left| {begin{array}{*{20}{c}}{x + 1}&{x + 2}& | vedclass

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

Question

(B) Let $\Delta = \left| {\begin{array}{*{20}{c}}{x + 1}&{x + 2}&{x + 4}\{x + 3}&{x + 5}&{x + 8}\{x + 7}&{x + 10}&{x + 14}\end{array}} ight|$.Appl

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(B) Let $\Delta = \left| {\begin{array}{*{20}{c}}{x + 1}&{x + 2}&{x + 4}\{x + 3}&{x + 5}&{x + 8}\{x + 7}&{x + 10}&{x + 14}\end{array}} ight|$.Applying $C_2 	o C_2 - C_1$ and $C_3 	o C_3 - C_2$:$\Delta = \left| {\begin{array}{*{20}{c}}{x + 1}&{1}&{2}\{x + 3}&{2}&{3}\{x + 7}&{3}&{4}\end{array}} ight|$.Applying $C_3 	o C_3 - C_2$:$\Delta = \left| {\begin{array}{*{20}{c}}{x + 1}&{1}&{1}\{x + 3}&{2}&{1}\{x + 7}&{3}&{1}\end{array}} ight|$.Applying $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_2$:$\Delta = \left| {\begin{array}{*{20}{c}}{x + 1}&{1}&{1}\{2}&{1}&{0}\{4}&{1}&{0}\end{array}} ight|$.Expanding along $C_3$:$\Delta = 1 	imes (2 - 4) = -2$.Thus,the value of the determinant is $-2$.

$\left| {\begin{array}{*{20}{c}}{x + 1}&{x + 2}&{x + 4}\\{x + 3}&{x + 5}&{x + 8}\\{x + 7}&{x + 10}&{x + 14}\end{array}} \right| = $

Similar Questions

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} = $

Without expanding,prove that $\Delta = \begin{vmatrix} x+y & y+z & z+x \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix} = 0$.

The value of $\left| \begin{array}{ccc} 41 & 42 & 43 \\ 44 & 45 & 46 \\ 47 & 48 & 49 \end{array} \right| = $

If $A$ is a square matrix of order $3$,then which of the following statements is true? (where $I$ is the identity matrix)

The value of $\left|\begin{array}{lll}1990 & 1991 & 1992 \\ 1991 & 1992 & 1993 \\ 1992 & 1993 & 1994\end{array}\right|$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module