A.Kaw矩阵代数初步学习笔记 4. Unary Matrix Operations

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“矩阵代数初步”(Introduction to MATRIX ALGEBRA)课程由Prof. A.K.Kaw(University of South Florida)设计并讲授。
PDF格式学习笔记下载（Academia.edu）
第4章课程讲义下载（PDF）

Summary

• Transpose
  Let $[A]$ be a $m\times n$ matrix. Then $[B]$ is the transpose of $[A]$ if $b_{ji} = a_{ij}$ for all $i$ and $j$. That is, the $i$-th row and the $j$-th column element of $[A]$ is the $j$-th row and $i$-th column element of $[B]$. Note that $[B]$ is a $n\times m$ matrix and is denoted by $[B] = [A]^{T}$. For example, $$[A] = \begin{bmatrix}1& 2& 3\\ 4& 5& 6\end{bmatrix}\Rightarrow [A]^{T} = \begin{bmatrix}1& 4\\ 2& 5\\ 3& 6\end{bmatrix}$$
• Symmetric matrix
  A square matrix $[A]$ with real elements where $a_{ij} = a_{ji}$ for $i = 1, \cdots, n$ and $j = 1, \cdots, n$ is called a symmetric matrix. That is, $[A]$ is a symmetric matrix if $[A] = [A]^{T}$. For example, $$[A] = \begin{bmatrix}1& 2& 3\\ 2& 4& 5\\ 3& 5& 7\end{bmatrix}$$
• Skew-symmetric matrix
  A $n\times n$ matrix is skew-symmetric if $a_{ij} = -a_{ji}$ for $i = 1, \cdots, n$ and $j = 1, \cdots, n$. That is, $[A]$ is a skew-symmetric matrix if $[A] = -[A]^{T}$. Note that the diagonal elements must be zero in a skew-symmetric matrix. For example, $$[A] = \begin{bmatrix}0& 2& 3\\ -2& 0& 5\\ -3& -5& 0\end{bmatrix}$$
• Trace of matrix
  The trace of a $n\times n$ matrix $[A]$ is the sum of the diagonal entries of $[A]$, that is, $$\text{tr}[A] = \sum_{i=1}^{n}a_{ii}$$ For example, $$[A] = \begin{bmatrix}1& 2& 3\\ 2& 4& 5\\ 3& 5& 7\end{bmatrix}\Rightarrow \text{tr}[A] = 1 + 4 +7=12$$
• Determinant
  Let $[A]$ be a $n\times n$ matrix.
  • The minor of entry $a_{ij}$ is denoted by $M_{ij}$ and is defined as the determinant of the $(n-1)\times(n-1)$ sub-matrix of $[A]$, where the sub-matrix is obtained by deleting the $i$-th row and $j$-th column of the matrix $[A]$. The determinant is then given by $$\det(A) = \sum_{j=1}^{n}(-1)^{i+j}a_{ij}M_{ij},\ \text{for any}\ i=1, 2, \cdots, n$$ or $$\det(A) = \sum_{i=1}^{n}(-1)^{i+j}a_{ij}M_{ij},\ \text{for any}\ j=1, 2, \cdots, n$$ For example, $$[A] = \begin{bmatrix}1& 2& 3\\ 2& 4& 5\\ 3& 5& 7\end{bmatrix}$$ $$\Rightarrow \det(A) =(-1)^{1+1}\cdot1\cdot\begin{vmatrix}4& 5 \\ 5& 7\end{vmatrix} + (-1)^{1+2}\cdot2\cdot\begin{vmatrix}2& 5 \\ 3& 7\end{vmatrix} + (-1)^{1+3}\cdot3\cdot\begin{vmatrix}2& 4 \\ 3& 5\end{vmatrix}$$ $$=(4\times7-5\times5) -2\times(2\times7-3\times5) + 3\times(2\times5 - 3\times4) = -1$$ Note that for a $2\times2$ matrix $[A] = \begin{bmatrix}a& b\\ c& d\end{bmatrix}$, $\det(A) = ad-bc$.
  • The number $(-1)^{i+j}M_{ij}$ is called the cofactor of $a_{ij}$ and is denoted by $C_{ij}$. The formula for the determinant can then be written as $$\det(A) = \sum_{j=1}^{n}a_{ij}C_{ij},\ \text{for any}\ i=1, 2, \cdots, n$$ or $$\det(A) = \sum_{i=1}^{n}a_{ij}C_{ij},\ \text{for any}\ j=1, 2, \cdots, n$$
  • If $[A]$ and $[B]$ are square matrices of same size, then $$\det(A\cdot B) = \det(A)\cdot\det(B)$$
  • $\det(A) = 0$ if
    • A row or a column is zero, or
    • A row (column) is proportional to another row (column).
  • If a row (column) is multiplied by $k$ to result in matrix $[B]$, then $$\det(B) = k\cdot\det(A)$$
  • If $[B] = k\cdot[A]$, then $$\det(B)=k^{n}\det(A)$$
  • If $[A]$ is a $n\times n$ upper or lower triangular matrix, then $$\det(A) = \prod_{i=1}^{n}a_{ii}$$
  • If $[B]$ is row-equivalent to $[A]$, then $$\begin{cases} R_i\leftrightarrow R_j: & \det(B) = -\det(A);\\ tR_i: & \det(B) = t\det(A);\\ R_i\rightarrow R_i+tR_j: &\det(B) = \det(A).\end{cases}$$

