12.20

1. Calculate the determinant of the following matrix using expansion by rows or columns: $A=\begin{pmatrix}1 & 2 & 3\\ 7 & 8 & 6\\ 0 & 2 & 0\end{pmatrix}$
  
  We consider expansion by the column $j=2$: $\det(A)=\sum_{i=1}^3(-1)^{i+2}\det(A(i|2))a_{i2}$
  
  Note that $\det(A(1|2)) = \det \begin{pmatrix} 7 & 6 \\ 0 & 0 \end{pmatrix} = 0$, $\det(A(2|2)) = \det \begin{pmatrix} 1 & 3 \\ 0 & 0 \end{pmatrix} = 0$, $\det(A(3|2)) = \det \begin{pmatrix} 1 & 3 \\ 7 & 6 \end{pmatrix} = -15$
  
  Therefore, $\det(A)=(-1)^{3+2}\det(A(3|2))a_{32}=30$ 2. $B=\begin{pmatrix}5 & 0 & 1\\ 2 & 0 & 3\\ 5 & 6 & 1\end{pmatrix}$
  
  We expand by row $i=3$:
  
  $\det(B)=\sum_{j=1}^3(-1)^{3+j}\det(A(3|j))a_{3j}=(-1)^{3+2}\det(A(3|2))a_{32}=-13\cdot6=-78$
Given $\lambda_1-\lambda_n \in \mathbb{R}$, we define the Vandermonde matrix by $V_n = \begin{pmatrix} 1 & 1 & \cdots & 1 \\ \lambda_1 & \lambda_2 & \cdots & \lambda_n \\ \lambda_1^2 & \lambda_2^2 & \cdots & \lambda_n^2 \\ \vdots & \vdots & \ddots & \vdots \\ \lambda_1^{n-1} & \lambda_2^{n-1} & \cdots & \lambda_n^{n-1} \end{pmatrix} \in M_n(\mathbb{R})$
1. Prove that $\det(V_n) = \prod_{1\leq i<j\leq n} (\lambda_j-\lambda_i)$ (by induction).
  
  $n=2$: $\det(V_2) = \det\begin{pmatrix}1 & 1\\ \lambda_1 & \lambda_2\end{pmatrix} = \lambda_2-\lambda_1 = \prod_{1\leq i<j\leq 2} (\lambda_j-\lambda_i)$ ✓
  
  Suppose $\det(V_n) = \prod_{1\leq i<j\leq n} (\lambda_j-\lambda_i)$, now let's see for $n+1$
  
  $\text{det}(V_{n+1})=\text{det}\left(\begin{array}{cccc}1 & 1 & \cdots & 1\\ \lambda_1 & \lambda_2 & \cdots & \lambda_{n+1}\\ \lambda_1^2 & \lambda_2^2 & \cdots & \lambda_{n+1}^2\\ \vdots & \vdots & \ddots & \vdots\\ \lambda_1^{n} & \lambda_2^{n} & \cdots & \lambda_{n+1}^{n}\end{array}\right)\overset{R_2-\lambda_1R_1,R_3-\lambda_1R_2}{=}\text{det}\left(\begin{array}{cccc}1 & 1 & \cdots & 1\\ 0 & \lambda_2-\lambda_1 & \cdots & \lambda_{n+1}-\lambda_1\\ 0 & \lambda_2(\lambda_2-\lambda_1) & \cdots & \lambda_{n+1}(\lambda_{n+1}-\lambda_1)\\ \vdots & \vdots & \ddots & \vdots\\ 0 & \lambda_2^{n-1}\left(\lambda_2-\lambda_1\right) & \cdots & \lambda_{n+1}^{n-1}(\lambda_{n+1}-\lambda_2)\end{array}\right)$
  
  expansion by 1st row $=(-1)^{1+1}\cdot1\cdot\text{det}\left(\begin{array}{cccc}\lambda_2-\lambda_1 & \cdots & \lambda_{n+1}-\lambda_1\\ \lambda_2(\lambda_2-\lambda_1) & \cdots & \lambda_{n+1}(\lambda_{n+1}-\lambda_1)\\ \vdots & \ddots & \vdots\\ \lambda_2^{n-1}(\lambda_2-\lambda_1) & \cdots & \lambda_{n+1}^{n-1}(\lambda_{n+1}-\lambda_1)\end{array}\right)$
  
  column operations $=\prod_{j=2}^{n+1}\left(\lambda_{j}-\lambda_1\right)\cdot\text{det}\left(\begin{array}{cccc}1 & \cdots & 1\\ \lambda_2 & \cdots & \lambda_{n+1}\\ \vdots & \ddots & \vdots\\ \lambda_2^{n-1} & \cdots & \lambda_{n+1}^{n-1}\end{array}\right)=\prod_{j=2}^{n+1}\left(\lambda_{j}-\lambda_1\right)\cdot\text{det}(V_{n})$ with scalars $\lambda_2,\ldots,\lambda_{n+1}$
  
