Is there is any matrix $C$ such that which satisfied $AC= \mathbb{I}_{3\times3}$?

Question

Is there is any matrix $C$ such that which satisfied $AC= \mathbb{I}_{3\times3}$ where $$A = \begin{bmatrix} 2&3 \\ 1& 2\\1&1\end{bmatrix}$$

I was thinking many times , but i could not get its

Any hints\ solution will be appreciated.

Answer 1

The matrix $C$ must be of form: $$C = \begin{bmatrix} a&c&e \\ b& d&f \end{bmatrix}$$

So we have $2a+3b=1$ , $a+2b = 0$ and $a+b=0$. But from last two you get $a=b=0$ so there is no $C$.

Answer 2

You can think of $A$ as a linear operator from $\mathbb{R}^2$ to $\mathbb{R}^3$. Because of the dimensions it cannot be surjective.

If there was a matrix $C$ such that $AC=I$ then we have $$A(C(x))=x$$ for every $x$, in particular the equation $A(y)=x$ is always solvable. But this is impossible since $A$ is not surjective.

Alternatively you could use brute force, write $$C=\begin{pmatrix} a & b & c \\ d & e & f \end{pmatrix}$$ and write down all the equations given by $AC=I$. You can show they have no solutions. The above method works in higher dimensions as well, as long as $A$ as more lines than columns.