Prove that $\Delta = \left| \begin{array}{ccc} a+bx & c+dx & p+qx \\ ax+b & cx+d & px+q \\ u & v & w \end{array} \right| = (1-x^2) \left| \begin{array}{ccc} a & c & p \\ b & d & q \\ u & v & w \end{array} \right|$
(A) Applying the row operation $R_1 \rightarrow R_1 - x R_2$ to the determinant $\Delta$,we get:
$\Delta = \left| \begin{array}{ccc} a+bx - x(ax+b) & c+dx - x(cx+d) & p+qx - x(px+q) \\ ax+b & cx+d & px+q \\ u & v & w \end{array} \right|$
$= \left| \begin{array}{ccc} a(1-x^2) & c(1-x^2) & p(1-x^2) \\ ax+b & cx+d & px+q \\ u & v & w \end{array} \right|$
Taking $(1-x^2)$ common from $R_1$:
$= (1-x^2) \left| \begin{array}{ccc} a & c & p \\ ax+b & cx+d & px+q \\ u & v & w \end{array} \right|$
Now,applying the row operation $R_2 \rightarrow R_2 - x R_1$ to the resulting determinant:
$= (1-x^2) \left| \begin{array}{ccc} a & c & p \\ (ax+b) - x(a) & (cx+d) - x(c) & (px+q) - x(p) \\ u & v & w \end{array} \right|$
$= (1-x^2) \left| \begin{array}{ccc} a & c & p \\ b & d & q \\ u & v & w \end{array} \right|$
Hence,the identity is proved.
$\Delta = \left| \begin{array}{ccc} a+bx - x(ax+b) & c+dx - x(cx+d) & p+qx - x(px+q) \\ ax+b & cx+d & px+q \\ u & v & w \end{array} \right|$
$= \left| \begin{array}{ccc} a(1-x^2) & c(1-x^2) & p(1-x^2) \\ ax+b & cx+d & px+q \\ u & v & w \end{array} \right|$
Taking $(1-x^2)$ common from $R_1$:
$= (1-x^2) \left| \begin{array}{ccc} a & c & p \\ ax+b & cx+d & px+q \\ u & v & w \end{array} \right|$
Now,applying the row operation $R_2 \rightarrow R_2 - x R_1$ to the resulting determinant:
$= (1-x^2) \left| \begin{array}{ccc} a & c & p \\ (ax+b) - x(a) & (cx+d) - x(c) & (px+q) - x(p) \\ u & v & w \end{array} \right|$
$= (1-x^2) \left| \begin{array}{ccc} a & c & p \\ b & d & q \\ u & v & w \end{array} \right|$
Hence,the identity is proved.
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