Linear Algebra/General = Particular + Homogeneous/Solutions

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Solutions[edit]

This exercise is recommended for all readers.
Problem 1

Solve each system. Express the solution set using vectors. Identify the particular solution and the solution set of the homogeneous system.

  1.  \begin{array}{*{2}{rc}r}
3x  &+  &6y  &=  &18  \\
x  &+  &2y  &=  &6
\end{array}
  2.  \begin{array}{*{2}{rc}r}
x  &+  &y   &=  &1  \\
x  &-  &y   &=  &-1
\end{array}
  3.  \begin{array}{*{3}{rc}r}
x_1  &   &     &+  &x_3   &=  &4  \\
x_1  &-  &x_2  &+  &2x_3  &=  &5  \\
4x_1  &-  &x_2  &+  &5x_3  &=  &17
\end{array}
  4.  \begin{array}{*{3}{rc}r}
2a   &+  &b    &-  &c     &=  &2  \\
2a   &   &     &+  &c     &=  &3  \\
a   &-  &b    &   &      &=  &0
\end{array}
  5.  \begin{array}{*{4}{rc}r}
x  &+  &2y   &-   &z   &    &    &=  &3  \\
2x  &+  &y    &    &    &+   &w   &=  &4  \\
x  &-  &y    &+   &z   &+   &w   &=  &1
\end{array}
  6.  \begin{array}{*{4}{rc}r}
x  &   &     &+   &z   &+   &w   &=  &4  \\
2x  &+  &y    &    &    &-   &w   &=  &2  \\
3x  &+  &y    &+   &z   &    &    &=  &7
\end{array}
Answer

For the arithmetic to these, see the answers from the prior subsection.

  1. The solution set is
    
\{\begin{pmatrix} 6 \\ 0 \end{pmatrix}+\begin{pmatrix} -2 \\ 1 \end{pmatrix}y
\,\big|\, y\in\mathbb{R}\}.
    Here the particular solution and the solution set for the associated homogeneous system are
    
\begin{pmatrix} 6 \\ 0 \end{pmatrix}
\quad\text{and}\quad
\{\begin{pmatrix} -2 \\ 1 \end{pmatrix}y
\,\big|\, y\in\mathbb{R}\}.
  2. The solution set is
    
\{\begin{pmatrix} 0 \\ 1 \end{pmatrix} \}.
    The particular solution and the solution set for the associated homogeneous system are
    
\begin{pmatrix} 0 \\ 1 \end{pmatrix}
\quad\text{and}\quad
\{\begin{pmatrix} 0 \\ 0 \end{pmatrix} \}
  3. The solution set is
    
\{\begin{pmatrix} 4 \\ -1 \\ 0 \end{pmatrix}+\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}x_3
\,\big|\, x_3\in\mathbb{R}\}.
    A particular solution and the solution set for the associated homogeneous system are
    
\begin{pmatrix} 4 \\ -1 \\ 0 \end{pmatrix}
\quad\text{and}\quad
\{\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}x_3
\,\big|\, x_3\in\mathbb{R}\}.
  4. The solution set is a singleton
    
\{\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\}.
    A particular solution and the solution set for the associated homogeneous system are
    
\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}
\quad\text{and}\quad
\{\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}t
\,\big|\, t\in\mathbb{R}\}.
  5. The solution set is
    
\{\begin{pmatrix} 5/3 \\ 2/3 \\ 0 \\ 0 \end{pmatrix}
+\begin{pmatrix} -1/3 \\ 2/3 \\ 1 \\ 0 \end{pmatrix}z
+\begin{pmatrix} -2/3 \\ 1/3 \\ 0 \\ 1 \end{pmatrix}w
\,\big|\, z,w\in\mathbb{R}\}.
    A particular solution and the solution set for the associated homogeneous system are
    
\begin{pmatrix} 5/3 \\ 2/3 \\ 0 \\ 0 \end{pmatrix}
\quad\text{and}\quad
\{\begin{pmatrix} -1/3 \\ 2/3 \\ 1 \\ 0 \end{pmatrix}z
+\begin{pmatrix} -2/3 \\ 1/3 \\ 0 \\ 1 \end{pmatrix}w
\,\big|\, z,w\in\mathbb{R}\}.
  6. This system's solution set is empty. Thus, there is no particular solution. The solution set of the associated homogeneous system is
    
\{\begin{pmatrix} -1 \\ 2 \\ 1 \\ 0 \end{pmatrix}z
+\begin{pmatrix} -1 \\ 3 \\ 0 \\ 1 \end{pmatrix}w
\,\big|\, z,w\in\mathbb{R}\}.
Problem 2

Solve each system, giving the solution set in vector notation. Identify the particular solution and the solution of the homogeneous system.

