“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。
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Summary
- Consistent and inconsistent system
A system of equations $$[A][X]=[B]$$ where $[A]$ is called the coefficient matrix, $[B]$ is called the right hand side vector and $[X]$ is called the solution vector. This system is consistent if there is a solution, and it is inconsistent if there is no solution. However, a consistent system of equations does not mean a unique solution, that is, a consistent system of equations may have a unique solution or infinite solutions. - Rank
- The rank of a matrix is defined as the order of the largest square sub-matrix whose determinant is not zero.
- For example, the matrix $$[A] = \begin{bmatrix}3& 1& 2\\ 2& 0& 5\\ 5& 1& 7 \end{bmatrix}$$ we know that $$\det(A) = (-1)^{1+2} \times 1 \times \begin{vmatrix}2& 5\\ 5& 7\end{vmatrix} + (-1)^{2+3}\times1\times\begin{vmatrix}3& 2\\ 2& 5\end{vmatrix} = 11 - 11 =0$$ Thus its rank will be less than 3. On the other hand, the determinant of the sub-matrix $\begin{bmatrix}3& 1\\ 2& 0\end{bmatrix}$ is $0-2 = -2\neq0$. Hence the rank of matrix $A$ is 2.
- A system of equations $$[A][X]=[B]$$ is consistent if the rank of $A$ is equal to the rank of the augmented matrix $[A|B]$.
- A system of equations $$[A][X]=[B]$$ is inconsistent if the rank of $A$ is less than the rank of the augmented matrix $[A|B]$.
- In practice, we can use elementary row operation to calculate the rank of a matrix. Or alternatively, directly find the result from the equivalent matrix of the augmented matrix of a system of equations.
- The rank of the coefficient matrix $[A]$ is same as the number of unknowns, then the solution is unique; if the rank of the coefficient matrix $[A]$ is less than the number of unknowns, then infinite solutions exist.
- A system of equations $$[A][X]=[B]$$
- It has Unique solution if rank$(A) =$ rank$(A|B)=$ number of unknowns;
- It has Infinite solutions if rank$(A)=$ rank$(A|B) < $ number of unknowns;
- It has No solution (i.e. inconsistent) if rank$(A) < $ rank$(A|B)$
- Inverse
- The inverse of a square matrix $[A]$, if existing, is denoted by $[A]^{-1}$ such that $$[A][A]^{-1}=[I]=[A]^{-1}[A]$$ where $[I]$ is the identity matrix. $[A]$ is called to be invertible or nonsingular.
- If $[A]$ and $[B]$ are two $n\times n$ matrices such that $[B][A] = [I]$, then these statements are also true:
- $[B]$ is the inverse of $[A]$
- $[A]$ is the inverse of $[B]$
- $[A]$ and $[B]$ are both invertible
- $[A][B]=[I]$
- $[A]$ and $[B]$ are both nonsingular
- all columns (rows) of $[A]$ and $[B]$ are linearly independent
- Given $$[A][X]=[B]$$ then $[X]=[A]^{-1}[B]$.
- The inverse of an invertible matrix can be found by $$[A]^{-1} = {1\over\det(A)}\text{adj}(A)$$ where $$\text{adj}(A) = \begin{bmatrix}C_{11}& C_{21}&\cdots& C_{n1} \\ \vdots & \vdots& &\vdots\\C_{1n}&C_{2n}&\cdots&C_{nn} \end{bmatrix}$$ where $C_{ij}$ are the cofactors of $a_{ij}$. This formula implies that $\det(A)\neq0$ if $[A]$ is invertible.
- The inverse of a square matrix is unique, if it exists. Since $$\begin{cases}[B][A] = [I]\\ [C][A]=[I]\end{cases}\Rightarrow [B][A][C] = [I][C] = [C]$$ $$\Rightarrow [B][I] = [C]\Rightarrow [B]=[C]$$
Selected Problems
1. For a set of equations $[A][X]=[B]$, a unique solution exists if ( ).
Solution:
rank$(A) =$ rank$(A|B)$ and rank$(A)=$ number of unknowns.
