How to adopt the Woodbury matrix identity to this matrix formula

Question

The Woodbury matrix identity is defined as follows: $$ {(A+UCV)}^{-1}=A^{-1}-A^{-1}U{(C^{-1}+VA^{-1}U)}^{-1}VA^{-1} $$

I want to use the Woodbury matrix identity theorem to change the following matrix formula $$ W={(XX^T+\lambda G)}^{-1}XY $$ into the following form $$ W=G^{-1} X {(X^TG^{-1}X+\lambda I)}^{-1}Y $$ The dimensions are as follows: $$ X\in R^{p\times n}\\ G\in R^{p\times p}\\ Y\in R^{n\times c} $$ Could anyone help give some hints?

UPDATE:

From the two formulas about $W$, we could get the following equations, thus the two $W$s should be equal:

Answer 1

I've found the solution as follows:

Answer 2

I'm not sure the different forms of W, as stated, are equivalent. For one thing, they do not appear to be equivalent when the matrices involved are replaced by scalars. To illustrate, let $X=a$, $G=b$ and $Y=c$. Then, $$ \begin{eqnarray*} W{}={}{(XX^T+\lambda G)}^{-1}XY &{}\implies{}&W{}={}\frac{ac}{a^2{}+{}\lambda b}\,,\newline \end{eqnarray*} $$ while $$ \begin{eqnarray*} W{}={}G^{-1} X {(X^TG^{-1}+\lambda I)}^{-1}Y &{}\implies{}&W{}={}\frac{ac}{a{}+{}\lambda b}\,.\newline \end{eqnarray*} $$

Furthermore, direct manipulation of the first posted equation with $W$ gives $$ W{}={}G^{-1} X {(X^TG^{-1}X+\lambda I)}^{-1}Y\,, $$ which is different from the second $W$ equation posted but, now, seems consistent (assuming, in addition, that $X$ is invertible).