Linear Algebra Section 3.3 HomeWork
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### 1.Label the following statements as true or false.
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**$(a)$ Any system of linear equations has at least one solution.**
False. Considerate the system $0x=2$, there is no solution.
**$(b)$ Any system of linear equations has at most one solution**
False. Considerate the system $0x=0$, there are infinite solutions.
**$(c)$ Any homogeneous system of linear equations has at least one solution.**
True. Consider any homogenous sytstem $Ax=0$, there will be at least one solution $x=0$
**$(d)$ Any system of n linear equations in n unknowns has at most one solution.**
False. Consider the system $0x=0$, there are infinte solutions.
**$(e)$ Any system of n linear equations in n unknowns has at least one solution.**
False. Considerate the system $0x=2$, there is no solution.
**$(f)$ If the homogeneous system corresponding to a given system of linear equations has a solution, then the given system has a solution.**
False. Considerate the system $0x=2$, there is no solution while the homogeneous system correpoding $0x=0$ has infinite solution.
**$(g)$ If the coefficient matrix of a homogeneous system of n linear equations in n unknowns is invertible, then the system has no nonzero
solutions.**
True. Consider the homogeneous sytstmen $Ax=0$, where $A$ is a $n\times n$ invertible matrix. The only solution will be $A^{-1}b$.
**$(h)$ The solution set of any system of m linear equations in n unknowns
is a subspace of $F^n$.**
False. Consider the system $x=2$, the solution set is $\{2\}$, which is not a subspace of $F^n$.
### 2.For each of the following homogeneous systems of linear equations, find the dimension of and a basis for the solution set.
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**$(a)$
\begin{align}
x_1 + 3x_2 &= 0 \\
2x_1 + 6x_2 &= 0
\end{align}**
sol:
First we write the system in matrix notation as follow:
\begin{align}
\underbrace{\begin{bmatrix}
1 & 3\\ 2 & 6
\end{bmatrix}}_{A}
\underbrace{\begin{bmatrix}
x_1 \\ x_2
\end{bmatrix}}_{\phantom{A}x\phantom{A}}&=\underbrace{
\begin{bmatrix} 0\\0
\end{bmatrix}}_{\phantom{A}0\phantom{A}}
\end{align}
Consider the matrix $A$:
\begin{equation}
\begin{bmatrix}
1 & 3\\ 2 & 6
\end{bmatrix}\xrightarrow[R_2: R_2-2R_1]{\hspace{2cm}}\begin{bmatrix}1 & 3\\0 & 0\end{bmatrix}
\end{equation}
Hence $rank(A)=1$.
$\because$ $rank(A)+Null(A)=2$,
$\therefore$ $Null(A)=2-1=1$ is the dimension of the solution set($\because$ the system is homogenous).
Now we consider the reduced matrix$\begin{bmatrix}1&3\\0&0\end{bmatrix}$,
we get the equation $x_1+3x_2=0$, it is trivial to see that $\begin{pmatrix}3\\-1\end{pmatrix}$ is a basis of the solution set.
**$(b)$
\begin{align}
x_1 + x_2 -x_3 &= 0 \\
4x_1 + x_2 -2x_3 &= 0
\end{align}**
sol:
First we write the system in matrix notation as follow:
\begin{align}
\underbrace{\begin{bmatrix}
1 & 1&-1\\ 4 & 1 &-2
\end{bmatrix}}_{A}
\underbrace{\begin{bmatrix}
x_1 \\ x_2 \\ x_3
\end{bmatrix}}_{\phantom{A}x\phantom{A}}&=\underbrace{
\begin{bmatrix} 0\\0
\end{bmatrix}}_{\phantom{A}0\phantom{A}}
\end{align}
Consider the matrix $A$:
\begin{align}
\begin{bmatrix}
1 & 1&-1\\ 4 & 1 &-2
\end{bmatrix}
\xrightarrow[C_1: C_1-C_2 \phantom{A} C_3:C3+C_2]{\hspace{2cm}}
\begin{bmatrix}
0 & 1 & 0\\3 & 1 & -1
\end{bmatrix}
\xrightarrow[C_3:C_3+\frac{1}{3}C_1]{\hspace{2cm}}
\begin{bmatrix}
0 & 1 & 0 \\ 3 & 1 & 0
\end{bmatrix}
\end{align}
Hence $Rank(A)=2$.