Selected Problems

1. Let $$[A] = \begin{bmatrix}25& 3& 6\\ 7& 9& 2\end{bmatrix}$$ Find $[A]^{T}$.

Solution:

$$[A]^{T} = \begin{bmatrix}25& 7\\ 3& 9\\ 6& 2\end{bmatrix}$$

2. If $[A]$ and $[B]$ are two $n\times n$ symmetric matrices, show that $[A]+[B]$ is also symmetric.

Solution:

Let $[C]=[A]+[B]$, so we have $$c_{ij} = a_{ij} + b_{ij} = a_{ji} + b_{ji} =c_{ji}$$ that is, $[C]=[C]^{T}$.

3. What is the trace of $$[A] = \begin{bmatrix}7& 2& 3& 4\\ -5& -5& -5& -5\\ 6& 6& 7& 9\\ -5& 2& 3& 10\end{bmatrix}$$

Solution:

$$\text{tr}[A] = 7-5+7+10=19$$

4. Find the determinant of $$[A] = \begin{bmatrix}10& -7& 0\\ -3& 2.099& 6\\ 5& -1& 5\end{bmatrix}$$

Solution:

$$\det(A)=(-1)^{1+1}\times10\times\begin{vmatrix}2.099& 6\\ -1& 5\end{vmatrix} + (-1)^{1+2}\times(-7)\times\begin{vmatrix}-3& 6\\ 5& 5\end{vmatrix}$$ $$=10\times(2.099\times5+1\times6) + 7\times(-15-30) = -150.05$$

5. What is the value of a $n\times n$ matrix $\det(3[A])$?

Solution:

$$\det(3[A]) = 3^n\det(A)$$

6. For a $5\times5$ matrix $[A]$, the first row is interchanged with the fifth row, what is the determinant of the resulting matrix $[B]$?

Solution:

The sign would be changed if interchaged two row (column). Thus $$\det(B) = -\det(A)$$

7. What is the determinant of $$[A] = \begin{bmatrix}0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\\ 1& 0& 0& 0\end{bmatrix}$$

Solution:

$$[A] = \begin{bmatrix}0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\\ 1& 0& 0& 0\end{bmatrix}\Rightarrow R_1\leftrightarrow R_4 \begin{bmatrix}1& 0& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\\ 0& 1& 0& 0\end{bmatrix}$$ $$\Rightarrow R_2\leftrightarrow R_3 \begin{bmatrix}1& 0& 0& 0\\ 0& 0& 0& 1\\ 0& 0& 1& 0\\0& 1& 0& 0\end{bmatrix}$$ $$\Rightarrow R_2\leftrightarrow R_4 \begin{bmatrix}1& 0& 0& 0\\0& 1& 0& 0\\ 0& 0& 1& 0\\0& 0& 0& 1\end{bmatrix}=[B]$$ Thus $\det(A) = (-1)^{3}\det(B)=-1$.

