# Linear Algebra/General = Particular + Homogeneous/Solutions

## Solutions

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. ${\begin{array}{*{4}{rc}r}x&&&+&z&+&w&=&4\\2x&+&y&&&-&w&=&2\\3x&+&y&+&z&&&=&7\end{array}}$

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. ${\begin{array}{*{5}{rc}r}a&+&2b&+&3c&+&d&-&e&=&1\\3a&-&b&+&c&+&d&+&e&=&3\end{array}}$

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. ${\begin{pmatrix}-1\\-4\\8\\-1\end{pmatrix}}$

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. ${\begin{pmatrix}2\\-1\\1\\1\end{pmatrix}}$

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. ${\begin{pmatrix}1&3\\4&12\end{pmatrix}}$

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

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.

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.)

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."?

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?

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.