Exercise 4.3.25

Question

Let $c_{jk}$ denote the cofactor of the row $j$, column $k$ entry of the matrix $A \in M_{n 	imes n}(F)$.\par 

(a) Prove that if $B$ is the matrix obtained from $A$ by replacing column $k$ by $e_j$, then $\det(B)= c_{jk}$.\par 

(b) Show that for $1 \ge j \ge n$, we have 
$$A \pmatrix {c_{j1} \ {c_{j2}} \ . \ . \ {c_{jn}}} = \det(A) \cdot e_j.$$
Hint: Apply Cramer's rule to $Ax = e_j$.

Jephian C.-H. Lin · Accepted Answer

\begin{enumerate}
\item Just expand along the $k$-th column.
\item It's better not to use a great theorem, such as Cramer's rule, to kill a small problem. We check each entry one by one. First, we have that 
$$\sum_{k=1}^n{A_{jk}c_{jk}}=\det(A)$$
 and so the $j$-th entry of the left term is $\det(A)$. Second, for $i\neq j$, we construct a matrix $B_i$ by replacing the $j$-th row of $A$ by the $i$-th row of $A$. Since $B_i$ has two identity rows, we have that $\det(B_i)=0$ for all $i\neq j$. Now we can calculate that 
$$\sum_{k=1}^n{A_{ik}c_{jk}}=\sum_{k=1}^n{B_{jk}c_{jk}}=\det(B)=0,$$
 for all $i\neq j$. So we get the desired conclusion.
\item Actually this matrix $C$ is the classical adjoint of matrix $A$ defined after this exercise. And this question is an instant result since 
$$AC=A\begin{pmatrix}c_{11}&c_{21}&\cdots &c_{n1}\c_{12}&c_{22}&\cdots &c_{n2}\\vdots &\vdots & &\vdots \c_{1n}&c_{2n}&\cdots &c_{nn}\end{pmatrix}$$
$$=A\begin{pmatrix}\det(A)&0&\cdots &0\0&\det(A)&\cdots &0\\vdots &\vdots &\ddots &\vdots \0&0&\cdots &\det(A)\end{pmatrix}$$
 by the previous exercise.
\item If $\det(A)\neq 0$, then we know $A$ is invertible. So we have 
$$A^{-1}=A^{-1}A[\det(A)]^{-1}C=[\det(A)]^{-1}C.$$
\end{enumerate}

Exercise 4.3.25

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Exercise 4.3.25

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