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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

$a = \frac{1}{3}, b 
eq \frac{7}{3}$

Vedclass · Answer

(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 $D$:$D = \begin{vmatrix} 2 & 1 & 1 \ 1 & -1 & 1 \ 1 & 1 & a \end{vmatrix} = 2(-a - 1) - 1(a - 1) + 1(1 + 1) = -2a - 2 - a + 1 + 2 = 1 - 3a$.Setting $D = 0$ gives $1 - 3a = 0$,so $a = \frac{1}{3}$.Next,calculate $D_z$ (or the constant determinant $D_3$):$D_3 = \begin{vmatrix} 2 & 1 & 5 \ 1 & -1 & 3 \ 1 & 1 & b \end{vmatrix} = 2(-b - 3) - 1(b - 3) + 5(1 + 1) = -2b - 6 - b + 3 + 10 = 4 - 3b$.Wait,let's re-evaluate $D_3$:$D_3 = 2(-b-3) - 1(b-3) + 5(1+1) = -2b - 6 - b + 3 + 10 = 4 - 3b$.Actually,let's use the augmented matrix approach for consistency:$R_1 	o R_1 - R_2$: $x + 2y = 2$$R_2 	o R_2 - R_3$: $-2y + (1-a)z = 3-b$Adding these: $x + (1-a)z = 5-b$.For no solution,$D=0 \implies a=1/3$. Substituting $a=1/3$ into the system:$2x+y+z=5$$x-y+z=3$$x+y+z/3=b$Subtracting equations,we find the condition for no solution is $a=1/3$ and $b 
eq 7/3$.

If the following system of linear equations
$2x + y + z = 5$
$x - y + z = 3$
$x + y + az = b$
has no solution,then :

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If the following system of linear equations$2x + y + z = 5$$x - y + z = 3$$x + y + az = b$has no solution,then :