Consider this system of equations
$$2x + 3y + z = 6$$
$$-x+y+2z = 7$$
$$ax+y+4z=b$$Find the values of $a$ and $b$ for which the system has an infinite number of solutions.
I am stuck struggling with the solution offered to this problem.
The first step is easy. The matrix of the coefficients of the equations must have zero determinant so that the solution is at least not unique, eg so $$\begin{vmatrix} 2 && 3 && 1 \\ -1 && 1 && 2 \\ a && 1 && 4 \end{vmatrix} = 0$$ This gives $5a + 15 = 0$, and so $a=-3$
However, the next step is a bit more confusing.
The solution expresses the system in the form $$\begin{pmatrix} x \\ y \\ z\end{pmatrix} = \frac{1}{5a+15} \begin{pmatrix} 2 && -11 && 5 \\ -2 && 11 && -5 \\ 2 && -11 && 5 \end{pmatrix} \begin{pmatrix} 6 \\ 7 \\ b\end{pmatrix} $$ They get this by left multiplying the system by the inverse of the matrix of coefficients.
The confusing part is that they then go and consider the final row of the system and set it equal to zero, eg $$\frac{1}{5a+15} (2 \times 6 + (-11) \times 7 + 5 \times b) = 0$$ giving $b=13$
My question is why does the final row being zero give us the infinite number of solutions?
Best Answer
Observe that if one of the equations in this system can be written as a linear combination of the other two equations, then the system reduces to two equations in three unknowns. Consequently, there is one degree of freedom, and thus, there are infinitely many solutions to the system. (Explicitly, each solution corresponds to setting the free variable equal to some constant.)
Ultimately, it suffices to find constants $c_1$ and $c_2$ such that $\langle a, 1, 4 \rangle = c_1 \langle 2, 3, 1 \rangle + c_2 \langle -1, 1, 2 \rangle.$ Of course, this can be achieved by solving the induced $2 \times 2$ system of equations $$\begin{cases} 3c_1 + c_2 = 1 \text{ and} \\ c_1 + 2 c_2 = 4. \end{cases}$$ Equivalently, one can obtain the reduced row-echelon form of the matrix $\begin{pmatrix} 3 & 1 & 1 \\ 1 & 2 & 4 \end{pmatrix}$ to find that $c_1 = -\frac 2 5$ and $c_2 = \frac{11} 5.$ From here, we find that $a = 2c_1 - c_2 = -3$ and $b = 6c_1 + 7c_2 = \frac{65} 5 = 13.$