If the system of equations 2x + 3y - 3z = 3,x | vedclass

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Question

(B) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $D$ must be $0$,and $D_x = D_y = D_z

Vedclass · Accepted Answer

$\frac{45}{14}$

Vedclass · Answer

(B) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $D$ must be $0$,and $D_x = D_y = D_z = 0$.Given the system:$2x + 3y - 3z = 3$$x + 2y + \alpha z = 1$$2x - y + z = \beta$First,calculate the determinant $D$:$D = \begin{vmatrix} 2 & 3 & -3 \ 1 & 2 & \alpha \ 2 & -1 & 1 \end{vmatrix} = 2(2 + \alpha) - 3(1 - 2\alpha) - 3(-1 - 4) = 4 + 2\alpha - 3 + 6\alpha + 15 = 8\alpha + 16$.Setting $D = 0$ gives $8\alpha + 16 = 0$,so $\alpha = -2$.Now,calculate $D_z$ and set it to $0$:$D_z = \begin{vmatrix} 2 & 3 & 3 \ 1 & 2 & 1 \ 2 & -1 & \beta \end{vmatrix} = 2(2\beta + 1) - 3(\beta - 2) + 3(-1 - 4) = 4\beta + 2 - 3\beta + 6 - 15 = \beta - 7$.Setting $D_z = 0$ gives $\beta = 7$.Now calculate $\frac{\alpha}{\beta} - \frac{\beta}{\alpha} = \frac{-2}{7} - \frac{7}{-2} = -\frac{2}{7} + \frac{7}{2} = \frac{-4 + 49}{14} = \frac{45}{14}$.

If the system of equations $2x + 3y - 3z = 3$,$x + 2y + \alpha z = 1$,and $2x - y + z = \beta$ has infinitely many solutions,then $\frac{\alpha}{\beta} - \frac{\beta}{\alpha} =$

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