The equation left| begin{array}{ccc} (1+x)^2 | vedclass

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(D) Let the given equation be $D_1 + D_2 = 0$. Note that $D_2$ is the transpose of a determinant $D_3$ where $D_3 = \left| \begin{array}{ccc} (1+x)^2

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has infinite number of solutions

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(D) Let the given equation be $D_1 + D_2 = 0$. Note that $D_2$ is the transpose of a determinant $D_3$ where $D_3 = \left| \begin{array}{ccc} (1+x)^2 & (1-x)^2 & 1-2x \ 2x+1 & 3x & 3x-2 \ x+1 & 2x & 2x-3 \end{array} ight|$. Since the value of a determinant remains unchanged by transposition,$D_2 = D_3$. Adding $D_1$ and $D_2$,we observe that the sum of the determinants simplifies significantly. Performing the operation $C_1 ightarrow C_1 + C_2$ in the combined determinant,we find that the resulting determinant is identically zero for all $x$. Thus,the equation $0 = 0$ holds true for all real and complex values of $x$. Therefore,the equation has an infinite number of solutions.

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

The equation $\left| \begin{array}{ccc} (1+x)^2 & (1-x)^2 & -(2+x^2) \\ 2x+1 & 3x & 1-5x \\ x+1 & 2x & 2-3x \end{array} \right| + \left| \begin{array}{ccc} (1+x)^2 & 2x+1 & x+1 \\ (1-x)^2 & 3x & 2x \\ 1-2x & 3x-2 & 2x-3 \end{array} \right| = 0$

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