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

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

Question

(D) For the system of equations to be consistent,the determinant of the coefficient matrix must be zero.The system is:$(\alpha + 1)^3 x + (\alpha + 2)

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(D) For the system of equations to be consistent,the determinant of the coefficient matrix must be zero.The system is:$(\alpha + 1)^3 x + (\alpha + 2)^3 y = (\alpha + 3)^3$$(\alpha + 1) x + (\alpha + 2) y = (\alpha + 3)$$x + y = 1$The determinant of the augmented matrix must be zero:$\begin{vmatrix} (\alpha + 1)^3 & (\alpha + 2)^3 & -(\alpha + 3)^3 \ \alpha + 1 & \alpha + 2 & -(\alpha + 3) \ 1 & 1 & -1 \end{vmatrix} = 0$Applying column operations $C_1 	o C_1 + C_2$ and $C_2 	o C_2 + C_3$ (or evaluating directly),we find:$6\alpha + 12 = 0$$6\alpha = -12$$\alpha = -2$Thus,the correct option is $D$.

If the system of equations $(\alpha + 1)^3 x + (\alpha + 2)^3 y - (\alpha + 3)^3 = 0$,$(\alpha + 1)x + (\alpha + 2)y - (\alpha + 3) = 0$,and $x + y - 1 = 0$ is consistent,what is the value of $\alpha$?

Similar Questions

Evaluate the determinant: $\left|\begin{array}{ll}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right|$

$\left|\begin{array}{ccc}x+2 & x+3 & x+5 \\ x+4 & x+6 & x+9 \\ x+8 & x+11 & x+15\end{array}\right|$ is equal to

If matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$,then $|A|^{-1}$ is equal to

Evaluate the determinant $\Delta = \begin{vmatrix} 1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0 \end{vmatrix}$.

If ${\Delta _1} = \left| {\begin{array}{{20}{c}}1&0\\a&b\end{array}} \right|$ and ${\Delta _2} = \left| {\begin{array}{{20}{c}}1&0\\c&d\end{array}} \right|$,then ${\Delta _2}{\Delta _1}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module