Is there an update rule for $$\left(\tilde{X}^T\tilde{X}+\alpha\cdot I\right)^{-1}$$ with $\tilde{X}=[X\;\; a]$ as a function of $A\triangleq (X^TX)^{-1}$, $X$ and $a$? I know that when $\alpha=0$ we have $$\left(\tilde{X}^T\tilde{X}\right)^{-1}= \left[ \begin{array}\; A+\frac{AX^Taa^TXA^T}{a^Ta-a^TXAX^Ta}&\; \frac{-AX^Ta}{a^Ta-a^TXAX^Ta}\\ \frac{-a^TXA^T}{a^Ta-a^TXAX^Ta} &\; \frac{1}{a^Ta-a^TXAX^Ta}\end{array}\right].$$ Is there a variation of this for when $\alpha\neq 0$?

Is there an update rule for $$\left(\tilde{X}^T\tilde{X}+\alpha\cdot I\right)^{-1}$$ with $\tilde{X}=[X\;\; a]$ as a function of $A\triangleq (X^TX)^{-1}$, $X$ and $a$? I know that when $\alpha=0$ we have $$\left(\tilde{X}^T\tilde{X}\right)^{-1}= \left[ \begin{array}\; A+\frac{AX^Taa^TXA^T}{a^Ta-a^TXAX^Ta}&\; \frac{-AX^Ta}{a^Ta-a^TXAX^Ta}\\ \frac{-a^TXA^T}{a^Ta-a^TXAX^Ta} &\; \frac{1}{a^Ta-a^TXAX^Ta}\end{array}\right].$$ Is there a variation of this for when $\alpha\neq 0$?

Update - this can be easily done using the Schur complement, by writing

$$\tilde{X}^T\tilde{X}= \left[ \begin{array}\; X^TX+ \alpha\cdot I &\; X^Ta\\ a^TX &\; a^Ta+\alpha\end{array}\right],$$ and then using the formula for the inverse in the Wiki page.