MHB Chris 's question at Yahoo Answers (Inverse of cA)

  • Thread starter Thread starter Fernando Revilla
  • Start date Start date
Click For Summary
If A is an invertible matrix and c is a non-zero scalar, then the matrix cA is also invertible. The inverse of cA can be expressed as (cA)^{-1} = (1/c)A^{-1}. This conclusion is derived using properties of scalar multiplication and matrix multiplication. The proof confirms that multiplying cA by its inverse yields the identity matrix. Thus, the relationship between cA and its inverse is established clearly.
Mathematics news on Phys.org
Hello Chris,

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 $(cA)^{-1}=\dfrac{1}{c}A^{-1}$.