If left|begin{array}{ccc}1+x & 1 & 1 1+y & 1 | vedclass

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

Question

(D) Let $\Delta = \left|\begin{array}{ccc}1+x & 1 & 1 \ 1+y & 1+2 y & 1 \ 1+z & 1+z & 1+3 z\end{array}ight|$.Applying $C_1 	o C_1 - C_3$ and $C_2

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) Let $\Delta = \left|\begin{array}{ccc}1+x & 1 & 1 \ 1+y & 1+2 y & 1 \ 1+z & 1+z & 1+3 z\end{array}ight|$.Applying $C_1 	o C_1 - C_3$ and $C_2 	o C_2 - C_3$:$\Delta = \left|\begin{array}{ccc}x & 0 & 1 \ y & 2y & 1 \ z-3z & -2z & 1+3z\end{array}ight| = \left|\begin{array}{ccc}x & 0 & 1 \ y & 2y & 1 \ -2z & -2z & 1+3z\end{array}ight|$.Taking $x, y, z$ common from $C_1, C_2, C_3$ respectively:$\Delta = xyz \left|\begin{array}{ccc}1+1/x & 1/x & 1/x \ 1/y+1 & 1/y+2 & 1/y \ 1/z+1 & 1/z+1 & 1/z+3\end{array}ight|$.Alternatively,expand the determinant directly:$\Delta = (1+x)[(1+2y)(1+3z) - (1+z)] - 1[(1+y)(1+3z) - (1+z)] + 1[(1+y)(1+z) - (1+2y)(1+z)]$.After simplification,$\Delta = x(2y)(3z) + x(2y) + x(3z) + (2y)(3z) = 6xyz + 2xy + 3xz + 6yz$.Factor out $xyz$: $\Delta = xyz(6 + \frac{2}{z} + \frac{3}{y} + \frac{6}{x})$.Comparing with $10kxyz(3 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z})$,we find $k=1$.

If $\left|\begin{array}{ccc}1+x & 1 & 1 \\ 1+y & 1+2 y & 1 \\ 1+z & 1+z & 1+3 z\end{array}\right| = 10 k x y z \left(3+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$,then $k = \text{ . . . . . . }$ (where $x, y, z \neq 0$ and $3+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \neq 0$).

Similar Questions

The number of real values of $t$ such that the system of homogeneous equations
$\begin{aligned}
t x+(t+1) y+(t-1) z &=0 \\
(t+1) x+t y+(t+2) z &=0 \\
(t-1) x+(t+2) y+t z &=0
\end{aligned}$
has non-trivial solutions is

If the area of a triangle with vertices $(2, 6)$,$(5, 4)$,and $(k, 4)$ is $35$ square units,then $k = \text{ . . . . . . }$.

The area of a triangle is $4$ sq. units,and its vertices are $(-2, 0)$,$(0, 4)$,and $(0, k)$. Find the value of $k$.

The solution set of $\left|\begin{array}{ccc}x & 3 & 5 \\ 2 & 6 & 10 \\ 7 & 21 & 35\end{array}\right|=0$ is . . . . . . .

If $\left[\begin{array}{rrr}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right]$ has no inverse,then the real value of $x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module