2. What is the rank of matrix $$[A] = \begin{bmatrix}4& 4& 4& 4\\ 4& 4& 4& 4\\ 4& 4& 4& 4\\ 4& 4& 4& 4\end{bmatrix}$$
Solution:
$$[A] = \begin{bmatrix}4& 4& 4& 4\\ 4& 4& 4& 4\\ 4& 4& 4& 4\\ 4& 4& 4& 4\end{bmatrix}\Rightarrow\begin{cases}R_2-R_1\\ R_3-R_1\\ R_4-R_1\\ {1\over4}R_1\end{cases} \begin{bmatrix}1& 1& 1& 1\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{bmatrix}$$ Thus the rank of $[A]$ is 1.
3. A $3\times4$ matrix can have a rank of at most ( )?
Solution:
Since there are no square sub-matrices of order 4 as it is a $3\times4$ matrix, the rank of this matrix is at most 3.
4. If $[A][X]=[B]$ has a unique solution, where the order of $[A]$ is $3\times3$, $[X]$ is $3\times1$, then the rank of $[A]$ is ( ).
Solution:
Since it has a unique solution, that is, the rank of $[A]$ equals to both of the rank of augmented matrix and the number of unknowns, which is 3.
5. Show if the following system of equations is consistent or inconsistent. If they are consistent, determine if the solution would be unique or infinite ones exist. $$\begin{bmatrix}1& 2& 5\\ 7& 3& 9\\ 8& 5& 14\end{bmatrix}\begin{bmatrix}x_1\\ x_2\\ x_3\end{bmatrix} = \begin{bmatrix}8\\19\\27\end{bmatrix}$$
Solution:
The augmented matrix is $$\begin{bmatrix}1& 2& 5 & 8\\ 7& 3& 9 & 19\\ 8& 5& 14 &27\end{bmatrix}\Rightarrow R_3-R_1-R_2 \begin{bmatrix}1& 2& 5 & 8\\ 7& 3& 9 & 19\\ 0& 0& 0 &0\end{bmatrix}$$ $$\Rightarrow R_2-7R_1 \begin{bmatrix}1& 2& 5 & 8\\ 0& -11& -26 & -37\\ 0& 0& 0 &0\end{bmatrix}$$ $$\Rightarrow -{1\over11}R_2 \begin{bmatrix}1& 2& 5 & 8\\ 0& 1&{26\over11} & {37\over11}\\ 0& 0& 0 &0\end{bmatrix}\Rightarrow R_1-2R_2 \begin{bmatrix}1& 0& {3\over11} & {14\over11}\\ 0& 1&{26\over11} & {37\over11}\\ 0& 0& 0 &0\end{bmatrix}$$ Thus the rank of both coefficient matrix and augmented matrix is 2, which is less than the number of unknowns, that is, it is consistent system and has infinite solutions.
6. Show if the following system of equations is consistent or inconsistent. If they are consistent, determine if the solution would be unique or infinite ones exist. $$\begin{bmatrix}1& 2& 5\\ 7& 3& 9\\ 8& 5& 14\end{bmatrix}\begin{bmatrix}x_1\\ x_2\\ x_3\end{bmatrix} = \begin{bmatrix}8\\19\\28\end{bmatrix}$$
Solution: The augmented matrix is $$\begin{bmatrix}1& 2& 5 & 8\\ 7& 3& 9 & 19\\ 8& 5& 14 &28\end{bmatrix}\Rightarrow R_3-R_1-R_2 \begin{bmatrix}1& 2& 5 & 8\\ 7& 3& 9 & 19\\ 0& 0& 0 &1\end{bmatrix}$$ The last row of the above matrix shows that it is an inconsistent system.
