Invertible matrix problem

Question

Given three $n \times n$ matrices $A$, $B$ and $C$. Prove that if $AB+AC$ is an invertible matrix then $A$ is also an invertible matrix.

How can this be possible? I found that $B=A^{-1}-C$ and when I plug it in it says $I=I$. What would that mean?

Answer 1

Remember, matrix addition commutes: $$AB+AC=A(B+C)$$ Here is one of the inversion identities: $$(AB)^{-1}=B^{-1}A^{-1}$$ Therefore, we have, if the premise is true: $$[A(B+C)]^{-1}=(B+C)^{-1}A^{-1}$$ Thus, for the premise to be true, $A^{-1}$ has to exist. It has to exist because it has been factored out as an explicit term.

If it doesn't exist, then the final equality is incorrect, which would then cascade back upwards to the premise. Proof by contradiction:

Assume: $A^{-1}$ does not exist. Therefore, for some invertible matrix $U$, the following (a matrix inversion identity) does not hold: $$U^{-1}A^{-1}=(AU)^{-1}$$ Let: $U$ be the sum of two other matrices: $B$, $C$. Then, the last component is equivalent to the following: $$U=B+C$$ Thus, the following does not hold (substitution): $$(AU)^{-1}=(A(B+C))^{-1}$$ Basically, we have "proven" that the given (or assumed) premise is false. Our logical argument lead to a contradiction of the premise, therefore, the assumption at the beginning about $A$ must be wrong.

Answer 2

We can prove the contrapositive by using determinants:

If $A$ is not invertible, then $AB + AC$ is not invertible.

Suppose that $A$ is not invertible. Then $\det A = 0$. But then since: $$ \det(AB + AC) = \det(A(B + C)) = (\det A)\cdot(\det(B+C)) = 0 \cdot \det(B + C) = 0 $$ it follows that $AB + AC$ is not invertible, as desired.