The system of equations x+y+z=6,x+2y+5z=9,x+5 | vedclass

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(A) For a system of linear equations to have no solution,the determinant of the coefficient matrix $D$ must be $0$,and at least one of the Cramer's ru

Vedclass · Accepted Answer

$\lambda=17, \mu 
eq 18$

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(A) For a system of linear equations to have no solution,the determinant of the coefficient matrix $D$ must be $0$,and at least one of the Cramer's rule determinants $(D_x, D_y, D_z)$ must be non-zero.First,calculate $D = \begin{vmatrix} 1 & 1 & 1 \ 1 & 2 & 5 \ 1 & 5 & \lambda \end{vmatrix} = 1(2\lambda - 25) - 1(\lambda - 5) + 1(5 - 2) = 2\lambda - 25 - \lambda + 5 + 3 = \lambda - 17$.Setting $D = 0$,we get $\lambda = 17$.Next,calculate $D_z = \begin{vmatrix} 1 & 1 & 6 \ 1 & 2 & 9 \ 1 & 5 & \mu \end{vmatrix} = 1(2\mu - 45) - 1(\mu - 9) + 6(5 - 2) = 2\mu - 45 - \mu + 9 + 18 = \mu - 18$.For no solution,$D = 0$ and $D_z 
eq 0$.Thus,$\lambda = 17$ and $\mu 
eq 18$.

The system of equations $x+y+z=6$,$x+2y+5z=9$,$x+5y+\lambda z=\mu$ has no solution if

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