a. The system represented by the matrix is

\(\begin{cases}x+2z=-1\\y-4z=-2\\0=0\end{cases}\)

b. This is a consistent system (because the last equation is true for all ordered triplets).

The system will not have a single solution, as it is, in effect, a system of TWO equations in THREE variables.

To solve, we take z to be any real number \(t\in\mathbb{R}\), and back-substitute into the other two equations:

\(\begin{cases}x+2(t)=-1\\x=-1-2t\end{cases}\)

\(\begin{cases}y-4(t)=-2\\y=-2+4t\end{cases}\)

Result:

a. \(\begin{cases}x+2z=-1\\y-4z=-2\\0=0\end{cases}\)

b. Consistent, solution set: \(\left\{(-1+2t, -2+4t,t),\ t\in\mathbb{R}\right\}\)

\(\begin{cases}x+2z=-1\\y-4z=-2\\0=0\end{cases}\)

b. This is a consistent system (because the last equation is true for all ordered triplets).

The system will not have a single solution, as it is, in effect, a system of TWO equations in THREE variables.

To solve, we take z to be any real number \(t\in\mathbb{R}\), and back-substitute into the other two equations:

\(\begin{cases}x+2(t)=-1\\x=-1-2t\end{cases}\)

\(\begin{cases}y-4(t)=-2\\y=-2+4t\end{cases}\)

Result:

a. \(\begin{cases}x+2z=-1\\y-4z=-2\\0=0\end{cases}\)

b. Consistent, solution set: \(\left\{(-1+2t, -2+4t,t),\ t\in\mathbb{R}\right\}\)