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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|线性代数代写linear algebra代考|DETERMINANT

The determinant is a function whose inputs are square matrices and whose outputs are scalars. We will eventually give a general definition of the determinant, but first we introduce it in some special cases.

Let us first establish our notation. For a square matrix $A$, the determinant of $A$ will be denoted by $|A|$ or $\operatorname{det}(A)$ (keep in mind that $|A|$ is a scalar).

• For a $1 \times 1$ matrix $A=[a],|A|=a$.
Example 2.37 If $A=[-6]$, then $|A|=-6$.
• For a $2 \times 2$ matrix $A=\left[\begin{array}{ll}a & b \ c & d\end{array}\right],|A|=\left|\begin{array}{ll}a & b \ c & d\end{array}\right|=a d-b c$. Example 2.38 $\left|\begin{array}{rr}1 & -3 \ 2 & 4\end{array}\right|=(1)(4)-(-3)(2)=10$
• For a $3 \times 3 \operatorname{matrix} A=\left[\begin{array}{lll}a & b & c \ d & e & f \ g & h & i\end{array}\right]$,
$$|A|-\left|\begin{array}{lll} a & b & c \ d & e & f \ g & h & i \end{array}\right|-(a)\left|\begin{array}{cc} e & f \ h & i \end{array}\right|-(b)\left|\begin{array}{ll} d & f \ g & i \end{array}\right|+(c)\left|\begin{array}{ll} d & e \ g & h \end{array}\right|$$
Notice that the $2 \times 2$ determinants in the definition of a $3 \times 3$ determinant are easy to remember. For instance, the $2 \times 2$ determinant next to $a$ is obtained by crossing out in the $3 \times 3$ determinant the row and column in which $a$ appears. For $b$ cross out the row and column in which $b$ appears, and the same for $c$.

## 数学代写|线性代数代写linear algebra代考|APPLICATIONS OF THE DETERMINANT

With the introduction of the determinant in the previous section we are now in a position to give alternative algorithms for solving systems of linear equations (assuming that there is a unique solution) and computing the inverse of a matrix (assuming that it exists). We will also introduce the cross product for elements of $\mathbb{R}^n$, since it can be defined and remembered using a mnemonic relating to the determinant. The following result, called Cramer’s Rule, achieves the first goal of this section. It is a method for finding the solution to a linear system having exactly one solution and is defined completely in terms of determinants.

Theorem $2.23$ Let $A X=B$ be a linear system in unknowns $x_1, x_2, \ldots, x_n$ with $A$ invertible. Set $A=\left[\begin{array}{llllll}c_1 & c_2 & \ldots & c_i & \ldots & c_n\end{array}\right]$, a column representation of $A$. Let $A_i=$ $\left[\begin{array}{llllll}c_1 & c_2 & \ldots & B & \ldots & c_n\end{array}\right]$, where the $i$ th column of $A$ has been replaced by $B$. Then the unique solution $\left[\begin{array}{llll}a_1 & a_2 & \ldots & a_n\end{array}\right]$ to $A X=B$ has the form $a_i=\left|A_i\right| /|A|$ for $i=$ $1,2, \ldots, n$.

Proof $2.27 B y$ Theorem 2.21, we know that $A X=B$ has a unique solution, say $X_0=\left[a_1, \ldots, a_n\right]\left(\right.$ so $\left.A X_0=B\right)$. Define
$$C_i=\left[e_1, \ldots, e_{i-1}, X_0, e_{i+1}, \ldots, e_n\right]=\left[\begin{array}{cccccc} 1 & 0 & & a_1 & & 0 \ 0 & 1 & & a_2 & & 0 \ 0 & 0 & \cdots & a_3 & \ldots & 0 \ \vdots & \vdots & & \vdots & & \vdots \ 0 & 0 & & a_n & & 1 \end{array}\right] .$$
Observe that $\left|C_i\right|=a_i$ (expand on row i). In the calculation which follows we appeal to Lemma 1.1.
Notice that
$$A C_i=\left[A e_1, \ldots, A e_{i-1}, A X_0, A e_{i+1}, \ldots, A e_n\right]$$
$$=\left[c_1, \ldots, c_{i-1}, B, c_{i+1}, \ldots, c_n\right]=A_i .$$
Therefore, $\left|A_i\right|=\left|A C_i\right|=|A|\left|C_i\right|=|A| a_i$ and solving for a $a_i$ yields
$$a_i=\frac{\left|A_i\right|}{|A|}$$

# 线性代数代考

## 数学代写|线性代数代写线性代数代考|行列式

• 对于$1 \times 1$矩阵$A=[a],|A|=a$ .
例2.37如果$A=[-6]$，则$|A|=-6$ .
• 对于$2 \times 2$矩阵$A=\left[\begin{array}{ll}a & b \ c & d\end{array}\right],|A|=\left|\begin{array}{ll}a & b \ c & d\end{array}\right|=a d-b c$ .
• 例2.38 $\left|\begin{array}{rr}1 & -3 \ 2 & 4\end{array}\right|=(1)(4)-(-3)(2)=10$
• 对于$3 \times 3 \operatorname{matrix} A=\left[\begin{array}{lll}a & b & c \ d & e & f \ g & h & i\end{array}\right]$，
$$|A|-\left|\begin{array}{lll} a & b & c \ d & e & f \ g & h & i \end{array}\right|-(a)\left|\begin{array}{cc} e & f \ h & i \end{array}\right|-(b)\left|\begin{array}{ll} d & f \ g & i \end{array}\right|+(c)\left|\begin{array}{ll} d & e \ g & h \end{array}\right|$$
注意$3 \times 3$行列式定义中的$2 \times 2$行列式很容易记住。例如，$a$旁边的$2 \times 2$行列式是通过在$3 \times 3$行列式中划掉$a$所在的行和列获得的。对于$b$，划掉$b$所在的行和列，对于$c$也一样。

## 数学代写|线性代数代写linear algebra代考|APPLICATIONS OF THE行列式

.

$$C_i=\left[e_1, \ldots, e_{i-1}, X_0, e_{i+1}, \ldots, e_n\right]=\left[\begin{array}{cccccc} 1 & 0 & & a_1 & & 0 \ 0 & 1 & & a_2 & & 0 \ 0 & 0 & \cdots & a_3 & \ldots & 0 \ \vdots & \vdots & & \vdots & & \vdots \ 0 & 0 & & a_n & & 1 \end{array}\right] .$$

$$A C_i=\left[A e_1, \ldots, A e_{i-1}, A X_0, A e_{i+1}, \ldots, A e_n\right]$$
$$=\left[c_1, \ldots, c_{i-1}, B, c_{i+1}, \ldots, c_n\right]=A_i .$$

$$a_i=\frac{\left|A_i\right|}{|A|}$$

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assignmentutor™您的专属作业导师
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