If the system of equations (k+1)^3 x + (k+2) | vedclass

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

Question

(B) The given system of equations is: $ (k+1)^3 x + (k+2)^3 y = (k+3)^3 $ $ (k+1) x + (k+2) y = k+3 $ $ x + y = 1 $ For the system to be consistent, t

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(B) The given system of equations is: $ (k+1)^3 x + (k+2)^3 y = (k+3)^3 $ $ (k+1) x + (k+2) y = k+3 $ $ x + y = 1 $ For the system to be consistent, the determinant of the augmented matrix must be zero. Let $ D = \begin{vmatrix} (k+1)^3 & (k+2)^3 & (k+3)^3 \ k+1 & k+2 & k+3 \ 1 & 1 & 1 \end{vmatrix} = 0 $. Applying column operations $ C_2 ightarrow C_2 - C_1 $ and $ C_3 ightarrow C_3 - C_1 $: $ D = \begin{vmatrix} (k+1)^3 & (k+2)^3 - (k+1)^3 & (k+3)^3 - (k+1)^3 \ k+1 & (k+2) - (k+1) & (k+3) - (k+1) \ 1 & 1 - 1 & 1 - 1 \end{vmatrix} = 0 $ $ D = \begin{vmatrix} (k+1)^3 & 3k^2 + 9k + 7 & 6k^2 + 24k + 26 \ k+1 & 1 & 2 \ 1 & 0 & 0 \end{vmatrix} = 0 $ Expanding along the third row: $ 1 \cdot [2(3k^2 + 9k + 7) - 1(6k^2 + 24k + 26)] = 0 $ $ 6k^2 + 18k + 14 - 6k^2 - 24k - 26 = 0 $ $ -6k - 12 = 0 $ $ -6k = 12 $ $ k = -2 $.

If the system of equations $ (k+1)^3 x + (k+2)^3 y = (k+3)^3 $, $ (k+1) x + (k+2) y = k+3 $, and $ x + y = 1 $ is consistent, then the value of $ k $ is:

Similar Questions

If $a \neq 1, b \neq -1, c \neq -1$ and the system of equations $x = a(y+z), y = b(z+x), z = c(x+y)$ has a non-trivial solution,then:

If $C = 2 \cos \theta$,then the value of the determinant $\Delta = \begin{vmatrix} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \end{vmatrix}$ is

If $x, y, z$ are all different and not equal to zero and $\left|\begin{array}{ccc}1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z\end{array}\right|=0$,then the value of $x^{-1}+y^{-1}+z^{-1}$ is equal to

If $q_1, q_2, q_3$ are roots of the equation $x^3 + 64 = 0$,then the value of $\left| \begin{array}{ccc} q_1 & q_2 & q_3 \\ q_2 & q_3 & q_1 \\ q_3 & q_1 & q_2 \end{array} \right|$ is

If $a^2 + b^2 + c^2 = -2$ and $f(x) = \left| \begin{array}{ccc} 1 + a^2x & (1 + b^2)x & (1 + c^2)x \\ (1 + a^2)x & 1 + b^2x & (1 + c^2)x \\ (1 + a^2)x & (1 + b^2)x & 1 + c^2x \end{array} \right|$,then $f(x)$ is a polynomial of degree

Vedclass Test Series

Exam Paper Generator

Online Exam Module