Consider the following system of equations:

Solution:

$D=\left|\begin{array}{ccc}1 & 2 & -3 \\ 2 & 6 & -11 \\ 1 & -2 & 7\end{array}\right|$

$=20-2(25)-3(-10)$

$=20-50+30=0$

$D_{1}=\left|\begin{array}{ccc}a & 2 & -3 \\ b & 6 & -11 \\ c & -2 & 7\end{array}\right|$

$=20 a-2(7 b+11 c)-3(-2 b-6 c)$

$=20 a-14 b-22 c+6 b+18 c$

$=20 a-8 b-4 c$

$=4(5 a-2 b-c)$

$D_{2}=\left|\begin{array}{ccc}1 & a & -3 \\ 2 & b & -11 \\ 1 & c & 7\end{array}\right|$

$=7 b+11 c-a(25)-3(2 c-b)$

$=7 b+11 c-25 a-6 c+3 b$

$=-25 a+10 b+5 c$

$=-5(5 a-2 b-c)$

$D_{3}=\left|\begin{array}{ccc}1 & 2 & a \\ 2 & 6 & b \\ 1 & -2 & c\end{array}\right|$

$=6 c+2 b-2(2 c-b)-10 a$

$=-10 a+4 b+2 c$

$=-2(5 a-2 b-c)$

for infinite solution

$\mathrm{D}=\mathrm{D}_{1}=\mathrm{D}_{2}=\mathrm{D}_{3}=0$

$\Rightarrow 5 a=2 b+c$