Answer
$30$ students in room A, $25$ students in room B, and $45$ students in room C.
Work Step by Step
Step 1. Assume there are $x$ number of students in room A, $y$ number of students in room B, and $z$ number of students in room C.
Step 2. Based on the conditions given, we have
$\begin{cases} x+y+z=100 \\ x/2+z/3=x/2+4y/5 \\ x/2+4y/5=y/5+2z/3 \end{cases}$
Step 3. Simplify the last two equations to get:
$\begin{cases} x+y+z=100 \\ 0x+12y-5z=0 \\ 15x+18y-20z=0 \end{cases}$
Step 4. Establish the augmented matrix of the system and use the Gauss Eliminations method:
$\begin{vmatrix} 1 & 1 & 1 & 100 \\0 & 12 & -5 & 0\\15 & 18 & -20 & 0 \end{vmatrix} \begin{array}( \\ \\R_3-15R_1\to R_3\\ \end{array}$
Step 5. Do the operations given on the right side of the matrix.
$\begin{vmatrix} 1 & 1 & 1 & 100 \\0 & 12 & -5 & 0\\0 & 3 & -35 & -1500 \end{vmatrix} \begin{array}( \\R_2-3R_3\to R_2 \\ \\ \end{array}$
Step 6. Do the operations given on the right side of the matrix.
$\begin{vmatrix} 1 & 1 & 1 & 100 \\0 & 0 & 100 & 4500\\0 & 3 & -35 & -1500 \end{vmatrix} \begin{array}( \\ \\ \\ \end{array}$
Step 7. Row 2 gives $z=45$, row 3 gives $3y-35z=-1500$ and $y=25$, and row 1 gives $x=30$
Step 8. The final results are: $30$ students in room A, $25$ students in room B, and $45$ students in room C.
