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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