7. Show if the following system of equations is consistent or inconsistent. If they are consistent, determine if the solution would be unique or infinite ones exist. $$\begin{bmatrix}1& 2& 5\\ 7& 3& 9\\ 8& 5& 13\end{bmatrix}\begin{bmatrix}x_1\\ x_2\\ x_3\end{bmatrix} = \begin{bmatrix}8\\19\\28\end{bmatrix}$$
Solution:
The augmented matrix is $$\begin{bmatrix}1& 2& 5 & 8\\ 7& 3& 9 & 19\\ 8& 5& 13 &28\end{bmatrix}\Rightarrow R_3-R_1-R_2 \begin{bmatrix}1& 2& 5 & 8\\ 7& 3& 9 & 19\\ 0& 0& -1 &1\end{bmatrix}$$ $$\Rightarrow R_2-7R_1\begin{bmatrix}1& 2& 5 & 8\\ 0& -11& -26 & -37\\ 0& 0& -1 &1\end{bmatrix}$$ $$\Rightarrow -{1\over11}R_2\begin{bmatrix}1& 2& 5 & 8\\ 0& 1& {26\over11}&{37\over11}\\ 0& 0& -1 &1\end{bmatrix}\Rightarrow \begin{cases}R_1-2R_2\\ R_2+{26\over11}R_3\end{cases} \begin{bmatrix}1& 0& {3\over11} & {14\over11}\\ 0& 1& 0 & {63\over11}\\ 0& 0& -1 &1\end{bmatrix}$$ $$\Rightarrow \begin{cases}R_1+{3\over11}R_3\\-R_3\end{cases} \begin{bmatrix}1& 0& 0 & {17\over11}\\ 0& 1& 0 & {63\over11}\\ 0& 0& 1 & -1\end{bmatrix}$$ That is, the system is consistent and it has unique solution.
8. For what value of $a$ will the following equation have $$\begin{cases}x_1+x_2+x_3=4\\ x_3=2\\ (a^2-4)x_1+x_3=a-2 \end{cases}$$ (A) Unique solution; (B) No solution; (C) Infinite solutions.
Solution:
The augmented matrix is $$\begin{bmatrix}1& 1& 1& 4\\ 0& 0& 1& 2\\ a^2-4& 0& 1& a-2\end{bmatrix}$$ If $a=2$, then $$\begin{bmatrix}1& 1& 1& 4\\ 0& 0& 1& 2\\ 0& 0& 1& 0\end{bmatrix}$$ The last row shows that it is inconsistent.\\ If $a=-2$, then $$\begin{bmatrix}1& 1& 1& 4\\ 0& 0& 1& 2\\ 0& 0& 1& -4\end{bmatrix}\Rightarrow R_3-R_2 \begin{bmatrix}1& 1& 1& 4\\ 0& 0& 1& 2\\ 0& 0& 0& -6\end{bmatrix}$$ The last row shows that it is inconsistent.\\ If $a\neq\pm2$, then $$\begin{bmatrix}1& 1& 1& 4\\ 0& 0& 1& 2\\ a^2-4& 0& 1& a-2\end{bmatrix} \Rightarrow {1\over a-2}R_3 \begin{bmatrix}1& 1& 1& 4\\ 0& 0& 1& 2\\ a+2 & 0& {1\over a-2}& 1\end{bmatrix}$$ From the second row we know that $x_2=2$, and deduce that other two unknowns are also unique, that is, it has unique solution. Thus, the system has unique solution if $a\neq\pm2$; the system has no solution if $a=\pm2$; and there is no possible to have infinite solutions.
