The determinant left| begin{array}{ccc} 4 + x | vedclass

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

Question

(B) Let $\Delta = \left| \begin{array}{ccc} 4 + x^2 & -6 & -2 \ -6 & 9 + x^2 & 3 \ -2 & 3 & 1 + x^2 \end{array} ight|$.Applying the column operati

Vedclass · Accepted Answer

$x^3$

Vedclass · Answer

(B) Let $\Delta = \left| \begin{array}{ccc} 4 + x^2 & -6 & -2 \ -6 & 9 + x^2 & 3 \ -2 & 3 & 1 + x^2 \end{array} ight|$.Applying the column operations $C_1 	o C_1 + x C_3$ and $C_2 	o C_2 - 3 C_3$:$\Delta = \left| \begin{array}{ccc} 4 + x^2 - 2x & -6 & -2 \ -6 + 3x & 9 + x^2 & 3 \ -2 + x + x^3 & 3 - 3 - 3x^2 & 1 + x^2 \end{array} ight|$ is not the most efficient path.Instead,expand the determinant directly or use row/column properties.After simplification,the determinant evaluates to $\Delta = x^2(x^2 + 14)$.Since $\Delta = x^2(x^2 + 14)$,it is divisible by $x$,$x^2$,and $(14 + x^2)$.It is not divisible by $x^3$ or $x^5$.

The determinant $\left| \begin{array}{ccc} 4 + x^2 & -6 & -2 \\ -6 & 9 + x^2 & 3 \\ -2 & 3 & 1 + x^2 \end{array} \right|$ for $x \neq 0$ is not divisible by:

Similar Questions

The rank of the matrix $A=\begin{bmatrix} 1 & 1 & 1 & 3 \\ 2 & 2 & -1 & 3 \\ 1 & 1 & -1 & 1 \end{bmatrix}$ is

If $A = \begin{bmatrix} 0 & 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 8 & 12 \\ 0 & 0 & 0 & 4 & 8 \end{bmatrix}$,then the rank of $A$ is

If $f(x) = \begin{vmatrix} x^3-x & 2e^{2x} & \sin x^2 \\ \cos(2x) & x+x^2 & e^{-x} \\ \tan 3x & \ln(1-2x) & x^2+x+1 \end{vmatrix}$,then $f'(0)$ is equal to:

The rank of the matrix $\left[ {\begin{array}{*{20}{c}}4&1&0&0\\3&0&1&0\\6&0&2&0\end{array}} \right]$ is:

If $A = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix}$ where $a = 7^x$,$b = 7^{7^x}$,$c = 7^{7^{7^x}}$,then $\int |A| \, dx$ (where $|A|$ is the determinant of the matrix $A$) is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module