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

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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 is given by the formula (cA)-1 = (1/c)A-1. This conclusion is derived using the properties of scalar multiplication and matrix multiplication, confirming that the product of cA and its inverse yields the identity matrix I.

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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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