Proving a matrix statement by induction

Question

$RTP : (A_1A_2A_3...A_{n-1}A_n)^{-1}=A_n^{-1}A_{n-1}^{-1}...A_3^{-1}A_2^{-1}A_1^{-1}$

Where $A$ is a square, invertible matrix

Solution : Lets show the base case, $n=1$, is true $$(A_1)^{-1}=A_1^{-1}$$

Which is true ( I'm assuming the brackets don't add any meaning here ). Next, assume true when $n = k$ $$(A_1A_2A_3...A_{k-1}A_k)^{-1}=A_k^{-1}A_{k-1}^{-1}...A_3^{-1}A_2^{-1}A_1^{1}$$

Now, show the statement is true when $n=k+1$. Multiply both sides by $A_{k+1}^{-1}$ in such a way that associativity is preserved ( I'm not sure how to word this ).

$$(A_1A_2A_3...A_{k-1}A_k)^{-1}A_{k+1}^{-1}=A_{k+1}^{-1}A_k^{-1}A_{k-1}^{-1}...A_3^{-1}A_2^{-1}A_1^{1}$$

Since we are simply multiplying both sides of an equation with the same quantity ( the sides are equal based on the induction assumption ), we can say that the statement is true for the $n=k+1$ case, and therefore is true for all $n \in N$. Is this proof valid?

Answer 1

I don't think you've justified that $(A_1\ldots A_k)^{-1} A_{k+1}^{-1} = A_{k+1}^{-1}A_k^{-1}\ldots A_1^{-1}$.

First show that $(AB)^{-1} = B^{-1}A^{-1}$, this is the case $k = 1$. Since inverses are unique you need only show that this is an inverse. Then for $k > 1$ we have $((A_1\ldots A_k)A_{k+1})^{-1} = A_{k+1}^{-1}(A_1\ldots A_k)^{-1}$ by the case $k = 1$. But by the inductive hypothesis $(A_1\ldots A_k)^{-1} = A_k^{-1}\ldots A_1^{-1}$ so the proof is complete.

Answer 2

Rewriting the claim as $A_1\cdots A_nA_n^{-1}\cdots A_1^{-1}=I$ makes the inductive step easier, since$$A_1\cdots A_kA_{k+1}A_{k+1}^{-1}A_k^{-1}\cdots A_1^{-1}=A_1\cdots A_kIA_k^{-1}\cdots A_1^{-1}=A_1\cdots A_kA_k^{-1}\cdots A_1^{-1}.$$