  1.  \begin{array}{*{3}{rc}r}
2x  &+  &y  &-  &z  &=  &1  \\
4x  &-  &y  &   &   &=  &3
\end{array}
  2.  \begin{array}{*{4}{rc}r}
x  &   &   &-  &z  &   &   &=  &1  \\
&   &y  &+  &2z &-  &w  &=  &3  \\
x  &+  &2y &+  &3z &-  &w  &=  &7
\end{array}
  3.  \begin{array}{*{4}{rc}r}
x  &-  &y  &+  &z  &   &   &=  &0  \\
&   &y  &   &   &+  &w  &=  &0  \\
3x  &-  &2y &+  &3z &+  &w  &=  &0  \\
&   &-y &   &   &-  &w  &=  &0
\end{array}
  4.  \begin{array}{*{5}{rc}r}
a  &+  &2b &+  &3c &+  &d  &-  &e  &=  &1  \\
3a  &-  &b  &+  &c  &+  &d  &+  &e  &=  &3
\end{array}
Answer

The answers from the prior subsection show the row operations.

  1. The solution set is
    
\{\begin{pmatrix} 2/3 \\ -1/3 \\ 0 \end{pmatrix}
+\begin{pmatrix} 1/6 \\ 2/3 \\ 1 \end{pmatrix}z
\,\big|\, z\in\mathbb{R}\}.
    A particular solution and the solution set for the associated homogeneous system are
    
\begin{pmatrix} 2/3 \\ -1/3 \\ 0 \end{pmatrix}
\quad\text{and}\quad
\{\begin{pmatrix} 1/6 \\ 2/3 \\ 1 \end{pmatrix}z
\,\big|\, z\in\mathbb{R}\}.
  2. The solution set is
    
\{\begin{pmatrix} 1 \\ 3 \\ 0 \\ 0 \end{pmatrix}
+\begin{pmatrix} 1 \\-2 \\ 1 \\ 0 \end{pmatrix}z
\,\big|\, z\in\mathbb{R}\}.
    A particular solution and the solution set for the associated homogeneous system are
    
\begin{pmatrix} 1 \\ 3 \\ 0 \\ 0 \end{pmatrix}
\quad\text{and}\quad
\{\begin{pmatrix} 1 \\ -2 \\ 1 \\ 0 \end{pmatrix}z
\,\big|\, z\in\mathbb{R}\}.
  3. The solution set is
    
\{\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}
+\begin{pmatrix} -1 \\ 0 \\ 1 \\ 0 \end{pmatrix}z
+\begin{pmatrix} -1 \\ -1 \\ 0 \\ 1 \end{pmatrix}w
\,\big|\, z,w\in\mathbb{R}\}.
    A particular solution and the solution set for the associated homogeneous system are
    
\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}
\quad\text{and}\quad
\{\begin{pmatrix} -1 \\ 0 \\ 1 \\ 0 \end{pmatrix}z
+\begin{pmatrix} -1 \\ -1 \\ 0 \\ 1 \end{pmatrix}w
\,\big|\, z,w\in\mathbb{R}\}.
  4. The solution set is
    
\{\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}
+\begin{pmatrix} -5/7 \\ -8/7 \\ 1 \\ 0 \\ 0 \end{pmatrix}c
+\begin{pmatrix} -3/7 \\ -2/7 \\ 0 \\ 1 \\ 0 \end{pmatrix}d
+\begin{pmatrix} -1/7 \\ 4/7 \\ 0 \\ 0 \\ 1 \end{pmatrix}e
\,\big|\, c,d,e\in\mathbb{R}\}.
    A particular solution and the solution set for the associated homogeneous system are
    
\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}
\quad\text{and}\quad
\{\begin{pmatrix} -5/7 \\ -8/7 \\ 1 \\ 0 \\ 0 \end{pmatrix}c
+\begin{pmatrix} -3/7 \\ -2/7 \\ 0 \\ 1 \\ 0 \end{pmatrix}d
+\begin{pmatrix} -1/7 \\ 4/7 \\ 0 \\ 0 \\ 1 \end{pmatrix}e
\,\big|\, c,d,e\in\mathbb{R}\}.
This exercise is recommended for all readers.
Problem 3

For the system


\begin{array}{*{4}{rc}r}
2x  &-  &y  &   &    &-  &w  &=  &3  \\
&   &y  &+  &z   &+  &2w &=  &2  \\
x  &-  &2y &-  &z   &   &   &=  &-1
\end{array}

which of these can be used as the particular solution part of some general solution?

