Question

The inverse of a skew-symmetric matrix (if it exists) is:
A.a symmetric matrix
B.a skew symmetric matrix
C.a diagonal matrix
D.none of a matrix is unique

Answer 1

Hint: In this question, we need to comment on the term known for the inverse of a skew-symmetric matrix. For this, we will use the property of skew symmetric matrix then apply the Reversal law of transpose of matrices.
Formula Used:
 $ A' = - A $ where $ A' $ is the transpose of $ A $
 $ (AB)' = B'A' $ (Reversal law of transpose of matrices).

Complete step-by-step answer:
We have, $ A' = - A - - - - (i) $
We know that, $ A{A^{ - 1}} = {A^{ - 1}}A = I - - - - (ii) $
where $ I $ is the identity matrix and $ {A^{ - 1}} $ is the inverse of $ A $
Now taking transpose on each side of equation (ii) we get,
 $ (A{A^{ - 1}})' = ({A^{ - 1}}A)' = (I)' $
Since , transpose of Identity matrix is Identity matrix so,
 $ (A{A^{ - 1}})' = ({A^{ - 1}}A)' = I $
Applying Reversal Law of transpose of matrices i.e. $ (AB)' = B'A' $ we get,
\[({A^{ - 1}})'A' = A'({A^{ - 1}})' = I\]
 $ ({A^{ - 1}})'( - A) = ( - A)({A^{ - 1}})' = I $ (Since from equation (i), $ A' = - A $ )
From the above equation we see that $ ({A^{ - 1}})'( - A) = I $ which means that $ ({A^{ - 1}})' $ is the inverse of $ ( - A) $ (Since , their product is identity matrix which implies that one is inverse of the other)
 $ \Rightarrow ({A^{ - 1}})' = - ({A^{ - 1}}) $
Now from the above equation, we see that $ {A^{ - 1}} $ satisfies equation 1 i.e. the condition of a skew symmetric matrix.
Hence, $ {A^{ - 1}} $ is skew symmetric i.e. the inverse of a skew symmetric matrix is a skew symmetric matrix
So, the correct answer is “Option B”.

Note: A matrix whose transpose is equal to the negative of the matrix itself is known as Skew-symmetric matrix while a matrix whose transpose is equal to the matrix itself is known as the symmetric matrix. Make sure to apply all the conditions of matrices very carefully. For better understanding of the question/solution, revise all laws related to the matrix.