I"m pretty certain this would need that $B$ likewise be invertible for it to it is in true.

You are watching: If a and b are invertible then ab is invertible

Is over there a quick means to disprove the if $A$ is invertible climate $AB$ is invertible?

The proof the if $A$ and also $B$ room invertible, climate $AB$ is invertible have the right to be done an ext elegantly if you recognize these two results:

$(1)$. $detAB = (det (A))*(det(B)).$$(2)$. A procession $B$ is invertible if and also only if $det(B) eq 0$.

**Proof**: expect that both $A$ and $B$ room invertible. Climate $det(A)
eq 0$ and also $det(B)
eq 0$. Now by $(1)$, $det(AB)
eq 0$, for this reason by $(2)$, $AB$ is invertible.

I don"t quite acquire your question since the insurance claim is false. What if $B$ is the zero matrix?

Implication of $(AB)^-1 implies det(A) eq 0$ is trivially true$$(AB)^-1 = C = B^-1A^-1 = C implies A^-1 = BC = B(AB)^-1$$

But the allude is, the converse walk not apply (if $A$ is invertible then $AB$ is invertible).

Consider the square matrices as direct maps $digitalrecordersreview.orgbbR^n o digitalrecordersreview.orgbbR^n$. These three conditions are equivalent: $A$ is invertible, $operatornameim A = digitalrecordersreview.orgbbR^n$, $operatornameker A = 0$.

Since $operatornameim abdominal subseteq operatornameim A$, we have actually $$ operatornameim abdominal = digitalrecordersreview.orgbbR^n Rightarrow operatornameim A = digitalrecordersreview.orgbbR^n. $$Due to the equivalent conditions mentioned above: $$ abdominal ext is invertible Rightarrow A ext is invertible. $$

You can use the 3rd condition to display that $B$ is additionally invertible.

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