I am trying to find the least-norm solution of the following set of equations
$$\begin{aligned} y+z &= -3\\ x+2y+z &= -2\\ -2x-3y-z &=1\end{aligned}$$
Using the expression for the least-norm solution
$$x=A^T(AA^T)^{-1}b$$
I do the following
$$A=\begin{pmatrix}0&1&1\\ 1&2&1\\ -2&-3&-1\end{pmatrix}$$
$$A^T=\begin{pmatrix}0&1&-2\\ 1&2&-2\\ 1&1&-1\end{pmatrix}$$
$$b=\begin{pmatrix}-3\\ -2\\ 1\end{pmatrix}$$
Now my issue comes when calculating $(AA^T)^{-1}$, the reason begin is when I take the inverse I get a determinant of zero which means that the least-norm solution is $$x=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix}$$ which I am not entirely sure is correct. I have even checked my calculations on an online calculator and still get zero determinant. Is this correct?
If $A\mathbf{x}=\mathbf{b}$ is consistent, then there exists $\mathbf{x}_0$ such that $(AA^T)\mathbf{x}_0 = \mathbf{b}$. Then the minimal solution to $A\mathbf{x}=\mathbf{b}$ is $A^T\mathbf{x}_0$.
Of course, if $AA^T$ is invertible, you can find $\mathbf{x}_0$ by calculating $(AA^T)^{-1}\mathbf{b}$; but if it is not invertible, then what you need is to find any solution to the system, and then multiply by $A^T$ to get the answer.
Here you have $$AA^T = \left(\begin{array}{rrr} 2 & 3 & -4\\ 3 & 6 & -9\\ -4 & -9 & 14 \end{array}\right).$$ So use Gaussian elimination to solve $$\left(\begin{array}{rrr|r} 2 & 3 & -4&-3\\ 3 & 6 & -9&-2\\ -4 & -9 & 14&1 \end{array}\right)$$ then take any solution to that system, multiply it by $A^T$ on the left, and you get the minimal solution to the original system.