$x + 2 y + 3 z = 6$
$2 x + 4 y + z = 7$
$3 x + 2 y + 9 z = 14$ .
Solve by Crammer's rule

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Solution

$x ∣ ∣ ∣ ∣ \begin{matrix} 1 & 2 & 3 2 & 4 & 1 3 & 2 & 9 \end{matrix} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \begin{matrix} 6 & 2 & 3 7 & 4 & 1 14 & 2 & 9 \end{matrix} ∣ ∣ ∣ ∣$

or $x D = D_{1}$

Operating $C_{1} - (C_{2} + C_{3})$ on $D_{1}$ , we get

$D_{1} = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 2 & 3 2 & 4 & 1 3 & 2 & 9 \end{matrix} ∣ ∣ ∣ ∣$ = D.

Hence xD = D or $x = 1$
Similarly $y = 1, z = 1$ .

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