$\because Rank(A)+Null(A)=3$
$\therefore Null(A)=3-2=1$ is the dimension of the solution set($\because$ the system is homogenous.）
Now we consider the matrix $A$:
\begin{align}\begin{bmatrix}
1 & 1&-1\\ 4 & 1 &-2
\end{bmatrix}
\xrightarrow[R_2:R_2-R1]{\hspace{2cm}}
\begin{bmatrix}1&1&-1\\3&0&-1\end{bmatrix}
\xrightarrow[R_1:R_1-R2]{\hspace{2cm}}
\begin{bmatrix}
-2 & 1&0\\ 3&0&-1
\end{bmatrix}\end{align}
We get the equation:
\begin{align}
-2x_1 + x_2+0x_3 &= 0 \\
3x_1 +0 x_2-x_3 &= 0
\end{align}
From the first equation we get $x_2=2x_1$,
and from the second equation we get $x_3=3x_1$.
It is trivial to see that $\begin{pmatrix}1\\2\\3\end{pmatrix}$ is a basis of the solution set.
**$(3)$
\begin{align}
x_1+ 2x_2 - x_3 &= 0\\
2x_1 + x_2 + x_3 &= 0
\end{align}**
sol:
First we write the system in matrix notation as follow:
\begin{align}\underbrace{\begin{bmatrix}
1 & 2 & -1 \\
2 & 1 & 1
\end{bmatrix}}_{A}
\underbrace{
\begin{bmatrix}
x_1 \\ x_2 \\ x_3
\end{bmatrix}
}_{x}=
\underbrace{
\begin{bmatrix}
0 \\ 0
\end{bmatrix}
}_{0}
\end{align}
Consider the matrix $A$:
\begin{align}
\begin{bmatrix}
1 & 2 & -1 \\
2 & 1 & 1
\end{bmatrix}
\xrightarrow[R_2:R_2-2R_1]{\hspace{2cm}}
\begin{bmatrix}
1 & 2 & -1 \\
0 & -3 & 3
\end{bmatrix}
\xrightarrow[R_1:R_1+\frac{1}{3}R_2]{\hspace{2cm}}
\begin{bmatrix}
1 & 1 & 0 \\
0 & -3 & 3
\end{bmatrix}
\end{align}
There are two independent columns in the reduced matrix$\begin{bmatrix}
1 & 1 & 0 \\
0 & -3 & 3
\end{bmatrix}$, thus $Rank(A)=2$.
$\because Rank(A)+Null(A)=3$
$\therefore Null(A)= 2-1=1$ is the dimension of the solution set.($\because$ the system is homogenous.）
Now we consider the reduced matrix:
\begin{align}
\begin{bmatrix}
1 & 1 & 0 \\
0 & -3 & 3
\end{bmatrix}
\end{align}
We get the following equations:
\begin{align}
x_1 +1 x_2 &= 0 \\
-3x_2 + 3x_3 &= 0
\end{align}
From the first equation we get $x_1=-x_2$,
and from the second equation we get $x_3=x_2$.
It is trival to see that$\begin{pmatrix}-1\\1\\1\end{pmatrix}$ is a basis for the solution set.
**$(4)$
\begin{align}
2x_1+x_2-x_3&=0\\
x_1-x_2+x_3&=0\\
x_1+2x_2-2x_3&=0
\end{align}**
sol:
First we write the system in matrix notation as follow:
\begin{align}
\underbrace{
\begin{bmatrix}
2&1&-1\\
1&-1&1\\
1&2&-2
\end{bmatrix}
}_{A}\underbrace{
\begin{bmatrix}
x_1\\x_2\\x_3
\end{bmatrix}}_{\phantom{A}x\phantom{A}}&=
\underbrace{
\begin{bmatrix}
0\\0\\0
\end{bmatrix}
}_{0}
\end{align}
Consider the matrix A:
\begin{align}\begin{bmatrix}
2&1&-1\\
1&-1&1\\
1&2&-2
\end{bmatrix}
\xrightarrow[R_2:R_2-\frac{1}{2}R_1,R_3:R_3-\frac{1}{2}R_1]{}
\begin{bmatrix}
2&1&-1\\
0&-3&3\\
0&0&0
\end{bmatrix}
\xrightarrow[R_2:R_2\times(-\frac{1}{3}),R_1=R_1+\frac{1}{3}R_2]{}
\begin{bmatrix}
2&0&0\\
0&1&-1\\
0&0&0
\end{bmatrix}
\end{align}
There are two indepedent columns,thus $rank(A)=2$.
$\because$
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