If the system of equations x+4y-z=lambda,7x+9 | vedclass

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

Question

(B) For the system of equations to have infinitely many solutions,the determinant of the coefficient matrix $\Delta$ must be $0$,and $\Delta_x = \Delt

Vedclass · Accepted Answer

$-3$

Vedclass · Answer

(B) For the system of equations to have infinitely many solutions,the determinant of the coefficient matrix $\Delta$ must be $0$,and $\Delta_x = \Delta_y = \Delta_z = 0$.First,calculate $\Delta = \begin{vmatrix} 1 & 4 & -1 \ 7 & 9 & \mu \ 5 & 1 & 2 \end{vmatrix} = 0$.Expanding along the first row: $1(18-\mu) - 4(14-5\mu) - 1(7-45) = 0$.$18 - \mu - 56 + 20\mu + 38 = 0$.$19\mu = 0 \Rightarrow \mu = 0$.Now,for $\Delta_x = 0$:$\Delta_x = \begin{vmatrix} \lambda & 4 & -1 \ -3 & 9 & 0 \ -1 & 1 & 2 \end{vmatrix} = 0$.Expanding along the first row: $\lambda(18-0) - 4(-6-0) - 1(-3+9) = 0$.$18\lambda + 24 - 6 = 0$.$18\lambda = -18 \Rightarrow \lambda = -1$.Finally,calculate $(2\mu + 3\lambda) = 2(0) + 3(-1) = -3$.

If the system of equations $x+4y-z=\lambda$,$7x+9y+\mu z=-3$,and $5x+y+2z=-1$ has infinitely many solutions,then $(2\mu+3\lambda)$ is equal to:

Similar Questions

The system of equations $x_1 - x_2 + x_3 = 2$,$3x_1 - x_2 + 2x_3 = -6$ and $3x_1 + x_2 + x_3 = -18$ has

If the system of equations $x + y + z = 6$,$x + 2y + 3z = 10$,and $x + 2y + \lambda z = 0$ has a unique solution,then $\lambda$ is not equal to:

The system of equations $x - 2y + 3z = 5$,$2x - 2y + z = 0$,and $-x + 2y - 3z = 6$ has

If the system of equations $\alpha x + y + z = 5$,$x + 2y + 3z = 4$,and $x + 3y + 5z = \beta$ has infinitely many solutions,then the ordered pair $(\alpha, \beta)$ is equal to:

If $AX=D$ represents the system of linear equations $3x-4y+7z+6=0$,$5x+2y-4z+9=0$ and $8x-6y-z+5=0$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module