For alpha, beta in R,suppose the system of li | vedclass

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(C) For the system to have infinitely many solutions,the determinant of the coefficient matrix must be zero: $\begin{vmatrix} 1 & -1 & 1 \ 2 & 2 & \a

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$x^2-18x+56=0$

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(C) For the system to have infinitely many solutions,the determinant of the coefficient matrix must be zero: $\begin{vmatrix} 1 & -1 & 1 \ 2 & 2 & \alpha \ 3 & -1 & 4 \end{vmatrix} = 0$ $1(8 + \alpha) - (-1)(8 - 3\alpha) + 1(-2 - 6) = 0$ $8 + \alpha + 8 - 3\alpha - 8 = 0$ $8 - 2\alpha = 0 \implies \alpha = 4$. Now,substitute $\alpha = 4$ into the system: $x - y + z = 5$ $2x + 2y + 4z = 8 \implies x + y + 2z = 4$ $3x - y + 4z = \beta$ Adding the first two equations: $(x - y + z) + (x + y + 2z) = 5 + 4 \implies 2x + 3z = 9$. For infinitely many solutions,the third equation must be a linear combination of the first two. Let $k_1(x - y + z) + k_2(x + y + 2z) = 3x - y + 4z$. Comparing coefficients: $k_1 + k_2 = 3$,$-k_1 + k_2 = -1$,$k_1 + 2k_2 = 4$. Solving $k_1 + k_2 = 3$ and $-k_1 + k_2 = -1$ gives $2k_2 = 2 \implies k_2 = 1$ and $k_1 = 2$. Thus,$\beta = 2(5) + 1(4) = 14$. The roots are $\alpha = 4$ and $\beta = 14$. The quadratic equation is $(x - 4)(x - 14) = x^2 - 18x + 56 = 0$.

For $\alpha, \beta \in R$,suppose the system of linear equations $x-y+z=5$,$2x+2y+\alpha z=8$,and $3x-y+4z=\beta$ has infinitely many solutions. Then $\alpha$ and $\beta$ are the roots of:

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