If the system of linear equations 3x + y + be | vedclass

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(B) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $\Delta$ must be $0$,and the determin

Vedclass · Accepted Answer

$31$

Vedclass · Answer

(B) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $\Delta$ must be $0$,and the determinants $\Delta_x, \Delta_y, \Delta_z$ must also be $0$.First,calculate $\Delta = \begin{vmatrix} 3 & 1 & \beta \ 2 & \alpha & -1 \ 1 & 2 & 1 \end{vmatrix} = 3(\alpha + 2) - 1(2 + 1) + \beta(4 - \alpha) = 3\alpha + 6 - 3 + 4\beta - \alpha\beta = 3\alpha + 4\beta - \alpha\beta + 3 = 0$.Next,calculate $\Delta_z = \begin{vmatrix} 3 & 1 & 3 \ 2 & \alpha & -3 \ 1 & 2 & 4 \end{vmatrix} = 3(4\alpha + 6) - 1(8 + 3) + 3(4 - \alpha) = 12\alpha + 18 - 11 + 12 - 3\alpha = 9\alpha + 19 = 0$.From $9\alpha + 19 = 0$,we get $\alpha = -\frac{19}{9}$.Substitute $\alpha$ into the equation $3\alpha + 4\beta - \alpha\beta + 3 = 0$:$3(-\frac{19}{9}) + 4\beta - (-\frac{19}{9})\beta + 3 = 0 \Rightarrow -\frac{19}{3} + 4\beta + \frac{19}{9}\beta + 3 = 0$.Multiply by $9$: $-57 + 36\beta + 19\beta + 27 = 0 \Rightarrow 55\beta = 30 \Rightarrow \beta = \frac{30}{55} = \frac{6}{11}$.Finally,calculate $22\beta - 9\alpha = 22(\frac{6}{11}) - 9(-\frac{19}{9}) = 2(6) + 19 = 12 + 19 = 31$.

If the system of linear equations $3x + y + \beta z = 3$,$2x + \alpha y - z = -3$,and $x + 2y + z = 4$ has infinitely many solutions,then the value of $22\beta - 9\alpha$ is:

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