If the system of equations (lambda-1) x+(lamb | vedclass

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(B) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $D$ must be $0$,and $D_x = D_y = D_z

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $D$ must be $0$,and $D_x = D_y = D_z = 0$.The determinant $D$ is given by:$D = \begin{vmatrix} \lambda-1 & \lambda-4 & \lambda \ \lambda & \lambda-1 & \lambda-4 \ \lambda+1 & \lambda+2 & -(\lambda+2) \end{vmatrix} = 0$Performing row operations $R_1 	o R_1 + R_2 + R_3$ or expanding the determinant,we find the condition for $D=0$ leads to $(\lambda-3)(2\lambda+1) = 0$.Thus,$\lambda = 3$ or $\lambda = -1/2$.Checking the consistency for $D_x = 0$ with $\lambda = 3$:$D_x = \begin{vmatrix} 5 & -1 & 3 \ 7 & 2 & -1 \ 9 & 5 & -5 \end{vmatrix} = 5(-10+5) + 1(-35+9) + 3(35-18) = 5(-5) - 26 + 3(17) = -25 - 26 + 51 = 0$.Since $D_x = 0$ at $\lambda = 3$,the system is consistent with infinitely many solutions.Therefore,$\lambda^2 + \lambda = 3^2 + 3 = 9 + 3 = 12$.

List-$I$	List-$II$
$(A)$ $\lambda=8, \mu \neq 15$	$1$. Infinitely many solutions
$(B)$ $\lambda \neq 8, \mu \in R$	$2$. No solution
$(C)$ $\lambda=8, \mu=15$	$3$. Unique solution

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

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