9. Find the cofactor matrix and the adjoint matrix of $$[A]=\begin{bmatrix}3& 4& 1\\ 2& -7& -1\\ 8& 1& 5\end{bmatrix}$$
Solution:
Firstly, find the cofactors of each $a_{ij}$: $$\begin{cases} C_{11} = M_{11} = -35+1 =-34\\ C_{12}=-M_{12}=-(10+8)=-18\\ C_{13}=M_{13}=2+56 =58\\ C_{21}=-M_{21}=-(20-1)=-19\\ C_{22}=M_{22}=15-8=7\\ C_{23}=-M_{23}=-(3-32)=29 \\ C_{31}=M_{31} = -4+7=4\\ C_{32}= -M_{32} = -(-3-2)=5\\ C_{33}=M_{33}=-21-8=-29 \end{cases}$$ Thus the cofactor matrix is $$\begin{bmatrix}-34& -18 & 58\\ -19& 7& 29\\ 3& 5& -29\end{bmatrix}$$ and the adjoint matrix is the transpose of the cofactor matrix, that is $$\begin{bmatrix}-34& -19& 3\\ -18& 7& 5\\ 58& 29& -29\end{bmatrix}$$
10. Find $[A]^{-1}$ of the matrix $$[A]=\begin{bmatrix}3& 4& 1\\ 2& -7& -1\\ 8& 1& 5\end{bmatrix}$$
Solution:
From the result of Question 9 and $[A]^{-1}={1\over\det(A)}\text{adj} (A)$, we have $$[A]^{-1}=-{1\over116}\begin{bmatrix}-34& -19& 3\\ -18& 7& 5\\ 58& 29& -29\end{bmatrix} = \begin{bmatrix}{17\over58}& {19\over116}& -{3\over116}\\ {9\over58}& -{7\over116}& -{5\over116} \\ -{1\over2}& -{1\over4}& {1\over4}\end{bmatrix}$$
11. Prove that if $[A]$ and $[B]$ are both invertible and are square matrices of same order, then $$([A][B])^{-1} = [B]^{-1}[A]^{-1}$$
Solution:
$$[A][B][B]^{-1}[A]^{-1}=[A][I][A]^{-1}=[A][A]^{-1}=[I]$$ and $$[B]^{-1}[A]^{-1}[A][B]=[B]^{-1}[I][B]=[B]^{-1}[B]=[I]$$ $$\Rightarrow ([A][B])^{-1} = [B]^{-1}[A]^{-1}$$
12. What is the inverse of a square diagonal matrix? Does it always exist?
Solution:
Since $$\begin{bmatrix}a_{11}&0&\cdots&0\\ 0& a_{22}&\cdots&0\\ \vdots& \vdots&\cdots&\vdots\\ 0 & 0 & \cdots & a_{nn} \end{bmatrix}\cdot \begin{bmatrix}b_{11}&0&\cdots&0\\ 0& b_{22}&\cdots&0\\ \vdots& \vdots&\cdots&\vdots\\ 0 & 0 & \cdots & b_{nn} \end{bmatrix}$$ $$= \begin{bmatrix} a_{11}b_{11} & 0 &\cdots &0\\ 0& a_{22}b_{22}&\cdots&0\\ \vdots& \vdots &\cdots &\vdots\\ 0 & 0 & \cdots & a_{nn}b_{nn} \end{bmatrix}$$ The inverse of a square matrix $$[A] = \begin{bmatrix}a_{11}&0&\cdots&0\\ 0& a_{22}&\cdots&0\\ \vdots& \vdots&\cdots&\vdots\\ 0 & 0 & \cdots & a_{nn} \end{bmatrix}$$ is hence $$[A]^{-1} = \begin{bmatrix}{1\over a_{11}} & 0 &\cdots&0\\ 0& {1\over a_{22}}&\cdots&0\\ \vdots & \vdots & \cdots & \vdots\\ 0 & 0 & \cdots & {1\over a_{nn}}\end{bmatrix}$$
13. $[A]$ and $[B]$ are square matrices. If $[A][B]=[0]$ and $[A]$ is invertible, show $[B]=[0]$.
