How would I prove that given a square matrix $A$ and non-zero scalar $c$ that $$(cA)^{-1}=c^{-1}A^{-1}$$
$\begingroup$
$\endgroup$
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
Assuming $A$ is invertible, so that $A^{-1}$ actually exists, you can simply check this directly. Is it true that $(cA)(c^{-1}A^{-1})=(c^{-1}A^{-1})(cA)=I$? Why?
$\begingroup$
$\endgroup$
we want to prove $cA$ has inverse matrix $c^{-1}A^{-1}$. suppose $cA$ has inverse matrix $B$, that is we want to show $B = c^{-1}A^{-1}$. Here is the proof.
Since $B$ is the inverse matrix, then $(cA)B = I$, $c(AB) = I$, $AB = \frac{1}{c}I$, finally we multiply both sides with $A^{-1}$ on the left, $A^{-1}AB = A^{-1}\frac{1}{c}I$, we get $IB = \frac{1}{c}A^{-1}I = \frac{1}{c}A^{-1}$
$\begingroup$
$\endgroup$
Multiplying a matrix $A$by a constant $c$ is the same as scaling every row of a matrix by $c$. You can then consider $c$ to be the $ n \times n $ matrix $c Id$, (so that every entry in the diagonal equals $c$ and $0$ everywhere else, and where $A$ is also $n \times n$) and then $cId$ is invertible for all $c \neq 0$, and then apply the result that the inverse of $AB$ is $B^{-1}A^{-1}$.