  1.  \begin{pmatrix} 0 \\ -3 \\ 5 \\ 0 \end{pmatrix}
  2.  \begin{pmatrix} 2 \\ 1 \\ 1 \\ 0 \end{pmatrix}
  3.  \begin{pmatrix} -1 \\ -4 \\ 8 \\ -1 \end{pmatrix}
Answer

Just plug them in and see if they satisfy all three equations.

  1. No.
  2. Yes.
  3. Yes.
This exercise is recommended for all readers.
Problem 4

Lemma 3.8 says that any particular solution may be used for \vec{p}. Find, if possible, a general solution to this system


\begin{array}{*{4}{rc}r}
x  &-  &y  &   &    &+  &w  &=  &4  \\
2x  &+  &3y &-  &z   &   &   &=  &0  \\
&   &y  &+  &z   &+  &w  &=  &4
\end{array}

that uses the given vector as its particular solution.

  1.  \begin{pmatrix} 0 \\ 0 \\ 0 \\ 4 \end{pmatrix}
  2.  \begin{pmatrix} -5 \\ 1 \\ -7 \\ 10 \end{pmatrix}
  3.  \begin{pmatrix} 2 \\ -1 \\ 1 \\ 1 \end{pmatrix}
Answer

Gauss' method on the associated homogeneous system gives


\left(\begin{array}{*{4}{c}|c}
1  &-1  &0  &1  &0  \\
2  &3   &-1 &0  &0  \\
0  &1   &1  &1  &0
\end{array}\right)
\;\xrightarrow[]{-2\rho_1+\rho_2}\;
\left(\begin{array}{*{4}{c}|c}
1  &-1  &0  &1  &0  \\
0  &5   &-1 &-2 &0  \\
0  &1   &1  &1  &0
\end{array}\right)
\;\xrightarrow[]{-(1/5)\rho_2+\rho_3}\;
\left(\begin{array}{*{4}{c}|c}
1  &-1  &0  &1  &0  \\
0  &5   &-1 &-2 &0  \\
0  &0   &6/5&7/5&0
\end{array}\right)

so this is the solution to the homogeneous problem:


\{\begin{pmatrix} -5/6 \\ 1/6 \\ -7/6 \\ 1 \end{pmatrix}w\,\big|\, w\in\mathbb{R}\}.
  1. That vector is indeed a particular solution, so the required general solution is
    
\{\begin{pmatrix} 0 \\ 0 \\ 0 \\ 4 \end{pmatrix}+
\begin{pmatrix} -5/6 \\ 1/6 \\ -7/6 \\ 1 \end{pmatrix}w\,\big|\, w\in\mathbb{R}\}.
  2. That vector is a particular solution so the required general solution is
    
\{\begin{pmatrix} -5 \\ 1 \\ -7 \\ 10 \end{pmatrix}+
\begin{pmatrix} -5/6 \\ 1/6 \\ -7/6 \\ 1 \end{pmatrix}w\,\big|\, w\in\mathbb{R}\}.
  3. That vector is not a solution of the system since it does not satisfy the third equation. No such general solution exists.
Problem 5

One of these is nonsingular while the other is singular. Which is which?

  1. \begin{pmatrix}
1  &3   \\
4  &-12
\end{pmatrix}
  2. \begin{pmatrix}
1  &3  \\
4  &12
\end{pmatrix}
Answer

The first is nonsingular while the second is singular. Just do Gauss' method and see if the echelon form result has non-0 numbers in each entry on the diagonal.

This exercise is recommended for all readers.
Problem 6

Singular or nonsingular?