Solution:
$$[A][B]=[0]\Rightarrow [A]^{-1}[A][B]=[A]^{-1}[0]\Rightarrow [I][B]=[0]$$ $$\Rightarrow [B]=[0]$$
14. If $[A][B][C]=[I]$, where $[A]$, $[B]$, and $[C]$ are of the same size, show that $[B]$ is invertible.
Solution:
We will show that $\det(B)\neq0$, which is equivalent to $[B]$ is invertible. $$\det(A)\det(B)\det(C) = \det(ABC)= \det([I]) = 1$$ $$\Rightarrow\begin{cases}\det(A)\neq0\\ \det(B)\neq0\\ \det(C)\neq0\end{cases}$$
15. Prove if $[B]$ is invertible, $$[A][B]^{-1}=[B]^{-1}[A]$$ if and only if $$[A][B]=[B][A]$$
Solution:
$$AB=BA\Rightarrow ABB^{-1}=BAB^{-1}\Rightarrow A=BAB^{-1}$$ $$\Rightarrow B^{-1}A=B^{-1}BAB^{-1}=IAB^{-1}=AB^{-1}$$ On the other hand $$AB^{-1}=B^{-1}A\Rightarrow AB^{-1}B=B^{-1}AB\Rightarrow A=B^{-1}AB$$ $$\Rightarrow BA=BB^{-1}AB=IAB=AB$$
16. For what value if $a$ does the linear system have $$\begin{cases}x+y=2\\ 6x+6y=a\end{cases}$$ (A) infinite solutions; (B) unique solution.
Solution:
It has infinite solution when $a=12$; it is impossible to have unique solution.
17. What is the rank of $$\begin{bmatrix}1& 2& 3\\ 4& 6& 7\\ 6& 10& 13\end{bmatrix}$$
Solution:
$$\begin{bmatrix}1& 2& 3\\ 4& 6& 7\\ 6& 10& 13\end{bmatrix}\Rightarrow\begin{cases}R_2-4R_1\\ R_3-6R_1\end{cases} \begin{bmatrix}1& 2& 3\\ 0& -2& -5\\ 0& -2& -5\end{bmatrix}$$ $$\Rightarrow\begin{cases}R_3-R_2\\ -{1\over2}R_2\end{cases} \begin{bmatrix}1& 2& 3\\ 0& 1& {5\over2}\\ 0& 0& 0\end{bmatrix}\Rightarrow R_1-2R_2\begin{bmatrix}1& 0& -2\\ 0& 1& {5\over2}\\ 0& 0& 0\end{bmatrix}$$ Thus the rank of this matrix is 2.
18. What is the rank of $$\begin{bmatrix}1& 2& 3 & 6\\ 4& 6& 7 & 17\\ 6& 10& 13 & 29\end{bmatrix}$$
Solution:
$$\begin{bmatrix}1& 2& 3 & 6\\ 4& 6& 7 & 17\\ 6& 10& 13 & 29\end{bmatrix}\Rightarrow \begin{cases}R_2-4R_1\\ R_3-6R_1\end{cases}\begin{bmatrix}1& 2& 3 & 6\\ 0& -2& -5 & -7\\ 0& -2 & -5 & -7\end{bmatrix}$$ $$\Rightarrow R_3-R_2\begin{bmatrix}1& 2& 3 & 6\\ 0& -2 & -5 & -7\\ 0& 0& 0 & 0\end{bmatrix}$$ The rank of this matrix is 2.
19. What is the rank of $$\begin{bmatrix}1& 2& 3 & 6\\ 4& 6& 7 & 18\\ 6& 10& 13 & 30\end{bmatrix}$$
Solution:
$$\begin{bmatrix}1& 2& 3 & 6\\ 4& 6& 7 & 18\\ 6& 10& 13 & 30\end{bmatrix}\Rightarrow \begin{cases}R_2-4R_1\\ R_3-6R_1\end{cases}\begin{bmatrix}1& 2& 3 & 6\\ 0& -2& -5 & -6\\ 0& -2 & -5 & -6\end{bmatrix}$$ $$\Rightarrow R_3-R_2\begin{bmatrix}1& 2& 3 & 6\\ 0& -2 & -5 & -6\\ 0& 0& 0 & 0\end{bmatrix}$$ The rank of this matrix is 2.
