If Delta=left|begin{array}{lll}1 & a & a^2 1 | vedclass

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

Question

(C) Given,$\Delta=\left|\begin{array}{lll}1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2\end{array}ight|$ and $\Delta_1=\left|\begin{array}{ccc}1 & 1 & 1

Vedclass · Accepted Answer

$\Delta_1=-\Delta$

Vedclass · Answer

(C) Given,$\Delta=\left|\begin{array}{lll}1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2\end{array}ight|$ and $\Delta_1=\left|\begin{array}{ccc}1 & 1 & 1 \ b c & c a & a b \ a & b & c\end{array}ight|$.First,we evaluate $\Delta$:$\Delta = (a-b)(b-c)(c-a)$.Now,for $\Delta_1$,we multiply and divide the determinant by $abc$ or perform row/column operations.Alternatively,multiply $R_1$ by $abc$,$R_2$ by $1$,and $R_3$ by $1$ is not efficient. Let us multiply $C_1$ by $a$,$C_2$ by $b$,and $C_3$ by $c$:$\Delta_1 = \frac{1}{abc} \left|\begin{array}{ccc}a & b & c \ abc & abc & abc \ a^2 & b^2 & c^2\end{array}ight| = \frac{abc}{abc} \left|\begin{array}{ccc}a & b & c \ 1 & 1 & 1 \ a^2 & b^2 & c^2\end{array}ight| = \left|\begin{array}{ccc}a & b & c \ 1 & 1 & 1 \ a^2 & b^2 & c^2\end{array}ight|$.Swapping $R_1$ and $R_2$ gives $-\left|\begin{array}{ccc}1 & 1 & 1 \ a & b & c \ a^2 & b^2 & c^2\end{array}ight|$.Swapping $R_2$ and $R_3$ gives $(-1)^2 \left|\begin{array}{ccc}1 & 1 & 1 \ a^2 & b^2 & c^2 \ a & b & c\end{array}ight|$.Actually,the standard result for $\Delta_1$ is $-(a-b)(b-c)(c-a)$.Thus,$\Delta_1 = -\Delta$.

If $\Delta=\left|\begin{array}{lll}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{array}\right|$ and $\Delta_1=\left|\begin{array}{ccc}1 & 1 & 1 \\ b c & c a & a b \\ a & b & c\end{array}\right|$,then

Similar Questions

For non-zero,real $a, b$ and $c$,if $\left| \begin{array}{ccc} \frac{a^2+b^2}{c} & c & c \\ a & \frac{b^2+c^2}{a} & a \\ b & b & \frac{c^2+a^2}{b} \end{array} \right| = \alpha abc$,then the value of $\alpha$ is

$\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2} - bc}\\1&b&{{b^2} - ac}\\1&c&{{c^2} - ab}\end{array}\,} \right| = $

If $|A|$ denotes the value of the determinant of the square matrix $A$ of order $3$,then $|-2A|=$

$\left|\begin{array}{ccc}a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b\end{array}\right|$ is equal to

$\left| \begin{array}{ccc} a - b & b - c & c - a \\ x - y & y - z & z - x \\ p - q & q - r & r - p \end{array} \right| = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module