For the system of linear equations:2x - y + 3 | vedclass

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(C) The determinant of the coefficient matrix is $\Delta = \begin{vmatrix} 2 & -1 & 3 \ 3 & 2 & -1 \ 4 & 5 & \alpha \end{vmatrix} = 2(2\alpha + 5) +

Vedclass · Accepted Answer

The system has infinitely many solutions for $\alpha = -6$ and $\beta = 9$.

Vedclass · Answer

(C) The determinant of the coefficient matrix is $\Delta = \begin{vmatrix} 2 & -1 & 3 \ 3 & 2 & -1 \ 4 & 5 & \alpha \end{vmatrix} = 2(2\alpha + 5) + 1(3\alpha + 4) + 3(15 - 8) = 4\alpha + 10 + 3\alpha + 4 + 21 = 7\alpha + 35 = 7(\alpha + 5)$.For a unique solution,$\Delta 
eq 0$,which means $\alpha 
eq -5$. Thus,the system has a unique solution for any $\beta$ when $\alpha 
eq -5$.For infinitely many solutions,we require $\Delta = \Delta_1 = \Delta_2 = \Delta_3 = 0$.Setting $\Delta = 0$ gives $\alpha = -5$.Calculating $\Delta_3 = \begin{vmatrix} 2 & -1 & 5 \ 3 & 2 & 7 \ 4 & 5 & \beta \end{vmatrix} = 2(2\beta - 35) + 1(3\beta - 28) + 5(15 - 8) = 4\beta - 70 + 3\beta - 28 + 35 = 7\beta - 63 = 7(\beta - 9)$.Setting $\Delta_3 = 0$ gives $\beta = 9$.When $\alpha = -5$ and $\beta = 9$,$\Delta = \Delta_1 = \Delta_2 = \Delta_3 = 0$,so the system has infinitely many solutions.Option $C$ states the system has infinitely many solutions for $\alpha = -6$ and $\beta = 9$,which is incorrect because $\Delta 
eq 0$ when $\alpha = -6$.

For the system of linear equations:
$2x - y + 3z = 5$
$3x + 2y - z = 7$
$4x + 5y + \alpha z = \beta$
Which of the following is $NOT$ correct?

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For the system of linear equations:$2x - y + 3z = 5$$3x + 2y - z = 7$$4x + 5y + \alpha z = \beta$Which of the following is $NOT$ correct?