An important concern in the study of heat transfer is to determine the steady-state temperature distribution of a thin plate when the temperature around the boundary is known. The figure represents a cross section of a metal beam, with negligible heat flow in the direction perpendicular to the plate. Let $$ T_{1}, \ldots, T_{4} $$ denote the temperatures at the four interior nodes of the figure. The temperature at a node is approximately equal to the average of the four nearest nodes-to the left, above, to the right, and below. Solve the system of equations below to find the temperatures $$ \mathrm{T}_{1}, \ldots, \mathrm{~T}_{4} $$. $$ \begin{array}{rl} 4 \mathrm{~T}_{1}-\mathrm{T}_{2}-\mathrm{T}_{4} & =35 \ -\mathrm{T}_{1}+4 \mathrm{~T}_{2}-\mathrm{T}_{3} & =60 \ -\mathrm{T}_{2}+4 \mathrm{~T}_{3}-\mathrm{T}_{4} & =70 \ -\mathrm{T} & \mathrm{~T}, \Lambda \mathrm{~T} \end{array} $$ Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. A. The unique solution of the system is $$ \left(T_{1}, T_{2}, T_{3}, T_{4}\right)=( $$ $$ \square $$ $$ \square $$ , $$ \square $$ , $$ \quad $$ ) ). (Type integers or decimals rounded to the nearest thousandth as needed.) B. The system has infinitely many solutions. C. The system has no solution.

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Solve the system of equations $$ 4T_{1}-T_{2}-T_{4}=35;-T_{1}+4T_{2}-T_{3}=60;-T_{2}+4T_{3}-T_{4}=70;-T_{1}+4T_{2}-T_{3}=60 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}4T_{1}-T_{2}-T_{4}=35\-T_{1}+4T_{2}-T_{3}=60\-T_{2}+4T_{3}-T_{4}=70\-T_{1}+4T_{2}-T_{3}=60\end{array}ight.$$
- step1: Simplify:
  $$\left\{ \begin{array}{l}4T_{1}-T_{2}-T_{4}=35\-T_{1}+4T_{2}-T_{3}=60\-T_{2}+4T_{3}-T_{4}=70\end{array}ight.$$
- step2: Solve the equation:
  $$\left\{ \begin{array}{l}T_{2}=-35+4T_{1}-T_{4}\-T_{1}+4T_{2}-T_{3}=60\-T_{2}+4T_{3}-T_{4}=70\end{array}ight.$$
- step3: Substitute the value of $$T_{2}:$$
  $$\left\{ \begin{array}{l}-T_{1}+4\left(-35+4T_{1}-T_{4}ight)-T_{3}=60\-\left(-35+4T_{1}-T_{4}ight)+4T_{3}-T_{4}=70\end{array}ight.$$
- step4: Simplify:
  $$\left\{ \begin{array}{l}15T_{1}-140-4T_{4}-T_{3}=60\35-4T_{1}+4T_{3}=70\end{array}ight.$$
- step5: Solve the equation:
  $$\left\{ \begin{array}{l}T_{3}=-200+15T_{1}-4T_{4}\35-4T_{1}+4T_{3}=70\end{array}ight.$$
- step6: Substitute the value of $$T_{3}:$$
  $$35-4T_{1}+4\left(-200+15T_{1}-4T_{4}ight)=70$$
- step7: Simplify:
  $$-765+56T_{1}-16T_{4}=70$$
- step8: Evaluate:
  $$-765-16T_{4}+56T_{1}=70$$
- step9: Move the expression to the right side:
  $$56T_{1}=70+765+16T_{4}$$
- step10: Add the numbers:
  $$56T_{1}=835+16T_{4}$$
- step11: Divide both sides:
  $$\frac{56T_{1}}{56}=\frac{835+16T_{4}}{56}$$
- step12: Divide the numbers:
  $$T_{1}=\frac{835+16T_{4}}{56}$$
- step13: Substitute the value of $$T_{1}:$$
  $$T_{3}=-200+15	imes \frac{835+16T_{4}}{56}-4T_{4}$$
- step14: Simplify:
  $$T_{3}=\frac{1325+16T_{4}}{56}$$
- step15: Substitute the value of $$T_{1}:$$
  $$T_{2}=-35+4	imes \frac{835+16T_{4}}{56}-T_{4}$$
- step16: Simplify:
  $$T_{2}=\frac{345+2T_{4}}{14}$$
- step17: Calculate:
  $$\left(T_{1},T_{2},T_{3},T_{4}ight) = \left(\frac{835+16T_{4}}{56},\frac{345+2T_{4}}{14},\frac{1325+16T_{4}}{56},T_{4}ight),T_{4} \in \mathbb{R}$$
- step18: Alternative Form:
  $$	extrm{Infinitely many solutions}$$
The system of equations has infinitely many solutions. Therefore, the correct choice is:

B. The system has infinitely many solutions.
The system has infinitely many solutions.

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