Vtyjkyuk

Sherman-Morrison formula (e.g., cf. in the matrix cookbook, sec. 3.2.4)

$$(A + bc^T)^-1 = A^-1 - fracA^-1 bc^T A^-11 + c^T A^-1 b .$$

Question/Clarification: If
$$(A + alpha bc^T)^-1 = A^-1 - alpha fracA^-1 bc^T A^-11 + alpha^2 c^T A^-1 b$$

Is it true? If yes, then how? If incorrect, please suggest the correct one.

If $alpha in mathbbR$, then

view the problem as

$$(A+(alpha b)c^T)^-1$$

that is whenever we see a $b$, we replace it by $alpha b$, then we obtain

$$(A + (alpha b)c^T)^-1 = A^-1 - fracA^-1 (alpha b)c^T A^-11 + alpha c^T A^-1 b =A^-1 - alpha fracA^-1 bc^T A^-11 + alpha cdot c^T A^-1 b$$

where all the inverse terms are assumed to exist.

Edit:

A numerical example:

octave:3> A = eye(2)
A =

Diagonal Matrix

 1 0
 0 1

octave:4> b = [1 ; 0]
b =

 1
 0

octave:5> c = b
c =

 1
 0

octave:6> alpha = 2
alpha = 2
octave:7> inv(A+alpha*b*c')-inv(A)+alpha * inv(A)*b*c' * inv(A)/(1+alpha * c' * inv(A)*b)
ans =

 -1.1102e-16 0.0000e+00
 0.0000e+00 0.0000e+00

octave:8> inv(A+alpha*b*c')-inv(A)+alpha * inv(A)*b*c' * inv(A)/(1+alpha*alpha * c' * inv(A)*b)
ans =

 -0.26667 0.00000
 0.00000 0.00000

You can use the Alternative verification. Moreover, consider $b$ and $c$ as $1times 1$ matrices (i.e. single numbers). Then:
$$(A + bc^T)^-1 = A^-1 - fracA^-1 bc^T A^-11 + c^T A^-1 b
Rightarrow
(1+bc)^-1=1-fracbc1+cb;$$
Similarly:
$$(A + alpha bc^T)^-1 = A^-1 - alpha fracA^-1 bc^T A^-11 + colorredalpha c^T A^-1 b Rightarrow \
(1 + alpha bc)^-1 = 1 - alpha fracbc1 + alpha cb$$
Or you can denote: $alpha b=d$, where $alpha$ is a scalar, $b,d$ are the vectors.