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(D) Let the common difference of the $A.P.$ be $d$. Then $a_n = a_1 + (n-1)d$. For matrix $A$,$|A| = \begin{vmatrix} a_1 & a_2 & a_3 \ a_4 & a_5 & a_

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All of the above.

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(D) Let the common difference of the $A.P.$ be $d$. Then $a_n = a_1 + (n-1)d$. For matrix $A$,$|A| = \begin{vmatrix} a_1 & a_2 & a_3 \ a_4 & a_5 & a_6 \ a_5 & a_6 & a_7 \end{vmatrix}$. Applying $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_2$,we get $R_2 = (3d, 3d, 3d)$ and $R_3 = (d, d, d)$. Since $R_2 = 3R_3$,the rows are linearly dependent,so $|A| = 0$. Thus,$A$ is singular. For the system of equations,the determinant of the coefficient matrix is $|C| = \begin{vmatrix} a_1 & a_2 & a_3 \ a_4 & a_5 & a_6 \ a_7 & a_8 & a_9 \end{vmatrix}$. Applying $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_2$,we get $R_2 = (3d, 3d, 3d)$ and $R_3 = (3d, 3d, 3d)$. Since $R_2 = R_3$,$|C| = 0$. Thus,the system has infinite solutions. For matrix $B$,$|B| = a_1^2 - (i a_2)^2 = a_1^2 + a_2^2$. Since $a_1, a_2$ are real and $a_1 
eq 0$,$|B| > 0$,so $B$ is non-singular. Therefore,all statements are correct.

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

Suppose $a_1, a_2, \dots$ are real numbers,with $a_1 \neq 0$. If $a_1, a_2, a_3, \dots$ are in $A.P.$,then:

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