8. Find the determinant of $$[A]=\begin{bmatrix}0& 0& 0\\ 2& 3& 5\\ 6& 9& 2\end{bmatrix}$$

Solution:

$\det(A)=0$ since the first row is zero.

9. Find the determinant of $$[A]=\begin{bmatrix}0& 0& 2& 3\\ 0& 2& 3& 5\\ 6& 7& 2& 3\\ 6.6& 7.7& 2.2& 3.3\end{bmatrix}$$

Solution:

Since $R_4 = 1.1R_3$, so $\det(A) = 0$.

10. Find the determinant of $$[A]=\begin{bmatrix}5& 0& 0& 0\\ 0& 3& 0& 0\\ 2& 5& 6& 0\\ 1& 2& 3& 9\end{bmatrix}$$

Solution:

This is a lower triangular matrix and hence $$\det(A) = 5\times3\times6\times9=810$$

11. Given the matrix $$[A]=\begin{bmatrix}125& 25& 5& 1\\ 512& 64& 8& 1\\ 1157& 89& 13& 1\\ 8& 4& 2& 1\end{bmatrix}$$ and $\det(A) = -32400$. Find the determinant of $$[A_1]=\begin{bmatrix}125& 25& 5& 1\\ 512& 64& 8& 1\\ 1141& 81& 9& -1\\ 8& 4& 2& 1\end{bmatrix};$$ $$[A_2]=\begin{bmatrix}125& 25& 1& 5\\ 512& 64& 1& 8\\ 1157& 89& 1& 13\\ 8& 4& 1& 2\end{bmatrix};$$ $$[A_3] = \begin{bmatrix} 125& 25& 5& 1\\ 1157& 89& 13& 1\\ 512& 64& 8& 1\\8& 4& 2& 1\end{bmatrix};$$ $$[A_4] = \begin{bmatrix} 125& 25& 5& 1\\ 1157& 89& 13& 1\\ 8& 4& 2& 1\\ 512& 64& 8& 1\end{bmatrix};$$ $$[A_5] = \begin{bmatrix} 125& 25& 5& 1\\ 512& 64& 8& 1\\ 1157& 89& 13& 1\\ 16& 8& 4& 2 \end{bmatrix}.$$

Solution:

$$[A]=\begin{bmatrix}125& 25& 5& 1\\ 512& 64& 8& 1\\ 1157& 89& 13& 1\\ 8& 4& 2& 1\end{bmatrix}\Rightarrow R_3-2R_4 \begin{bmatrix}125& 25& 5& 1\\ 512& 64& 8& 1\\ 1141& 81& 9& -1\\ 8& 4& 2& 1\end{bmatrix}=[A_1]$$ Thus $\det(A_1) = \det(A) =-32400$. $$[A]=\begin{bmatrix}125& 25& 5& 1\\ 512& 64& 8& 1\\ 1157& 89& 13& 1\\ 8& 4& 2& 1\end{bmatrix}\Rightarrow C_3\leftrightarrow C_4 \begin{bmatrix}125& 25& 1& 5\\ 512& 64& 1& 8\\ 1157& 89& 1& 13\\ 8& 4& 1& 2\end{bmatrix} = [A_2]$$ Thus $\det(A_2)=-\det(A)=32400$. $$[A]=\begin{bmatrix}125& 25& 5& 1\\ 512& 64& 8& 1\\ 1157& 89& 13& 1\\ 8& 4& 2& 1\end{bmatrix}\Rightarrow R_2\leftrightarrow R_3\begin{bmatrix} 125& 25& 5& 1\\ 1157& 89& 13& 1\\ 512& 64& 8& 1\\8& 4& 2& 1\end{bmatrix}= [A_3]$$ Thus $\det(A_3) = -\det(A) = 32400$. $$[A]=\begin{bmatrix}125& 25& 5& 1\\ 512& 64& 8& 1\\ 1157& 89& 13& 1\\ 8& 4& 2& 1\end{bmatrix}\Rightarrow \begin{cases} R_2\leftrightarrow R_3\\ R‘_3\leftrightarrow R_4\end{cases} \begin{bmatrix} 125& 25& 5& 1\\ 1157& 89& 13& 1\\ 8& 4& 2& 1\\ 512& 64& 8& 1\end{bmatrix} = [A_4]$$ Thus $\det(A_4) = (-1)^2\det(A) = -32400$. $$[A]=\begin{bmatrix}125& 25& 5& 1\\ 512& 64& 8& 1\\ 1157& 89& 13& 1\\ 8& 4& 2& 1\end{bmatrix} \Rightarrow 2R_4\begin{bmatrix} 125& 25& 5& 1\\ 512& 64& 8& 1\\ 1157& 89& 13& 1\\ 16& 8& 4& 2 \end{bmatrix} = [A_5]$$ Thus $\det(A_5) = 2\det(A) = -64800$.

