Find cofactors of adj(A) in terms of |A|

Question

Let $A= \begin{pmatrix}a_{11} & a_{12} & a_{13} & a_{14}\\ a_{21} & a_{22} &a_{23} & a_{24} \\a_{31} & a_{32} & a_{33} & a_{34}\\a_{41}&a_{42}&a_{43}&a_{44} \end{pmatrix}$

Let $A'$ be a matrix formed by using corresponding cofactors of $a_{ii}$'s

Let $A'= \begin{pmatrix}A_{11} & A_{12} & A_{13} & A_{14}\\ A_{21} & A_{22} &A_{23} & A_{24} \\A_{31} & A_{32} & A_{33} & A_{34}\\A_{41}&A_{42}&A_{43}&A_{44} \end{pmatrix}$

where $A_{ii}$ is cofactor of $a_{ii}$.

Prove that cofactor of $A_{11}=a_{11}\left(det(A)\right)^2$.

Answer 1

The adjugate matrix is the transpose of $A'$. It satisfies the property that $$ \DeclareMathOperator{\adj}{adj} \adj(A)A=\det(A)I $$ where $I$ is the identity matrix of the appropriate size, say $n$. Assuming $\det(A)\ne0$, we conclude that $$ \det(\adj(A))\det(A)=(\det(A))^n $$ Therefore we have $$ \adj(\adj(A))\adj(A)=(\det(A))^{n-1}I $$ whence your relation.

For the case $\det(A)=0$, it takes a bit of care, but it's valid as well.