# 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.