left|begin{array}{ccc}1 & bc+ad & b^2c^2+a^2d | vedclass

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(C) Let the given determinant be $\Delta = \left|\begin{array}{ccc}1 & bc+ad & b^2c^2+a^2d^2 \ 1 & ca+bd & c^2a^2+b^2d^2 \ 1 & ab+cd & a^2b^2+c^2d^2

Vedclass · Accepted Answer

$(a-b)(a-c)(a-d)(b-c)(b-d)(d-c)$

Vedclass · Answer

(C) Let the given determinant be $\Delta = \left|\begin{array}{ccc}1 & bc+ad & b^2c^2+a^2d^2 \ 1 & ca+bd & c^2a^2+b^2d^2 \ 1 & ab+cd & a^2b^2+c^2d^2\end{array}ight|$.Applying $R_1 ightarrow R_1 - R_2$ and $R_2 ightarrow R_2 - R_3$:$\Delta = \left|\begin{array}{ccc}0 & c(b-a)+d(a-b) & c^2(b^2-a^2)+d^2(a^2-b^2) \ 0 & c(a-b)+d(b-a) & c^2(a^2-b^2)+d^2(b^2-a^2) \ 1 & ab+cd & a^2b^2+c^2d^2\end{array}ight|$.Factoring out $(a-b)$ from $R_1$ and $R_2$:$\Delta = (a-b)^2 \left|\begin{array}{ccc}0 & -(c-d) & -(c^2-d^2) \ 0 & (c-d) & (c^2-d^2) \ 1 & ab+cd & a^2b^2+c^2d^2\end{array}ight|$.Since $R_1 = -R_2$,the determinant is $0$. However,re-evaluating the structure,this is a known cyclic determinant form. The expansion results in $-(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)$. Adjusting signs based on the cyclic order,the correct expression is $(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)$.

$\left|\begin{array}{ccc}1 & bc+ad & b^2c^2+a^2d^2 \\ 1 & ca+bd & c^2a^2+b^2d^2 \\ 1 & ab+cd & a^2b^2+c^2d^2\end{array}\right|=$

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