Show if A is invertible that A^-1 = 1/detA. I'm sort of stuck on this.
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Originally Posted by lord12 Show if A is invertible that A^-1 = 1/detA. I'm sort of stuck on this. We know that $\displaystyle \det(AB)=\det(A)\det(B)$ using this and the property that $\displaystyle AA^{-1}=I$ we get $\displaystyle \det(AA^{-1})=\det(I)$ $\displaystyle \det(A)\det(A^{-1})=1 \iff \det(A)=\frac{1}{\det(A^{-1})}$
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