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

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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.
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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 $(cA)^{-1}=\dfrac{1}{c}A^{-1}$.
 
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