1. Left inverse of matrix

Consider the matrix $\mathbf{C} = \begin{pmatrix}1 & -1\\ 1&1 \\ 0&1 \end{pmatrix}$

Find a left-inverse of C.

Attempt:

$\begin{pmatrix}x1 & y1 & z1\\ x2&y2 &z2 \end{pmatrix}\begin{pmatrix}1 & -1\\ 1&1\\ 0&1 \end{pmatrix} = \begin{pmatrix}1 &0\\ 0&1\end{pmatrix}$
$\begin{pmatrix}x1 + y1 & -x1 +y1 + z1 \\ x2 + y2& -x2 + y2 + z2 \end{pmatrix}= \begin{pmatrix}1 &0\\ 0&1\end{pmatrix}$

I know that you suppose to somehow use gauss reduction, and thought about gauss reducing:

$x1 + y1 = 1$
$-x1 + y1 + z1 = 0$

and

$x2 + y2 = 0$
$-x2 + y2 + z2 = 1$

but didnt get any luck

2. Only square matrices have inverses.

To solve a matrix equation of the form

$\mathbf{Y} = \mathbf{AX}$ for $\mathbf{X}$ where $\mathbf{A}$ is not square, you need to premultiply both sides by $\mathbf{A}^T$ first.

$\mathbf{Y} = \mathbf{AX}$

$\mathbf{A}^T\mathbf{Y} = \mathbf{A}^T\mathbf{AX}$.

Now $\mathbf{A}^T\mathbf{A}$ is a square matrix, so you can premultiply both sides by $(\mathbf{A}^T\mathbf{A})^{-1}$.

$\mathbf{A}^T\mathbf{Y} = \mathbf{A}^T\mathbf{AX}$

$(\mathbf{A}^T\mathbf{A})^{-1}\mathbf{A}^T\mathbf{Y} = (\mathbf{A}^T\mathbf{A})^{-1}\mathbf{A}^T\mathbf{AX}$

$(\mathbf{A}^T\mathbf{A})^{-1}\mathbf{A}^T\mathbf{Y} = \mathbf{IX}$

$(\mathbf{A}^T\mathbf{A})^{-1}\mathbf{A}^T\mathbf{Y} = \mathbf{X}$.

3. Originally Posted by Prove It
Only square matrices have inverses.
Why is this so? In my notes it says that a matrix that is not square can have left-inverses or right-inverses, but it can't have both.

4. You have 6 variables and 4 equations.
This system has very many solutions.
To limit number of solutions two variables may be given some values.
Let z1=0 and z2=0

$
\begin{pmatrix}1/2 & 1/2 & 0\\ -1/2 & 1/2 & 0 \end{pmatrix}
$