Answer: The matrix has determinant \(0\), so it has no inverse.

The general solution of the system is \(\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\1\\-2\end{pmatrix}+t\begin{pmatrix}2\\-2\\3\end{pmatrix}\), where \(t\in\mathbb{R}\).

Let \(A=\begin{pmatrix}1&4&2\\3&0&-2\\3&-3&-4\end{pmatrix}\).

First show that the matrix has no inverse. Its determinant is

\(\det A=\begin{vmatrix}1&4&2\\3&0&-2\\3&-3&-4\end{vmatrix}\).

Expanding along the first row gives

\(\det A=1\begin{vmatrix}0&-2\\-3&-4\end{vmatrix}-4\begin{vmatrix}3&-2\\3&-4\end{vmatrix}+2\begin{vmatrix}3&0\\3&-3\end{vmatrix}\).

Now

\(\begin{vmatrix}0&-2\\-3&-4\end{vmatrix}=0(-4)-(-2)(-3)=-6\),

\(\begin{vmatrix}3&-2\\3&-4\end{vmatrix}=3(-4)-(-2)(3)=-12+6=-6\),

and

\(\begin{vmatrix}3&0\\3&-3\end{vmatrix}=3(-3)-0(3)=-9\).

Therefore

\(\det A=1(-6)-4(-6)+2(-9)=-6+24-18=0\).

Since \(\det A=0\), the matrix \(A\) is singular, so it has no inverse.

Now solve the system

\(x+4y+2z=0,\quad 3x-2z=4,\quad 3x-3y-4z=5\).

From the second equation, \(3x-2z=4\), so \(2z=3x-4\) and \(z=\frac{3x-4}{2}\).

Substitute this into the first equation:

\(x+4y+2\left(\frac{3x-4}{2}\right)=0\).

Hence \(x+4y+3x-4=0\), so \(4x+4y=4\), and therefore \(x+y=1\).

Thus \(y=1-x\). Put \(x=2t\). Then \(y=1-2t\).

Using \(3x-2z=4\), we get

\(3(2t)-2z=4\).

So \(6t-2z=4\), hence \(-2z=4-6t\), and therefore \(z=3t-2\).

So the general solution is

\(x=2t,\quad y=1-2t,\quad z=-2+3t\).

In vector form,

\(\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\1\\-2\end{pmatrix}+t\begin{pmatrix}2\\-2\\3\end{pmatrix}\), where \(t\in\mathbb{R}\).

Finally, check by substitution:

\(x+4y+2z=2t+4(1-2t)+2(-2+3t)=0\),

\(3x-2z=3(2t)-2(-2+3t)=4\),

and

\(3x-3y-4z=3(2t)-3(1-2t)-4(-2+3t)=5\).

Therefore the general solution is correct.