Prove that Delta = left| begin{array}{ccc} a+ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

Vedclass · Answer

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

Prove that $\Delta = \left| \begin{array}{ccc} a+bx & c+dx & p+qx \\ ax+b & cx+d & px+q \\ u & v & w \end{array} \right| = (1-x^2) \left| \begin{array}{ccc} a & c & p \\ b & d & q \\ u & v & w \end{array} \right|$

Similar Questions

If $x, y, z$ are all positive and are the $p$-th, $q$-th, and $r$-th terms of a geometric progression respectively, then the value of the determinant $\left|\begin{array}{lll} \log x & p & 1 \\ \log y & q & 1 \\ \log z & r & 1 \end{array}\right|$ equals:

If $\Delta = \begin{vmatrix} a & b & c \\ x & y & z \\ p & q & r \end{vmatrix}$,then $\begin{vmatrix} ka & kb & kc \\ kx & ky & kz \\ kp & kq & kr \end{vmatrix}$ =

If $\Delta = \begin{vmatrix} a + x & b & c \\ b & x + c & a \\ c & a & x + b \end{vmatrix}$,which of the following is a factor for the above determinant?

By using properties of determinants,show that:
$\left|\begin{array}{lll}x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y\end{array}\right|=(x-y)(y-z)(z-x)(x y+y z+z x)$

Without expanding the determinant,prove that $\left|\begin{array}{lll}a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b\end{array}\right|=\left|\begin{array}{lll}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right|$

Vedclass Test Series

Exam Paper Generator

Online Exam Module