If I have actually a solitary matrix A the is non-singular, how deserve to I prove the determinant of its train station = \$frac1det(A)\$?

I understand that \$(A)(A^-1) = I\$, however I to be not certain what to perform with the knowledge.

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\$egingroup\$ The equation around determinants stop for all \$A\$ and \$B\$, so that holds if \$B\$ is miscellaneous particular. \$endgroup\$
\$egingroup\$ since \$det(AB)=det Adet B\$, letting \$B=A^-1\$ offers \$det I=1=det Adet(A^-1)\$, therefore \$det(A^-1)=frac1det A\$. \$endgroup\$
first of all we recognize that

\$\$det(A cdot B)=det(A) imesdet(B)\$\$

also we know that

\$\$A imes A^-1=I\$\$

we understand that

\$\$det(A cdot A^-1)=det(I)\$\$

or

\$\$det(A) imesdet(A^-1)=det(I)\$\$

Can you proceed from this? asking yourself: what is \$det(I)\$?) take the example of the \$3 imes 3\$ identification matrix:

\$\$ i = eginpmatrix1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 endpmatrix\$\$

\$\$ det(I) = 1\$\$

So,

\$\$det(A) imesdet(A^-1)=1\$\$

or

\$\$oxeddet(A^-1)=frac1det(A)\$\$

But avoid

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exactly how to prove the determinant deserve to take any kind of real worth using only this definition of the determinant?

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