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 \(s \in \mathbb{R}\).

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 \(t\in \mathbb{R}\).

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 \(\begin{bmatrix} z\\w\end{bmatrix} = \begin{bmatrix} 2 \\ 1 - i\end{bmatrix}\).

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*}

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 \(\frac{1}{3}\).

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.

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 \( y = \frac{av - cu}{ad - bc}.\)

Thus, \(x = \frac{u}{a} - \frac{b}{a}\left(\frac{-cu + av}{ad - bc}\right) = \frac{u}{a} - \frac{-bcu+abv}{a(ad - bc)} = \frac{adu-bcu}{a(ad-bc)} - \frac{-bcu+abv}{a(ad - bc)} = \frac{adu-abv}{a(ad-bc)} = \frac{du - bv}{ad-bc}.\)

In summary, there is a unique solution given by \(x = \frac{du - bv}{ad-bc}\) and \( y = \frac{av - cu}{ad - bc}.\)