Henry Ernest Dudeney/Modern Puzzles/13 - Find the Coins/Working

Working for Modern Puzzles by Henry Ernest Dudeney: $13$ -- Find the Coins

The simultaneous equations in matrix form:

$\begin {bmatrix} 4 & -4 & -4 \\ -2 & 6 & -2 \\ -1 & -1 & 7 \end {bmatrix} \begin {bmatrix} a \\ b \\ c \end {bmatrix} = \begin {bmatrix} 120 \\ 120 \\ 120 \end {bmatrix}$

when converted to reduced echelon form, gives:

$\begin {bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {bmatrix} \begin {bmatrix} a \\ b \\ c \end {bmatrix} = \begin {bmatrix} 195 \\ 105 \\ 60 \end {bmatrix}$

Proof

\(\ds \)

\(\)

\(\ds \paren {\begin {array} {ccc{{|}}c} 4 & -4 & -4 & 120 \\ -2 & 6 & -2 & 120 \\ -1 & -1 & 7 & 120 \end {array} }\)

\(\ds \)

\(\leadsto\)

\(\ds \paren {\begin {array} {ccc{{|}}c} 1 & -1 & -1 & 30 \\ -1 & 3 & -1 & 60 \\ -1 & -1 & 7 & 120 \end {array} }\)

$r_1 \to r_1 / 4$, $r_2 \to r_2 / 2$

\(\ds \)

\(\leadsto\)

\(\ds \paren {\begin {array} {ccc{{|}}c} 1 & -1 & -1 & 30 \\ 0 & 2 & -2 & 90 \\ 0 & -2 & 6 & 150 \end {array} }\)

$r_2 \to r_2 + r_1$, $r_3 \to r_3 + r_1$

\(\ds \)

\(\leadsto\)

\(\ds \paren {\begin {array} {ccc{{|}}c} 1 & -1 & -1 & 30 \\ 0 & 1 & -1 & 45 \\ 0 & -1 & 3 & 75 \end {array} }\)

$r_2 \to r_2 / 2$, $r_3 \to r_3 / 2$

\(\ds \)

\(\leadsto\)

\(\ds \paren {\begin {array} {ccc{{|}}c} 1 & -1 & -1 & 30 \\ 0 & 1 & -1 & 45 \\ 0 & 0 & 2 & 120 \end {array} }\)

$r_3 \to r_3 + r_2$

\(\ds \)

\(\leadsto\)

\(\ds \paren {\begin {array} {ccc{{|}}c} 1 & -1 & -1 & 30 \\ 0 & 1 & -1 & 45 \\ 0 & 0 & 1 & 60 \end {array} }\)

$r_3 \to r_3 / 2$

\(\ds \)

\(\leadsto\)

\(\ds \paren {\begin {array} {ccc{{|}}c} 1 & -1 & 0 & 90 \\ 0 & 1 & 0 & 105 \\ 0 & 0 & 1 & 60 \end {array} }\)

$r_1 \to r_1 + r_3$, $r_2 \to r_2 + r_3$

\(\ds \)

\(\leadsto\)

\(\ds \paren {\begin {array} {ccc{{|}}c} 1 & 0 & 0 & 195 \\ 0 & 1 & 0 & 105 \\ 0 & 0 & 1 & 60 \end {array} }\)

$r_1 \to r_1 + r_2$

$\blacksquare$