20. How many solutions does the following system of equations have $$\begin{bmatrix}1& 2& 3\\ 4& 6& 7\\ 6& 10& 13\end{bmatrix}\begin{bmatrix}a\\ b\\ c\end{bmatrix} = \begin{bmatrix}6\\ 17\\ 29\end{bmatrix}$$
Solution:
From the previous questions 17, 18, we know that the rank of the coefficient matrix equals to the rank of the augmented matrix, which is 2. And it is less than the number of unknowns which is 3. Thus this system has infinite solutions.
21. How many solutions does the following system of equations have $$\begin{bmatrix}1& 2& 3\\ 4& 6& 7\\ 6& 10& 13\end{bmatrix}\begin{bmatrix}a\\ b\\ c\end{bmatrix} = \begin{bmatrix}6\\ 18\\ 30\end{bmatrix}$$
Solution:
From the previous questions 17, 19, we know that the rank of the coefficient matrix equals to the rank of the augmented matrix, which is 2. And it is less than the number of unknowns which is 3. Thus this system has infinite solutions.
22. Find the second column of the inverse of $$\begin{bmatrix}1& 2& 0\\ 4& 5& 0\\ 0& 0& 13\end{bmatrix}$$
Solution:
The second column of the product is $\begin{bmatrix}0 \\ 1 \\ 0 \end{bmatrix}$, which is the product of the given matrix and the second column of its inverse, say $\begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix}$. Thus we have $$\begin{bmatrix}1& 2& 0\\ 4& 5& 0\\ 0& 0& 13\end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix} =\begin{bmatrix}0 \\ 1 \\ 0 \end{bmatrix}$$ $$\Rightarrow \begin{cases}x_1+2x_2=0\\ 4x_1+5x_2=1\\ 13x_3=0\end{cases}\Rightarrow \begin{cases}x_1={2\over3}\\ x_2=-{1\over3}\\ x_3=0\end{cases}\Rightarrow \begin{bmatrix}{2\over3}\\ -{1\over3}\\ 0\end{bmatrix}$$
23. Write out the inverse of $$\begin{bmatrix}1& 0& 0& 0\\ 0& 2& 0& 0\\ 0& 0& 4& 0\\ 0& 0& 0& 5\end{bmatrix}$$
Solution:
$$\begin{bmatrix}1& 0& 0& 0\\ 0& {1\over2}& 0& 0\\ 0& 0& {1\over4}& 0\\ 0& 0& 0& {1\over5}\end{bmatrix}$$
24. Solve $[A][X]=[B]$ for $[X]$ if $$[A]^{-1}=\begin{bmatrix}10& -7& 0\\ 2& 2& 5\\ 2& 0& 6\end{bmatrix}$$ and $$[B]=\begin{bmatrix}7 \\ 2.5\\ 6.012\end{bmatrix}$$
Solution:
$$[A][X]=[B]$$ $$\Rightarrow [X]=[A]^{-1}[B]= \begin{bmatrix}10& -7& 0\\ 2& 2& 5\\ 2& 0& 6\end{bmatrix}\cdot\begin{bmatrix}7 \\ 2.5\\ 6.012\end{bmatrix} = \begin{bmatrix}52.5 \\ 49.06\\ 50.072 \end{bmatrix}$$
25. Let $[A]$ be a $3\times3$ matrix. Suppose $$[X]=\begin{bmatrix}7\\2.5\\6.012\end{bmatrix}$$ is a solution to the homogeneous set of equations $[A][X]=[0]$. Does $[A]$ have an inverse?
Solution:
$$[A][X]=[0]\Rightarrow [X]=[A]^{-1}[0]=[0]$$ which contradicts to the value of $[X]$. Thus $[A]$ is not invertible.
