇒12(A+A′)=12⎡⎢⎣000000000⎤⎥⎦=⎡⎢⎣000000000⎤⎥⎦
12(A−A′)=12⎛⎜⎝⎡⎢⎣0ab−a0c−b−c0⎤⎥⎦−⎡⎢⎣0−a−ba0−cbc0⎤⎥⎦⎞⎟⎠
⇒12(A−A′)=12⎡⎢⎣02a2b−2a02c−2b−2c0⎤⎥⎦=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣12×012×(2a)12×(2b)12×(−2a)12×012×(2c)12×(−2b)12×(−2c)12×0⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦
⇒12(A−A′)=⎡⎢⎣0ab−a0c−b−c0⎤⎥⎦
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