Adjugate and Cramer's rule

Elementary matrices	Recall that the row operations (a)-(c) correspond to elementary matrices, say $E_a$, $E_b$, and $E_c$ (and recall how we constructed $E_a$, $E_b$, and $E_c$ in Lecture summary No.8). Since $\det I_n = 1$, we obtain $\det E_a = \det I = 1$, $\det E_b = -\det I = -1$, and $\det E_c = c\cdot\det I = c$. Then the properties of determinant over row operations can be summarized in $$\det E A = (\det E)(\det A) \mbox{ for every elementary matrix $E$. }$$ (6)
Invertible matrix SECTION 3.2 THEOREM 4; p.194 Examples 3-4	If $A$ is invertible, there is a series of elementary matrices $E_1, E_2, \ldots, E_p$ so that $E_p \cdots E_2 E_1 A = I_n$. By applying (6) repeatedly, we obtain $(\det E_p)\cdots (\det E_2)(\det E_1)(\det A) = 1$, and therefore ${\det A \neq 0}$. Similarly if $A$ is not invertible, we obtain ${\det A = 0}$ (why?). Together we conclude that $$\det A \neq 0$$    if and only if $A$ is invertible.
Multiplication THEOREM 6; p.196 Example 5	If $A$ is invertible, we can express $A = E_1^{-1} \cdots E_p^{-1}$ (why?). Since $E_i^{-1}$'s are also elementary matrices, by (6) we obtain $$\det AB = \det E_1^{-1} \cdots E_p^{-1} B = (\det E_1^{-1} \cdots E_p^{-1}) (\det B) = (\det A) (\det B)$$ Thus, we have shown that $\det AB = (\det A) (\det B)$. [If $A$ is not invertible, neither is $A B$ (why?); thus, ${\det AB = (\det A) (\det B) = 0}$.]
Adjugate SECTION 3.3 THEOREM 8; p.203 Example 3	Let $C_{ij}$ be the $(i,j)$-cofactor of $A$. Then we define the adjugate `` adj$A$'' of $A$ by    adj$$A = \begin{bmatrix} C_{11} & C_{21} & \cdots & C_{n1} \\ C_{12}... ... \vdots & \vdots & & \vdots \\ C_{1n} & C_{2n} & \cdots & C_{nn} \end{bmatrix}$$ By applying the Laplace expansions and the property (e) of determinants (that is, $\det [ \mathbf{a}_1 \ldots \mathbf{a}_k \ldots \mathbf{a}_{k} \ldots \mathbf{a}_n] = 0$), we can find that $$( A) \begin{bmatrix}a_{11} & a_{12} & \cdots & a_{1n} a_{21} & ... ...dots & 0 \vdots & \vdots & & \vdots 0 & 0 & \cdots & \det A \end{bmatrix}$$ (7) If $A$ is invertible (that is, $\det A \neq 0$), (7) immediately implies that $$A^{-1} = \frac{1}{\det A} A.$$ (8)
Cramer's rule THEOREM 7; p.201 Examples 1-2	By using (8) we can express the solution to the matrix equation  $A \mathbf{x} = \mathbf{b}$ by $$\mathbf{x} = \frac{1}{\det A} \begin{bmatrix} C_{11} & C_{21} & \... ...s \\ \det [ \mathbf{a}_1 \mathbf{a}_2 \ldots \mathbf{b} ] \end{bmatrix}$$ Let $A_i(\mathbf{b})$ be the matrix produced by replacing the $i$th column by $\mathbf{b}$. Then the Cramer's rule gives the solution  $\mathbf{x} = A^{-1} \mathbf{b}$ entry-wise in the following form: $$x_i = \frac{\det A_i(\mathbf{b})}{\det A}$$    for $i=1,\ldots,n$.
Exercise.	Prove (or explain) three why's above. Exercises 27-36, 37-38, 39-40 from Section 3.2, and Exercises 7-13, 15 from Section 3.3.