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(D) 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

$a = 3, b 
eq 13$

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(D) 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 the determinant of the coefficient matrix $D$:$D = \begin{vmatrix} 2 & 3 & 6 \ 1 & 2 & a \ 3 & 5 & 9 \end{vmatrix} = 2(18 - 5a) - 3(9 - 3a) + 6(5 - 6) = 36 - 10a - 27 + 9a - 6 = 3 - a$.For $D = 0$,we must have $a = 3$.Next,calculate the determinant $D_z$ (or check the consistency of the augmented matrix):$D_z = \begin{vmatrix} 2 & 3 & 8 \ 1 & 2 & 5 \ 3 & 5 & b \end{vmatrix} = 2(2b - 25) - 3(b - 15) + 8(5 - 6) = 4b - 50 - 3b + 45 - 8 = b - 13$.For the system to have no solution when $D = 0$,we require $D_z 
eq 0$,which implies $b - 13 
eq 0$,or $b 
eq 13$.Thus,the system has no solution when $a = 3$ and $b 
eq 13$.

The values of $a$ and $b$,for which the system of equations $2x + 3y + 6z = 8$,$x + 2y + az = 5$,and $3x + 5y + 9z = b$ has no solution,are:

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