# 线性代数代考_linear algebra代考_Determinant and Inverse

## 线性代数代考_linear algebra代考_DEFINITION OF DETERMINANT

The concepts of linear algebra introduced so far are not sufficient to give a direct definition of the determinant. Hence, we shall introduce the determinant with an indirect definition and then we will compute it in some special cases, which will prove to be the most significant for us, and finally we will arrive at an algorithmic method to compute it in general.

Let $A$ be a square matrix, i.e. an $n \times n$ matrix. We want to associate to it a real number, called determinant of $A$, which is calculated starting from the elements of the matrix $A$. The definition we give is apparently not a constructive one, however, we will see that, starting from simple rules, we can calculate the determinant of a matrix.

Before we begin, it is necessary to introduce the definition of the identity matrix.
Definition 7.1.1 The identity matrix or identity matrix of order $n$, is the $n \times n$ matrix having all the elements of the main diagonal equal to the number 1 , while the remaining elements are equal to 0 . Usually it is indicated with $I_r$, or with $I$, if there are no ambiguities.
For example, the identity matrix of order 3 is:
$$I=\left(\begin{array}{lll} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array}\right) .$$
Definition 7.1.2 The determinant of a square matrix $A$ of order $n$ is a real number, denoted by $\operatorname{det}(A)$, with the following properties:

1. If the $j$-th row of $A$ is the sum of two elements $\mathbf{u}$ and $\mathbf{v}$ of $\mathbb{R}^n$, then the determinant of $A$ is the sum of the determinants of the two matrices obtained by replacing the $j$-th row of $A$ with $\mathbf{u}$ and $\mathbf{v}$, respectively.

## 线性代数代考_linear algebra代考_CALCULATING THE DETERMINANT

For $2 \times 2$ and $3 \times 3$ matrices, there are simple formulas for the calculation of the determinant.
We begin by examining the case of $2 \times 2$ matrices. Let $A$ be the matrix:
$$A=\left(\begin{array}{ll} a_{11} & a_{12} \ u_{21} & u_{22} \end{array}\right) .$$
We proceed with the algorithm we explained above, considering two cases.
Case 1 . Suppose first that $a_{11} \neq 0$. In this case, we make the following elementary operation:

• $2 \mathrm{nd}$ row $\rightarrow 2 \mathrm{nd}$ row $-\frac{a_{21}}{a_{11}} \cdot 1$ st row.
In this way, by Proposition 7.1.4 (b), the determinant does not change and we obtain the triangular matrix:
$$\left(\begin{array}{cc} a_{11} & a_{12} \ 0 & a_{22}-\frac{a_{21}}{a_{11}} \cdot a_{12} \end{array}\right) .$$
Now we simply take the product of the diagonal coefficients and we have that:
$$\operatorname{det}(A)=a_{11} \cdot\left(a_{22}-\frac{a_{21}}{a_{11}} \cdot a_{12}\right)=a_{11} a_{22}-a_{12} a_{21} .$$
Case 2. Suppose $a_{11}=0$. We exchange the first and the second row; the determinant changes sign, and we get:
$$\left(\begin{array}{cc} a_{21} & a_{22} \ 0 & a_{12} \end{array}\right) .$$

# 线性代数代考

## 线性代数代考_linear algebra代考_DEFINITION OF DETERMINANT

$$I=\left(\begin{array}{lllllllll} 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \end{array}\right) .$$

1. 如果 $j$-第行 $A$ 是两个元嫊的和 $\mathbf{u}$ 和 $\mathbf{v}$ 的 $\mathbb{R}^n$ ，那么行列式 $A$ 是通过替换得到的两个矩阵的行列式之和 $j$-第 行 $A$ 和 $\mathbf{u}$ 和 $\mathbf{v}$ ，分别。

## 线性代数代考_linear algebra代考_CALCULATING THE DETERMINANT

$$A=\left(\begin{array}{llll} a_{11} & a_{12} & u_{21} & u_{22} \end{array}\right) .$$

• 2nd排 $\rightarrow 2 \mathrm{nd}$ 排 $-\frac{a_{21}}{a_{11}} \cdot 1$ 街道。
这样，根据命题7.1.4(b)，行列式不变，得到三角矩阵:
$$\left(\begin{array}{lll} a_{11} & a_{12} 0 & a_{22}-\frac{a_{21}}{a_{11}} \cdot a_{12} \end{array}\right) .$$
现在我们简单地取对角系数的乘积，我们有:
$$\operatorname{det}(A)=a_{11} \cdot\left(a_{22}-\frac{a_{21}}{a_{11}} \cdot a_{12}\right)=a_{11} a_{22}-a_{12} a_{21} .$$
情况 2 . 假设 $a_{11}=0$. 我们交换第一行和第二行; 行列式改变符号，我们得到:

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