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

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$(1, 3)$

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(C) 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 $\Delta$:$\Delta = \begin{vmatrix} \alpha & 1 & 1 \ 1 & 2 & 3 \ 1 & 3 & 5 \end{vmatrix} = 0$$\alpha(10 - 9) - 1(5 - 3) + 1(3 - 2) = 0$$\alpha(1) - 2 + 1 = 0 \Rightarrow \alpha - 1 = 0 \Rightarrow \alpha = 1$.Now,substitute $\alpha = 1$ into the system and calculate $\Delta_x$ (or use the augmented matrix approach). For infinitely many solutions,the augmented matrix $[A|B]$ must have a rank less than $3$.Using the augmented matrix:$\begin{bmatrix} 1 & 1 & 1 & | & 5 \ 1 & 2 & 3 & | & 4 \ 1 & 3 & 5 & | & \beta \end{bmatrix}$$R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$:$\begin{bmatrix} 1 & 1 & 1 & | & 5 \ 0 & 1 & 2 & | & -1 \ 0 & 2 & 4 & | & \beta - 5 \end{bmatrix}$$R_3 	o R_3 - 2R_2$:$\begin{bmatrix} 1 & 1 & 1 & | & 5 \ 0 & 1 & 2 & | & -1 \ 0 & 0 & 0 & | & \beta - 5 - 2(-1) \end{bmatrix}$For infinitely many solutions,the last row must be all zeros,so $\beta - 5 + 2 = 0 \Rightarrow \beta - 3 = 0 \Rightarrow \beta = 3$.Thus,the ordered pair is $(\alpha, \beta) = (1, 3)$.

If the system of equations $\alpha x + y + z = 5$,$x + 2y + 3z = 4$,and $x + 3y + 5z = \beta$ has infinitely many solutions,then the ordered pair $(\alpha, \beta)$ is equal to:

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