Exercise 7.2.1

\(\mathbb{S_1}\)

\(\begin{eqnarray*} x + 2 y - z &=& 3\\ y - z &=& 1\\ z &=& 2 \end{eqnarray*}\)

Substituting \(z = 2\) into the second equation gives \(y - 2 = 1\), so \(y = 3\).

Substituting these two values into equation \(1\) yields \(x + 6 - 2 = 3\), so \(x = -1\).

Thus, the solution is \((-1,3,2)\).

There is a geometrical dimension to this system of equations.

In a three-dimensional \(x,y,z\) coordinate system, the set of all points satisfying the first equation is a plane. The same is true of the second and third equations. All three planes intersect at the point \((-1,3,2)\).