26. Is the set of vectors $$\vec{A}=\begin{bmatrix}1\\ 1\\ 1\end{bmatrix},\ \vec{B}=\begin{bmatrix}1\\ 2\\ 5\end{bmatrix},\ \vec{C}=\begin{bmatrix}1\\ 4\\ 25\end{bmatrix}$$ linearly independent?
Solution:
If the rank of the vectors is 3, then it would be independent set of vectors. $$\begin{bmatrix}1& 1& 1\\ 1& 2& 4\\ 1& 5& 25\end{bmatrix} \Rightarrow \begin{bmatrix}1& 1& 1\\ 0& 1& 3\\ 0& 4& 24\end{bmatrix} \Rightarrow \begin{bmatrix}1& 0& -2\\ 0& 1& 3\\ 0& 0& 12\end{bmatrix} \Rightarrow \begin{bmatrix}1& 0& -2\\ 0& 1& 3\\ 0& 0& 1\end{bmatrix}$$ whose rank is 3. Thus they are independent vectors.
27. What is the rank of the set of vectors $$\vec{A}=\begin{bmatrix}1\\ 1\\ 1\end{bmatrix},\ \vec{B}=\begin{bmatrix}1\\ 2\\ 5\end{bmatrix},\ \vec{C}=\begin{bmatrix}1\\ 3\\ 6\end{bmatrix}$$
Solution:
$$\begin{bmatrix}1& 1& 1\\ 1& 2& 3\\ 1& 5& 6\end{bmatrix}\Rightarrow \begin{bmatrix}1& 1& 1\\ 0& 1& 2\\ 0& 4& 5\end{bmatrix} \Rightarrow \begin{bmatrix}1& 0& -1\\ 0& 1& 2\\ 0& 0& -3\end{bmatrix} \Rightarrow \begin{bmatrix}1& 0& -1\\ 0& 1& 2\\ 0& 0& 1\end{bmatrix}$$ Thus the rank of the vectors is 3.
28. What is the rank of $$\vec{A}=\begin{bmatrix}1\\ 1\\ 1\end{bmatrix},\ \vec{B}=\begin{bmatrix}2\\ 2\\ 4\end{bmatrix},\ \vec{C}=\begin{bmatrix}3\\ 3\\ 5\end{bmatrix}$$
Solution:
$$\begin{bmatrix}1& 2& 3\\ 1& 2& 3\\ 1& 4& 5\end{bmatrix}\Rightarrow \begin{bmatrix}1& 2& 3\\ 0& 0& 0\\ 0& 2& 2\end{bmatrix} \Rightarrow \begin{bmatrix}1& 0& 1\\ 0& 0& 0\\ 0& 1& 1\end{bmatrix} $$ Thus the rank of the vectors is 2.
29. The set of equations $$\begin{bmatrix}1& 2& 5\\ 2& 3& 7\\ 5& 8& 19\end{bmatrix}\begin{bmatrix}x_1\\ x_2\\ x_3\end{bmatrix}= \begin{bmatrix}18\\ 26\\ 70\end{bmatrix}$$ has ( ) solution(s).
Solution:
$$\begin{bmatrix}1& 2& 5 & 18\\ 2& 3& 7 & 26\\ 5& 8& 19 & 70\end{bmatrix} \Rightarrow \begin{bmatrix}1& 2& 5 & 18\\ 0& -1& -3 & -10\\ 0& -2& -6 & -20\end{bmatrix} \Rightarrow \begin{bmatrix}1& 2& 5 & 18\\ 0& -1& -3 & -10\\ 0& 0&0 & 0\end{bmatrix}$$ The rank of the coefficient matrix equals to the augmented matrix, which is 2. But it is less than the number of unknowns which is 3. Thus it has infinite solutions.
30. Does $\begin{bmatrix}6& 7\\ 12& 14\end{bmatrix}$ have an inverse?
Solution:
Since the determinant of this matrix is $6\times14-12\times7=0$, thus it does not have inverse.