Question

1. Show that the system of equations do not have a solution unless a + c = 2b.
3x + 4y + 5z = a, 4x + 5y + 6z = b, 5x + 6y + 72 = 0​

Answer 1

Answer:

Show that the system of equations do not have a solution unless a + c = 2b.

3x + 4y + 5z = a, 4x + 5y + 6z = b, 5x + 6y + 7z = c​

$\triangle=\left|\begin{array}{ccc}3&4&5\\4 &5&6\\5&6&7\end{array}\right|$

$\triangle=\left|\begin{array}{ccc}8&10&12\\4 &5&6\\5&6&7\end{array}\right|\:R_1\implies\:R_1+R_3$

$\triangle=\left|\begin{array}{ccc}0&0&0\\4 &5&6\\5&6&7\end{array}\right|\:R_1\implies\:R_1-2R_2$

$\triangle=0$

$\triangle_x=\left|\begin{array}{ccc}a&4&5\\b&5&6\\c&6&7\end{array}\right|$

$\triangle_x=\left|\begin{array}{ccc}a+c&10&12\\b&5&6\\c&6&7\end{array}\right|\:R_1\implies\:R_1+R_3$

$\triangle_x=\left|\begin{array}{ccc}a+c-2b&0&0\\b&5&6\\c&6&7\end{array}\right|\:R_1\implies\:R_1-2R_2$

$\triangle_x=(a+c-2b)(35-36)$

$\triangle_x=-(a+c-2b)$

$\triangle_x\neq\:0\implies\:a+c-2b\neq\:0\:\implies\:a+c\beq\:2b$

$\text{The given system of equations do not have solution if }\:a+c\neq\:b$