If the system of equations x+2y-3z=2,2x+lambd | vedclass

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

Question

(D) For a system of linear equations to have infinitely many solutions,the determinant $D$ must be $0$,and $D_1 = D_2 = D_3 = 0$.$D = \begin{vmatrix}

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(D) For a system of linear equations to have infinitely many solutions,the determinant $D$ must be $0$,and $D_1 = D_2 = D_3 = 0$.$D = \begin{vmatrix} 1 & 2 & -3 \ 2 & \lambda & 5 \ 14 & 3 & \mu \end{vmatrix} = 1(\lambda\mu - 15) - 2(2\mu - 70) - 3(6 - 14\lambda) = 0$$\lambda\mu - 15 - 4\mu + 140 - 18 + 42\lambda = 0 \Rightarrow \lambda\mu + 42\lambda - 4\mu + 107 = 0$.Now,consider $D_3 = \begin{vmatrix} 1 & 2 & 2 \ 2 & \lambda & 5 \ 14 & 3 & 33 \end{vmatrix} = 1(33\lambda - 15) - 2(66 - 70) + 2(6 - 14\lambda) = 0$$33\lambda - 15 + 8 + 12 - 28\lambda = 0 \Rightarrow 5\lambda + 5 = 0 \Rightarrow \lambda = -1$.Substitute $\lambda = -1$ into the equation for $D = 0$:$(-1)\mu + 42(-1) - 4\mu + 107 = 0 \Rightarrow -5\mu + 65 = 0 \Rightarrow \mu = 13$.Thus,$\lambda + \mu = -1 + 13 = 12$.

If the system of equations $x+2y-3z=2$,$2x+\lambda y+5z=5$,$14x+3y+\mu z=33$ has infinitely many solutions,then $\lambda+\mu$ is equal to:

Similar Questions

The system of linear equations $x + y + z = 2$,$2x + y - z = 3$,and $3x + 2y + kz = 4$ has a unique solution if

If the system of equations $x + 2y - 3z = 1$,$(k + 3)z = 3$,and $(2k + 1)x + z = 0$ is inconsistent,then the value of $k$ is

The system of equations $4x + y + 2z = 5$, $x - 5y + 3z = 10$, and $9x - 3y + 7z = 20$ has

The following system of linear equations is given: $2x + 3y + 2z = 9$,$3x + 2y + 2z = 9$,and $x - y + 4z = 8$. Which of the following statements is true?

If $A = \begin{bmatrix} -1 & 2 \\ 2 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 3 \\ 1 \end{bmatrix}$,$AX = B$,then $X = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module