- Thread starter
- #1

- Jan 29, 2012

- 661

Show that if A is an invertible matrix, then (cA) is invertible and (cA)-1=1/cA-1? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

- Thread starter Fernando Revilla
- Start date

- Thread starter
- #1

- Jan 29, 2012

- 661

Show that if A is an invertible matrix, then (cA) is invertible and (cA)-1=1/cA-1? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

- Thread starter
- #2

- Jan 29, 2012

- 661

Suppose $c\neq 0$. Then, using the well known properties $(\lambda M) N=M(\lambda N)=\lambda (MN)$ and $\lambda(\mu M)=(\lambda\mu)M$ : $$(cA)\left(\frac{1}{c}A^{-1}\right)=\frac{1}{c}\left((cA)A^{-1}\right)=\frac{1}{c}\left(c\;\left(AA^{-1}\right)\right)=\left(\frac{1}{c}\cdot c\right)I=1I=I$$ So, $cA$ is invertivle and $(cA)^{-1}=\dfrac{1}{c}A^{-1}$.