If A is a non singular matrix show that $(xA)^{-1} = \frac{1}{x}A^{-1}$

Question

If A is a non singular matrix show that $(xA)^{-1} = \frac{1}{x}A^{-1}$

I was trying:

because A is not singular then there exist $A^{-1}$ such that $A^{-1}A = I$ let A be a matrix such that $A=[a_{ij}]_{m\times n}$ and $(xA) = [xa_{ij}]_{m\times n}$ then:

$(xA)^{-1} = ([xa_{ij}])^{-1}_{m \times n} = \frac{1}{x}A^{-1}$

if this proof is correct can you help me improve it else why is this proof wrong?

Answer 1

Since you have an explicit expression for the inverse, you just need to check that it is actually the inverse.

To do that you just have to show that $(xA)(\frac{1}{x}A^{-1}) = I$, which follows directly from distributivity of the product of a matrix and a scalar.