Finding the number of solutions in a system of equations

Question

Given:

$$M = \begin{bmatrix}1&0&7&0&3\\0&1&6&0&4\\0&0&0&1&2\\0&0&0&0&0\end{bmatrix}$$ How many solutions does this system of equations have? I am given four options:

A) One

B) Infinite

C) No solution

D) Three

I selected D but it was wrong as I simply counted the constants that are not zero, but obviously thats an incorrect way of finding the number of solutions. Does it have anything to do with leading and free variables?

Answer 1

I'm assuming that the last column are the right hand sides of the equations and the 4 columns to its left are the coefficients of the 4 unknowns.

The last equation is $0=0$, no contradiction. Only three relevant equations left:

We only have three pivots and column 3 has no pivot, so $x_3$ is a free variable. The other equations can then be back-substituted and we get infinitely many solutions in terms of the free parameter $x_3$. So B) is right.

D) can never be right for linear equations over the reals, it's always 0,1 or infinity (when we have a solution space of positive dimension, as here: a line).