12. Find the determinant of $$[A] = \begin{bmatrix}25& 5& 1\\ 64& 8& 1\\ 144& 12& 5\end{bmatrix}$$

Solution:

$$\det(A) = (-1)^{1+3}a_{13}M_{13}+(-1)^{2+3}a_{23}M_{23} + (-1)^{3+3}a_{33}M_{33}$$ $$ = \begin{vmatrix}64& 8\\ 144& 12\end{vmatrix} - \begin{vmatrix}25& 5\\ 144& 12\end{vmatrix} + 5\times \begin{vmatrix}25& 5\\ 64& 8\end{vmatrix} = -564$$

13. Show that if $[A][B]=[I]$, where $[A]$, $[B]$ and $[I]$ are matrices of $n\times n$ size and $[I]$ is an identity matrix, then $\det(A)\neq0$ and $\det(B)\neq0$.

Solution: $$\det(A)\det(B)=\det(AB) =\det(I) = 1$$ $$\Rightarrow \det(A)\neq0,\ \det(B)\neq0.$$

14. If the determinant of a $4\times4$ matrix $[A]$ is given as 20, then what is the determinant of $5[A]$?

Solution:

$$\det(k[A])=k^n\det(A)$$ $$\Rightarrow \det(5[A]) = 5^4\det(A) = 625\times20=12500$$

15. If the matrix product $[A][B][B]$ is defined, what is $([A][B][C])^{T}$?

Solution:

$$([A][B])^{T}=[B]^{T}[A]^{T}$$ $$\Rightarrow ([A][B][C])^{T}=[C]^{T}([A][B])^{T}=[C]^{T}[B]^{T}[A]^{T}$$

16. The determinant of the matrix $$[A] = \begin{bmatrix}25& 5& 1\\ 0& 3& 8\\ 0& 9& a\end{bmatrix}$$ is 50. What is the value of $a$?

Solution:

$$\det(A) = 25\times\begin{vmatrix}3& 8\\ 9& a\end{vmatrix} = 25\times(3a-72)=50$$ $$\Rightarrow a={74\over3}$$

17. $[A]$ is a $5\times 5$ matrix and a matrix $[B]$ is obtained by the row operations of replacing Row1 with Row3, and then Row3 is replaced by a linear combination of $2\times$Row3$+4\times$Row2. If $\det(A)=17$, then what is the value of $\det(B)$?

Solution:

The process is $$[A]\Rightarrow R_1\leftrightarrow R_3 \Rightarrow 2R_3\Rightarrow R_3+4R_2\Rightarrow [B]$$ Thus $$\det(B) = (-1)\times2\cdot\det(A) = -34$$

A.Kaw矩阵代数初步学习笔记 4. Unary Matrix Operations

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原文地址：http://www.cnblogs.com/zhaoyin/p/4145705.html