  1. 
\begin{pmatrix}
1  &2  \\
1  &3
\end{pmatrix}
  2. 
\begin{pmatrix}
1  &2  \\
-3  &-6
\end{pmatrix}
  3. 
\begin{pmatrix}
1  &2  &1  \\
1  &3  &1
\end{pmatrix}   (Careful!)
  4. 
\begin{pmatrix}
1  &2  &1  \\
1  &1  &3  \\
3  &4  &7
\end{pmatrix}
  5. 
\begin{pmatrix}
2  &2  &1  \\
1  &0  &5  \\
-1  &1  &4
\end{pmatrix}
Answer
  1. Nonsingular:
    \begin{array}{rcl}
&\xrightarrow[]{-\rho_1+\rho_2}
&\begin{pmatrix}
1  &2  \\
0  &1
\end{pmatrix}
\end{array}
    ends with each row containing a leading entry.
  2. Singular:
    \begin{array}{rcl}
&\xrightarrow[]{3\rho_1+\rho_2}
&\begin{pmatrix}
1  &2  \\
0  &0
\end{pmatrix}
\end{array}
    ends with row  2 without a leading entry.
  3. Neither. A matrix must be square for either word to apply.
  4. Singular.
  5. Nonsingular.
This exercise is recommended for all readers.
Problem 7

Is the given vector in the set generated by the given set?

  1.  \begin{pmatrix} 2 \\ 3 \end{pmatrix},  \{\begin{pmatrix} 1 \\ 4 \end{pmatrix},
\begin{pmatrix} 1 \\ 5 \end{pmatrix}\}
  2.  \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix},  \{\begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix},
\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\}
  3.  \begin{pmatrix} 1 \\ 3 \\ 0 \end{pmatrix},  \{\begin{pmatrix} 1 \\ 0 \\ 4 \end{pmatrix},
\begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix},
\begin{pmatrix} 3 \\ 3 \\ 0 \end{pmatrix},
\begin{pmatrix} 4 \\ 2 \\ 1 \end{pmatrix}\}
  4.  \begin{pmatrix} 1 \\ 0 \\ 1 \\ 1 \end{pmatrix},  \{\begin{pmatrix} 2 \\ 1 \\ 0 \\ 1 \end{pmatrix},
\begin{pmatrix} 3 \\ 0 \\ 0 \\ 2 \end{pmatrix}\}
Answer

In each case we must decide if the vector is a linear combination of the vectors in the set.

  1. Yes. Solve
    
c_1\begin{pmatrix} 1 \\ 4 \end{pmatrix}+c_2\begin{pmatrix} 1 \\ 5 \end{pmatrix}=\begin{pmatrix} 2 \\ 3 \end{pmatrix}
    with
    \begin{array}{rcl}
\left(\begin{array}{*{2}{c}|c}
1  &1  &2  \\
4  &5  &3
\end{array}\right)
&\xrightarrow[]{-4\rho_1+\rho_2}
&\left(\begin{array}{*{2}{c}|c}
1  &1  &2  \\
0  &1  &-5
\end{array}\right)
\end{array}
    to conclude that there are c_1 and c_2 giving the combination.
  2. No. The reduction
    
\left(\begin{array}{*{2}{c}|c}
2  &1  &-1 \\
1  &0  &0  \\
0  &1  &1
\end{array}\right)
\;\xrightarrow[]{-(1/2)\rho_1+\rho_2}\;
\left(\begin{array}{*{2}{c}|c}
2  &1     &-1 \\
0  &-1/2  &1/2  \\
0  &1     &1
\end{array}\right)
\;\xrightarrow[]{2\rho_2+\rho_3}\;
\left(\begin{array}{*{2}{c}|c}
2  &1     &-1 \\
0  &-1/2  &1/2  \\
0  &0     &2
\end{array}\right)
    shows that
    
c_1\begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}+c_2\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}
=\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}
    has no solution.
  3. Yes. The reduction
    
\left(\begin{array}{*{4}{c}|c}
1  &2  &3  &4  &1  \\
0  &1  &3  &2  &3  \\
4  &5  &0  &1  &0
\end{array}\right)
\;\xrightarrow[]{-4\rho_1+\rho_3}\;
\left(\begin{array}{*{4}{c}|c}
1  &2  &3  &4  &1  \\
0  &1  &3  &2  &3  \\
0  &-3 &-12&-15&-4
\end{array}\right)
\;\xrightarrow[]{3\rho_2+\rho_3}\;
\left(\begin{array}{*{4}{c}|c}
1  &2  &3  &4  &1  \\
0  &1  &3  &2  &3  \\
0  &0  &-3 &-9 &5
\end{array}\right)
    shows that there are infinitely many ways
    
