If left|begin{array}{ccc}a^{2} & b c & c^{2}+ | vedclass

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

Question

(D) Let $\Delta = \left|\begin{array}{ccc}a^{2} & b c & c^{2}+a c \ a^{2}+a b & b^{2} & c a \ a b & b^{2}+b c & c^{2}\end{array}ight|$.Taking $a$

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) Let $\Delta = \left|\begin{array}{ccc}a^{2} & b c & c^{2}+a c \ a^{2}+a b & b^{2} & c a \ a b & b^{2}+b c & c^{2}\end{array}ight|$.Taking $a$ common from $C_1$, $b$ common from $C_2$, and $c$ common from $C_3$:$\Delta = abc \left|\begin{array}{ccc}a & c & \frac{c^2+ac}{c} \ a+b & b & a \ b & b+c & c\end{array}ight| = abc \left|\begin{array}{ccc}a & c & c+a \ a+b & b & a \ b & b+c & c\end{array}ight|$.Applying $C_1 	o C_1 + C_2 - C_3$:$\Delta = abc \left|\begin{array}{ccc}a+c-(c+a) & c & c+a \ a+b+b-a & b & a \ b+b+c-c & b+c & c\end{array}ight| = abc \left|\begin{array}{ccc}0 & c & c+a \ 2b & b & a \ 2b & b+c & c\end{array}ight|$.Applying $R_3 	o R_3 - R_2$:$\Delta = abc \left|\begin{array}{ccc}0 & c & c+a \ 2b & b & a \ 0 & c & c-a\end{array}ight|$.Expanding along $C_1$:$\Delta = abc \cdot (-2b) \cdot \left|\begin{array}{cc}c & c+a \ c & c-a\end{array}ight| = -2ab^2c \cdot (c(c-a) - c(c+a)) = -2ab^2c \cdot (c^2 - ac - c^2 - ac) = -2ab^2c \cdot (-2ac) = 4a^2b^2c^2$.Comparing with $k a^2b^2c^2$, we get $k = 4$.

If $\left|\begin{array}{ccc}a^{2} & b c & c^{2}+a c \\ a^{2}+a b & b^{2} & c a \\ a b & b^{2}+b c & c^{2}\end{array}\right|=k a^{2} b^{2} c^{2}$, then $k=$

Similar Questions

If $A$ is a square matrix of order $n$ and $A = kB$,where $k$ is a scalar,then $|A|=$

Using properties of determinants,prove that:
$\left|\begin{array}{ccc}\alpha & \alpha^{2} & \beta+\gamma \\ \beta & \beta^{2} & \gamma+\alpha \\ \gamma & \gamma^{2} & \alpha+\beta\end{array}\right|=(\beta-\gamma)(\gamma-\alpha)(\alpha-\beta)(\alpha+\beta+\gamma)$

If ${D_r} = \left| \begin{array}{ccc} {2^{r - 1}} & {2 \cdot 3^{r - 1}} & {4 \cdot 5^{r - 1}} \\ x & y & z \\ {2^n} - 1 & {3^n} - 1 & {5^n} - 1 \end{array} \right|$,then the value of $\sum\limits_{r = 1}^n {D_r} = $

$\left| {\begin{array}{ccc} 1 & 1+ac & 1+bc \\ 1 & 1+ad & 1+bd \\ 1 & 1+ae & 1+be \end{array}} \right| = $

By using properties of determinants,show that:
$\left|\begin{array}{ccc}x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4\end{array}\right|=(5x+4)(4-x)^{2}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module