Consider the system of linear equations x+y+z | vedclass

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(D) The coefficient matrix $A$ is given by $\begin{bmatrix} 1 & 1 & 1 \ 1 & 2 & \lambda^2 \ 1 & 3 & \lambda \end{bmatrix}$.For the system to have a

Vedclass · Accepted Answer

System has a unique solution if $\lambda 
eq 1$ and $\mu 
eq 13$.

Vedclass · Answer

(D) The coefficient matrix $A$ is given by $\begin{bmatrix} 1 & 1 & 1 \ 1 & 2 & \lambda^2 \ 1 & 3 & \lambda \end{bmatrix}$.For the system to have a unique solution,$\det(A) 
eq 0$.$\det(A) = 1(2\lambda - 3\lambda^2) - 1(\lambda - \lambda^2) + 1(3 - 2) = 2\lambda - 3\lambda^2 - \lambda + \lambda^2 + 1 = -2\lambda^2 + \lambda + 1 = -(2\lambda+1)(\lambda-1)$.Thus,$\det(A) = 0$ when $\lambda = 1$ or $\lambda = -1/2$.If $\lambda 
eq 1$ and $\lambda 
eq -1/2$,the system has a unique solution for any $\mu$.If $\lambda = 1$,the equations become $x+y+z=5$,$x+2y+z=9$,and $x+3y+z=\mu$. Subtracting the first from the second gives $y=4$. Subtracting the second from the third gives $y=\mu-9$. Thus,$4 = \mu-9 \Rightarrow \mu=13$. If $\mu=13$,there are infinite solutions. If $\mu 
eq 13$,there is no solution.Option $D$ states the system has a unique solution if $\lambda 
eq 1$ and $\mu 
eq 13$. This is incorrect because even if $\lambda 
eq 1$,if $\lambda = -1/2$,the system may not have a unique solution regardless of $\mu$.

Consider the system of linear equations $x+y+z=5$,$x+2y+\lambda^2 z=9$,and $x+3y+\lambda z=\mu$,where $\lambda, \mu \in R$. Then,which of the following statements is $NOT$ correct?

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