System of equations with infinitely many solutions

Question

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?

Answer 1

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 $b = 6c_1 + 7c_2 = \frac{65} 5 = 13.$

Answer 2

Another way of getting b is through SVD (or equivalently eigenvalues, but I think SVD is a bit cleaner). If you're trying to solve Ax=y, then plugging in the SVD of A, you have USV'x=y, or SV'x=U'y. Now, the final entry of the left hand side is going to be zero because that singular value is zero, so the final entry of the right hand side had better be zero as well if you're going to have a solution. That gives you that the dot product of the last column of U with the right hand side is zero, and from there it's straightforward to plug in the numbers. I wouldn't want to do it by hand, but it's almost trivial on a computer.