Prove that $A^{-1}=\frac{adj(A)}{|A|}$

Question

Prove that $$A^{-1}=\frac{adj(A)}{|A|}$$

Can the above condition be proved without actually putting values and solving it. I consulted many books. Only way they proved it is by solving it by putting values and then proving that $A^{-1}A=I_n$.

Answer 1

As suggested by @Brian Moehring, first show that $$(\text{adj}A)\cdot A=I_n\cdot |A|=A\cdot (\text{adj}A)$$ You can check the proof here.

Now, we've $$A\left(\frac{1}{|A|}\text{adj}A\right)=I_n=\left(\frac{1}{|A|}\text{adj}A\right)A$$

By the definition of inverse of a matrix,

A square matrix $A$ of order $n$ is invertible if there exists a square matrix $B$ if the same order such that $$AB=I_n=BA$$ In such case, we say that the inverse of $A$ is $B$ and we write, $A^{-1}=B$.

We get $$A^{-1}=\frac{1}{|A|}\text{adj}A$$