Find all the solutions to the system
\(\begin{array}{r}
x_1 - 2x_2 + x_3 = 1 \\
x_1 - x_2 - x_3 = 0
\end{array}\)
defined over \(\mathbb{R}\).

We use the first equation to write \(x_1\) in terms of the other variables:
\(x_1 = 1 + 2x_2 - x_3\). Substituting \(x_1\) into the second equation,
we get \[(1 + 2x_2 - x_3) -x_2 - x_3 = 0.\]
Simplifying gives
\[x_2 - 2x_3 = -1.\]
So we can set \(x_3\) to any real number \(s\) and obtain
\(x_2 = -1 + 2s\).
Hence, \(x_1 = 1 + 2(-1+2s) - s = -1 + 3s\).
In summary, the solutions are given by
\(\begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix} =
\begin{bmatrix} -1+3s\\ -1 + 2s\\s\end{bmatrix}\)
for all \(s \in \mathbb{R}\).

Example 2

For what values of \(a\) does the system
\(\begin{array}{r}
x - y + z = 1 \\
-x - y - 3z = 0 \\
y + z = a
\end{array}\)
defined over \(\mathbb{R}\) have no solution?

The first equation gives \(x = 1 + y - z\).
Substituting this into the second equation gives
\(-(1+y-z) - y - 3z = 0\), or equivalently, \(-2y - 2z = 1\).
Multiplying both sides by \(-\frac{1}{2}\), we obtain
\(y + z = -\frac{1}{2}\).
Note that this equation has the same left-hand side as the third equation.
Thus, if \(a\neq -\frac{1}{2}\), the system has no solution.
But if \(a = -\frac{1}{2}\), then
\(\begin{bmatrix} x \\ y \\ z\end{bmatrix}
= \begin{bmatrix}
\frac{1}{2} - 2t
\\ -\frac{1}{2} - t \\ t
\end{bmatrix}\)
is a solution for each \(t\in \mathbb{R}\).

Example 3

Find all the solutions to the system
\(
\begin{split}
z - 2w & = 2i \\
z + iw & = 3+i
\end{split}
\)
defined over \(\mathbb{C}\).

We use the first equation to write \(z\) in terms of \(w\):
\(z = 2i + 2w\). Substituting this into the second equation gives
\[(2i + 2w) + iw = 3+i.\]
Simplifying gives
\[ (2+i)w = 3 - i.\]
Hence,
\[ w = \frac{3-i}{2+i} =
\frac{(3-i)(2-i)}{(2+i)(2-i)}
=\frac{6-3i-2i-1}{2^2 - i^2}
= \frac{5-5i}{5} = 1-i.\]
Since \(z = 2i + 2w\), we have that \(z = 2i + 2(1-i) = 2\).
In summary, the system has the unique solution
\(\begin{bmatrix} z\\w\end{bmatrix} =
\begin{bmatrix} 2 \\ 1 - i\end{bmatrix}\).

Example 4

Let \(A = \begin{bmatrix} 2 & -1 & 3 \\ 4 & 5 & 0\end{bmatrix}\),
\(x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\),
\(b = \begin{bmatrix} 8 \\ 9\end{bmatrix}\).
Write out the system \(Ax = b\) in full.

Note that \(Ax\) is the tuple
\(\begin{bmatrix} 2x_1-x_2 + 3x_3 \\ 4x_1 + 5x_2 + 0x_3 \end{bmatrix}\).
Hence, the system written out in full is
\begin{align*}
2x_1 - x_2 + 3x_3 & = 8 \\
4x_1 + 5x_2 & = 9.
\end{align*}

Example 5

Show that one can transform the system
\(\begin{array}{r}
x - 2y = 3 \\
x + y = 0
\end{array}\)
to the system
\(\begin{array}{r}
y = -1 \\
x - 2y = 3
\end{array}\)
using elementary operations.

A question like this can be solved by trial and error while paying attention
to what is given.
First note that the second equation in the second system is the same as the
first equation in the first system.
So if we perform the elementary operation of interchanging the positions
of the two equations in the first system, we obtain
\begin{eqnarray*}
x + y & = & 0 \\
x - 2y & = & 3.
\end{eqnarray*}

Now, the coefficient of \(x\) is zero in
the first equation of the target system.
So we want to eliminate \(x\) by adding \(-1\) times the second equation
of the system above to the first equation to obtain
\begin{eqnarray*}
3 y & = & -3 \\
x - 2y & = & 3.
\end{eqnarray*}

We can now transform this system to the target
system by multiplying the first equation by \(\frac{1}{3}\).

Example 6

Let \(x = 2t -1\) and \(y = -t + 1\).
For what values of \(t\) is the pair \(x,y\) a solution to the system
\(\begin{array}{r}
x + y = 1 \\
2x - y = 2
\end{array}\)?

Substituting \(x\) and \(y\) into the system, we obtain system,
\begin{eqnarray*}
(2t-1) + (-t+1) = 1 \\
2(2t-1) - (-t+1) = 2
\end{eqnarray*}

Simplifying the left-hand sides gives
\begin{eqnarray*}
t = 1 \\
5t - 3= 2
\end{eqnarray*}

Adding \(3\) to both sides of the second equation gives
\begin{eqnarray*}
t = 1 \\
5t = 5
\end{eqnarray*}

So \(t = 1\) is the only solution.

Example 7

Let \(a,b,c,d,u,v \in \mathbb{R}\).
Assuming that \(a \neq 0\) and \(ad - bc \neq 0\), solve
the following system for \(x\) and \(y\) using the method of substitution.
\begin{eqnarray*}
ax + by & = & u \\
cx + dy & = & v
\end{eqnarray*}

Since \(a \neq 0\), the first equation gives \(x = \frac{u}{a} - \frac{b}{a}y\).
Substituting this into the
second equation gives
\[c\left(\frac{u}{a} - \frac{b}{a}y\right) + dy = v.\]
Simplifying the left-hand side gives
\[\frac{cu}{a} + \frac{ad-bc}{a}y = v.\]

Subtracting \(\frac{cu}{a}\) from both sides gives
\[\frac{ad-bc}{a}y = v - \frac{cu}{a}.\]

Since \(a\neq 0\) and \(ad-bc\neq 0\),
we can divide both sides by \(\frac{a}{ad-bc}\)
to obtain
\( y = \frac{av - cu}{ad - bc}.\)