  $\text{IH}=\prod_{j=2}^{n+1}\left(\lambda_{j}-\lambda_1\right)\cdot\prod_{2\leq i\leq j\leq n+1}\left(\lambda_{j}-\lambda_{i}\right)=\prod_{1\leq i\leq j\leq n+1}\left(\lambda_{j}-\lambda_{i}\right)$ 2. Let $\lambda \in \mathbb{R} \setminus \{0\}$ and consider the sequence given by $\text{exp}: \mathbb{N} \rightarrow \mathbb{R}, \quad \text{exp}(\lambda)_n = \lambda^{n-1}.$
  
  Prove that $S = \{ \text{exp}(\lambda_1), \text{exp}(\lambda_2), \ldots, \text{exp}(\lambda_n) \} \subset \mathbb{R}^{\mathbb{N}} \text{ is L.I. }$iff $\lambda_i \neq \lambda_j \; \forall i \neq j.$
  
  We have that $S=\{(1,\lambda_1,\lambda_1^2,\ldots,\lambda_1^{n-1},\ldots),(1,\lambda_2,\lambda_2^2,\ldots,\lambda_2^{n-1},\ldots),\ldots,(1,\lambda_{n},\lambda_{n}^2,\ldots,\lambda_{n}^{n-1},\ldots)\}.$
  
  Therefore, $S$ is L.I.
  
  \[ \;\iff\;\begin{pmatrix}1 & \lambda_1 & \lambda_1^2 & \cdots & \lambda_1^{n-1}\\ 1 & \lambda_2 & \lambda_2^2 & \cdots & \lambda_2^{n-1}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & \lambda_{n} & \lambda_{n}^2 & \cdots & \lambda_{n}^{n-1}\end{pmatrix}=V_{n}\text{ is invertible for all }n\in\mathbb{N} \]
  
  \[ \;\iff\;\text{det}(V_{n})\neq0\;\forall n\in\mathbb{N}\;\iff\;\prod_{1\leq i<j\leq n}(\lambda_{j}-\lambda_{i})\neq0\;\iff\;\lambda_{j}\neq\lambda_{i}\;\forall i\neq j \]
Find the inverse of $A=\begin{pmatrix}1 & 2 & 1\\ 0 & 2 & 1\\ 2 & 3 & 1\end{pmatrix}$using the adjoint.

$\det(A) = 2 \cdot 0 + 4 \cdot 1 - 4 \cdot 3 - 0 = -1.$

$\text{adj}(A):$ we need to find the cofactors of $A$

$c_{ij} = (-1)^{i+j} \det(A(i,j)) \rightarrow C = \begin{pmatrix} c_{11} & c_{12} & c_{13} \\ c_{21} & c_{22} & c_{23} \\ c_{31} & c_{32} & c_{33} \end{pmatrix}$

$c_{11}=(-1)^2\det\begin{pmatrix}2 & 1\\ 3 & 1\end{pmatrix}=-1$

$c_{21}=(-1)^3\det\begin{pmatrix}2 & 1\\ 3 & 1\end{pmatrix}=1$

$c_{31}=(-1)^4\det\begin{pmatrix}2 & 1\\ 2 & 1\end{pmatrix}=0$

and so on... Then $C = \begin{pmatrix} -1 & 2 & -4 \\ 1 & -1 & 1 \\ 0 & -1 & 2 \end{pmatrix} \text{ (matrix of cofactors of } A\text{)}$

Therefore, $A^{-1}=\frac{1}{\det(A)}\text{adj}(A)=\frac{1}{-1}C^{T}=-\begin{pmatrix}-1 & 1 & 0\\ 2 & -1 & -1\\ -4 & 1 & 2\end{pmatrix}=\begin{pmatrix}1 & -1 & 0\\ -2 & 1 & 1\\ 4 & -1 & -2\end{pmatrix}$

$$

$$

$$

$$

$$

$$

‍