\{\begin{pmatrix} c_1 \\ c_2 \\ c_3 \\ c_4 \end{pmatrix}=
\begin{pmatrix} -10 \\ 8 \\ -5/3 \\ 0 \end{pmatrix}+
\begin{pmatrix} -9 \\ 7 \\ -3 \\ 1 \end{pmatrix}c_4
\,\big|\, c_4\in\mathbb{R}\}
    to write
    
\begin{pmatrix} 1 \\ 3 \\ 0 \end{pmatrix}=
c_1\begin{pmatrix} 1 \\ 0 \\ 4 \end{pmatrix}+
c_2\begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix}+
c_3\begin{pmatrix} 3 \\ 3 \\ 0 \end{pmatrix}+
c_4\begin{pmatrix} 4 \\ 2 \\ 1 \end{pmatrix}.
  4. No. Look at the third components.
Problem 8

Prove that any linear system with a nonsingular matrix of coefficients has a solution, and that the solution is unique.

Answer

Because the matrix of coefficients is nonsingular, Gauss' method ends with an echelon form where each variable leads an equation. Back substitution gives a unique solution.

(Another way to see the solution is unique is to note that with a nonsingular matrix of coefficients the associated homogeneous system has a unique solution, by definition. Since the general solution is the sum of a particular solution with each homogeneous solution, the general solution has (at most) one element.)

Problem 9

To tell the whole truth, there is another tricky point to the proof of Lemma 3.7. What happens if there are no non-" 0=0 " equations? (There aren't any more tricky points after this one.)

Answer

In this case the solution set is all of  \mathbb{R}^n , and can be expressed in the required form


\{c_1\begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix}
+c_2\begin{pmatrix} 0 \\ 1 \\ \vdots \\ 0 \end{pmatrix}
+\cdots
+c_n\begin{pmatrix} 0 \\ 0 \\ \vdots \\ 1 \end{pmatrix}
\,\big|\, c_1,\ldots,c_n\in\mathbb{R}\}.
This exercise is recommended for all readers.
Problem 10

Prove that if  \vec{s} and  \vec{t} satisfy a homogeneous system then so do these vectors.

  1.  \vec{s}+\vec{t}
  2.  3\vec{s}
  3.  k\vec{s}+m\vec{t} for  k,m\in\mathbb{R}

What's wrong with: "These three show that if a homogeneous system has one solution then it has many solutions— any multiple of a solution is another solution, and any sum of solutions is a solution also— so there are no homogeneous systems with exactly one solution."?

Answer

Assume  \vec{s},\vec{t}\in\mathbb{R}^n and write


\vec{s}=\begin{pmatrix} s_1 \\ \vdots \\ s_n \end{pmatrix}
\quad\mbox{and}\quad
\vec{t}=\begin{pmatrix} t_1 \\ \vdots \\ t_n \end{pmatrix}.

Also let  a_{i,1}x_1+\cdots+a_{i,n}x_n=0 be the  i -th equation in the homogeneous system.

  1. The check is easy:
    \begin{array}{rcl}
a_{i,1}(s_1+t_1)+\cdots+a_{i,n}(s_n+t_n)
&=
&(a_{i,1}s_1+\cdots+a_{i,n}s_n)
+(a_{i,1}t_1+\cdots+a_{i,n}t_n)         \\
&=
&0+0.
\end{array}
  2. This one is similar:
    
a_{i,1}(3s_1)+\cdots+a_{i,n}(3s_n)
=3(a_{i,1}s_1+\cdots+a_{i,n}s_n)
=3\cdot 0=0.
  3. This one is not much harder:
    \begin{array}{rcl}
a_{i,1}(ks_1+mt_1)+\cdots+a_{i,n}(ks_n+mt_n)
&=
&k(a_{i,1}s_1+\cdots+a_{i,n}s_n)
+m(a_{i,1}t_1+\cdots+a_{i,n}t_n)         \\
&=
&k\cdot 0+m\cdot 0.
\end{array}

What is wrong with that argument is that any linear combination of the zero vector yields the zero vector again.

Problem 11

Prove that if a system with only rational coefficients and constants has a solution then it has at least one all-rational solution. Must it have infinitely many?

Answer

First the proof.

Gauss' method will use only rationals (e.g.,  -(m/n)\rho_i+\rho_j ). Thus the solution set can be expressed using only rational numbers as the components of each vector. Now the particular solution is all rational.

There are infinitely many (rational vector) solutions if and only if the associated homogeneous system has infinitely many (real vector) solutions. That's because setting any parameters to be rationals will produce an all-rational solution.