Piles of matchsticks

Let $a, b, c$ be the original number of matches on each pile and $A, B, C$ be the current number of matches on each pile.

First step-

Take $B$ matches from the first pile

$A = A - B \implies A = a - b$

then move them to the second pile

$B = B + B \implies 2 \times b$ .

Second step-

Take $C$ matches from the second pile

$B = B - C \implies B = 2 \times b - c$

then move them to the third pile

$C = C + C \implies C = 2 \times c$ .

Third step-

Take $A$ matches from the third pile

$C = C - A \implies C = 2 \times c - (a - b)$

then take them to the first pile

$A = A + A \implies A = 2 \times (a - b)$ .

Next step-

The total number of matches is $48$

$a + b + c = 48$ and $A + B + C = 48$

and since $A = B = C$

then,

$A = B = C = \frac{48}{3} = 16$ .

Last step-

Simply the three equations.

$\begin{bmatrix} 2 & -2 & 0\\ 0 & 2 & -1\\ -1 & 1 & 2 \end{bmatrix} \begin{bmatrix} a\\ b\\ c \end{bmatrix} = \begin{bmatrix} 16\\ 16\\ 16 \end{bmatrix}$

Three equations for three variables. lots of possible way to solve from this point but here's one way.

From eqn1,

$2a - 2b = 16 \implies a = 8 + b$

from eqn2,

$2b - c = 16 \implies c = 2b - 16$

substitute to eqn3,

$-a + b + 2c = 16 \implies -(8 + b) + b + 2(2b - 16) = 16 \implies -8 - b + b + 4b - 32 = 16 \implies \fbox{b = 14}$

using this, get $a$ and $c$ by substituting (b) to eqn1 and eqn2,

$a = 8 + b \implies a = 8 + (14) \implies \fbox{a = 22}$

$c = 2b - 16 \implies c = 2(14) - 16 \implies \fbox{c = 12}$ .