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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