# Determining that the Inverse of a Matrix is Equal to

Here's the question in two parts.

Part One.

Suppose $$A$$ is an invertible matrix.
Explain why $$A^{T}A$$ is also invertible.

Part Two.

Then show that $$A^{-1} = (A^{T}A)^{-1}A^{T}.$$

I understand part one. Because of the fact that $$det(A^{T}A) = det(A^{T}) * det(A)$$,

and because the determinant of the a matrix's transpose is equal to the determinant of the original matrix, you can determine that $$A^{T}A$$ is transposable.

I don't understand where exactly to begin with the second part.

• Hint: $(AB)^{-1}=B^{-1}A^{-1}$ – Anurag A Oct 16 '18 at 17:30

Because if $$A$$ and $$B$$ are invertible $$n\times n$$ matrices then $$AB$$ is also invertible and $$(AB)^{-1}=B^{-1}A^{-1}$$. The proof is very simple, just multiply: $$(AB)(B^{-1}A^{-1})=A(BB^{-1})A^{-1}=AA^{-1}=I$$.
So now you have $$(A^TA)^{-1}A^T=A^{-1}(A^T)^{-1}A^T=A^{-1}$$.
Hint: Show that $$A^{-1} A = I$$ where $$A^{-1}$$ is as given in the question.
$$A^{-1} A = ((A^\top A)^{-1} A^\top) A = (A^\top A)^{-1} (A^\top A) = I$$.
The basic fact (cf. If $$AB=I$$ then $$BA=I$$) is that if $$X$$ is a square matrix whose entries are taken from a field, then $$X^{-1}=Y$$ iff $$YX=I$$. That is, $$X$$ has a two-sided inverse $$Y$$ if and only if $$Y$$ is a left-inverse of $$X$$.
So, to show that $$\underbrace{A^{-1}}_{X^{-1}} = \underbrace{(A^TA)^{-1}A^T}_Y$$, it suffices to prove that $$\underbrace{(A^TA)^{-1}A^T}_Y\underbrace{A}_X=I$$.