First slide

Applications of determinant

Question

The inverse of a skew-symmetric matrix (if it exists) is

Moderate

A

a symmetric matrix
B

a skew-symmetric matrix
C

a diagonal matrix
D

none of these

Solution

We have $A^{'} = - A$
Now, $A A^{- 1} = A^{- 1} A = I_{n}$
$\begin{array}{l} \Rightarrow {(A A^{- 1})}^{'} = {(A^{- 1} A)}^{'} = {(I_{n})}^{'} \\ \Rightarrow {(A^{- 1})}^{'} A^{'} = A^{'} {(A^{- 1})}^{'} = I_{n} \\ \Rightarrow {(A^{- 1})}^{'} (- A) = (- A) {(A^{- 1})}^{'} = I_{n} \end{array}$
Thus, ${(A^{- 1})}^{'} = - (A^{- 1})$ [inverse of a matrix is unique]

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