While studying Linear Algebra from Hoffman Kunze, I am unable to understand few arguments given in lesson- Determinants. As my Institute is closed, so I have no help other than asking questions here :
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Questions : (1) How can formula (5-21) be summarized in (5-23) in matrix form .
I have question in paragraph before Theorem 4 : Authors write that if inverse of det(A) exists in K then $A^{-1}$ = $(det A)^{-1} $ adj(A) is inverse of A. To prove it I need to show that $A^{-1}$ A= I =$A A^{-1} $ but if I use A from (adj A) A = det(A) I ( see 5-23) then $(adj(A))^{-1}$ must exist. How to be sure that $(adj(A))^{-1}$ must exist?
Can anyone please help.
For your first question, since $C^T=\operatorname{adj} A$, from $(5-21)$ we obtain $$ \left((\operatorname{adj} A)A\right)_{jk}=(C^TA)_{jk}=\sum_{i=1}^n (C^T)_{ji}A_{ik}=\sum_{i=1}^n C_{ij}A_{ik}=\delta_{jk}\det A. $$ Therefore $(\operatorname{adj} A)A=(\det A)I$.
For your second question, you don't need to prove that $\operatorname{adj} A$ is invertible. As stated in the second paragraph on p.160, by definition, $A$ is called invertible if there exists a matrix $B$ such that $AB=BA=I$. And if this is the case, $B$ is called the inverse of $A$ and we denote $B$ by $A^{-1}$. Now, if $(\det A)^{-1}$ exists and if we put $B=(\det A)^{-1}\operatorname{adj}(A)$, then by $(5-23)$ and $(5-25)$, we have $BA=AB=I$. Hence $A$ is invertible and its inverse is $B$.