If A, B are two non-singular matrices of orde | vedclass

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(C) Given $A$ and $B$ are non-singular matrices of order $3$ and $|B|=k$.$A$. $|k^{-1} A^{-1}| = (k^{-1})^3 |A^{-1}| = \frac{1}{k^3 |A|} = \frac{1}{|B

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$A-I, B-V, C-II, D-IV$

Vedclass · Answer

(C) Given $A$ and $B$ are non-singular matrices of order $3$ and $|B|=k$.$A$. $|k^{-1} A^{-1}| = (k^{-1})^3 |A^{-1}| = \frac{1}{k^3 |A|} = \frac{1}{|B|^3 |A|}$. Thus, $A-III$.$B$. $|	ext{Adj}(A^{-1})| = |A^{-1}|^{3-1} = |A^{-1}|^2 = \frac{1}{|A|^2}$. Thus, $B-V$.$C$. Given $BAB^{-1} = I$, then $A = B^{-1}IB = B^{-1}B = I$ is not necessarily true, but $BAB^{-1} = I \Rightarrow A = B^{-1}B = I$ is wrong. Actually, $BAB^{-1} = I \Rightarrow A = B^{-1}B = I$ is not implied. However, $BAB^{-1} = I \Rightarrow BA^kB^{-1} = (BAB^{-1})^k = I^k = I$. Wait, the options suggest $BA^kB^{-1} = B \frac{	ext{Adj}(B)}{|B|} = I$. Since $B 	ext{Adj}(B) = |B|I$, then $B \frac{	ext{Adj}(B)}{|B|} = I$. Thus, $C-II$.$D$. $	ext{Adj}(	ext{Adj}(A^{-1})) = |A^{-1}|^{3-2} (A^{-1}) = |A^{-1}| A^{-1} = \frac{1}{|A|} A^{-1}$. Thus, $D-IV$.Therefore, the correct match is $A-III, B-V, C-II, D-IV$.

List-$I$	List-$II$
$A$. $\|k^{-1} A^{-1}\|$	$I$. $BA^k + A^kB$
$B$. $\|\text{Adj}(A^{-1})\|$	$II$. $\frac{B\text{Adj}(B)}{\|B\|}$
$C$. $BAB^{-1} = I \Rightarrow BA^kB^{-1} =$	$III$. $\frac{1}{\|B\|^3\|A\|}$
$D$. $\text{Adj}(\text{Adj}(A^{-1})) =$	$IV$. $\frac{1}{\|A\|}(A^{-1})$
	$V$. $\frac{1}{\|A\|^2}$

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