# Left/right inverse matrix question

For what values of $$a,b,c$$ does a left and/or right inverse exist for $$A=\begin{bmatrix} 1 & a \\ 2 & b \\ 3 & c \end{bmatrix}$$ exist?

We know that a left inverse matrix $$X$$ exists such that $$XA=I_2$$ where $$I_2$$ is the $$2\times 2$$ identity matrix so $$X$$ is a $$2\times 3$$ matrix. What do we do next? Thanks.

• How do you know that a left inverse exists? What is that left inverse if $\ a=1\$, $\ b=2\$ and $\ c=3\$? – lonza leggiera Apr 5 at 0:32
• Hint: if the equation $xA=b$ has a solution, what is the relationship between $A$ and $b$? – amd Apr 5 at 0:35
• The fact that "a matrix has left Inverse iff its linear map is an injection and has right inverse iff its linear map is a surjection" may help. – Kamyar Mirzavaziri Apr 5 at 0:43
• what have you tried? – Manuel Pena Apr 6 at 0:07

Here we can view $$A$$ as a map from $$\mathcal{R}^2$$ to $$\mathcal{R}^3$$.

Having a left inverse is impossible if $$A$$ is not an one to one mapping.

Image of the mapping $$A$$ is the column space of $$A$$.

If the columns space is 1-dimentional then $$A$$ is not one to one (because by rank-nullity theorem, kernel of $$A$$ is no trivial).

So to have a left inverse we need $$[a,b,c]$$ not a multiple of $$[1,2,3]$$ to guaranty that image of $$A$$ is not 1-dimentional.

Having a right inverse is impossible as there are no mapping $$X$$ that is one to one from $$\mathcal{R}^3$$ to $$\mathcal{R}^2$$.

• So would that be the opposite if $A=\begin{bmatrix} 1 & 2 & 3\\ a & b & c \end{bmatrix}$ meaning a left inverse is impossible in this case? – squenshl Apr 5 at 1:38
• In this case, $A$ could have right inverse but can't have left inverse and rank of row space is the same as rank of column space. so making $[a b c]$ not a multiple of $[1 2 3]$ would work. – user 6663629 Apr 5 at 1:45