Question

A system of equations is given below, 𝑡𝑥 + 2𝑦 + 3𝑧 = 𝑎 2𝑥 + 3𝑦 − 𝑡𝑧 = 𝑏 3𝑥 + 5𝑦 + (𝑡 + 1)𝑧 = 𝑐 Where 𝑡 is an integer and 𝑎, 𝑏, 𝑐 are real constants. The system does not have a unique solution, but it is consistent. Show that 𝑎 + 𝑏 = 𝑐.

Answer 1

By Cramer's rule, the system has many solutions if

$\Delta=\begin{vmatrix} t & 2&3 \\ 2 & 3&-t\\ 3&5&t+1 \end{vmatrix}=t(8t+3)-2\cdot (5t+2)+3=8t^2-7t-1=0$

$t=\frac{7\pm \sqrt{49+32}}{16}$

$t_1=1,t_2=-1/8$

$a+b=tx+2y+3z+2x+3y-tz=x(t+2)+5y+z(3-t)$

So, when $t=1$ , then:

$a+b=3x+5y+2z=c$