The system of equations x + y + z = 6,x + 2y | vedclass

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Question

(B) The system of equations is given by:$x + y + z = 6$$x + 2y + 3z = 10$$x + 2y + \lambda z = \mu$The coefficient matrix $A$ is:$A = \begin{bmatrix}

Vedclass · Accepted Answer

$\lambda = 3, \mu 
e 10$

Vedclass · Answer

(B) The system of equations is given by:$x + y + z = 6$$x + 2y + 3z = 10$$x + 2y + \lambda z = \mu$The coefficient matrix $A$ is:$A = \begin{bmatrix} 1 & 1 & 1 \ 1 & 2 & 3 \ 1 & 2 & \lambda \end{bmatrix}$The determinant $|A| = 1(2\lambda - 6) - 1(\lambda - 3) + 1(2 - 2) = 2\lambda - 6 - \lambda + 3 = \lambda - 3$.For the system to have no solution or infinitely many solutions,we must have $|A| = 0$,which implies $\lambda = 3$.When $\lambda = 3$,the system becomes:$x + y + z = 6$$x + 2y + 3z = 10$$x + 2y + 3z = \mu$Subtracting the second equation from the third,we get $0 = \mu - 10$. If $\mu 
e 10$,this leads to a contradiction $(0 = 	ext{non-zero constant})$,meaning the system has no solution.Thus,the condition for no solution is $\lambda = 3$ and $\mu 
e 10$.

The system of equations $x + y + z = 6$,$x + 2y + 3z = 10$,and $x + 2y + \lambda z = \mu$ has no solution for:

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