If A = BC and B is invertible, then how does reducing "B to I" also reduce "A to C"?

Question

If $A = B*C$, where $B$ is an inverse, use row-ops to reduces "$B$ to $I$" also shows that it will reduce "$A$ .. $C$".

Big-Hint: Represent the row operations by a sequence of elementary matrices.

Answer 1

Row operations on a matrix can be thought of as left-multiplying by elementary matrices.

For example, if $B = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, then

$$\begin{pmatrix} 1 & 0 \\ -\frac{c}{a} & 1 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a & b \\ 0 & d-\frac{bc}{a} \end{pmatrix}$$ is a row operation -- specifically, the second row less $\frac{c}{a}$ times the first row.

Let $\{P_1,\ldots,P_n\}$ be a sequence of matrices representing row operations such that $P_n P_{n-1} \cdots P_1 B = I$.

Then, $$\begin{align*} P_n P_{n-1}\cdots P_1 A &= P_n P_{n-1}\cdots P_1 BC \\ &= IC\\ &= C \end{align*}$$