If $A$ is a non-singular matrix, then
- ${ A }^{ -1 }$ is symmetric if $A$ is symmetric
- ${ A }^{ -1 }$ is skew-symmetric if $A$ is symmetric
- $\left| { A }^{ -1 } \right| =\left| A \right| $
- $\left| { A }^{ -1 } \right| ={ \left| A \right| }^{ -1 }$
Reveal answer
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A,D
Correct answer
Explanation
Since $\left| A \right| \neq 0$, therefore ${ A }^{ -1 }$ exists.
Now, $A{ A }^{ -1 }=I={ A }^{ -1 }A$
$\Rightarrow \left( A{ A }^{ -1 } \right) '=I'=\left( { A }^{ -1 }A \right) '\Rightarrow \left( { A }^{ -1 } \right) 'A'=I=A'\left( { A }^{ -1 } \right) '\quad \quad \quad \left( \because A'=A \right) $
$\Rightarrow \left( { A }^{ -1 } \right) 'A=I=A\left( { A }^{ -1 } \right) '\Rightarrow { A }^{ -1 }=\left( { A }^{ -1 } \right) '\Rightarrow { A }^{ -1 }$ is symmetric
Also, since $\left| A \right| \neq 0,\therefore { A }^{ -1 }$ exists such that
$A{ A }^{ -1 }=I={ A }^{ -1 }A\Rightarrow \left| A{ A }^{ -1 } \right| =\left| I \right| $
$\Rightarrow \left| A \right| \left| { A }^{ -1 } \right| =1\quad \quad \left( \because \left| AB \right| =\left| A \right| \left| B \right| \right) $
$\displaystyle \Rightarrow \left| { A }^{ -1 } \right| =\frac { 1 }{ \left| A \right| } $