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General=Particular+Homogeneous

Every solution set splits into one particular solution plus the unrestricted combinations that solve the associated homogeneous system.

Linear Algebra, General=Particular+Homogeneous

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12 exercises

Exercise 1

Solve this system. Then solve the associated homogeneous system.

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Exercise 2

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

  • (a)  3x+6y=18x+2y=6\begin{array}{rcrcr}3x&+&6y&=&18\\ x&+&2y&=&6\end{array}     (b)  x+y=1x−y=−1\begin{array}{rcrcr}x&+&y&=&1\\ x&-&y&=&-1\end{array}     (c)  x1+x3=4x1−x2+2x3=54x1−x2+5x3=17\begin{array}{rcrcrcr}x_{1}&&&+&x_{3}&=&4\\ x_{1}&-&x_{2}&+&2x_{3}&=&5\\ 4x_{1}&-&x_{2}&+&5x_{3}&=&17\end{array}     (d)  2a+b−c=22a+c=3a−b=0\begin{array}{rcrcrcr}2a&+&b&-&c&=&2\\ 2a&&&+&c&=&3\\ a&-&b&&&=&0\end{array}     (e)  x+2y−z=32x+y+w=4x−y+z+w=1\begin{array}{rcrcrcrcr}x&+&2y&-&z&&&=&3\\ 2x&+&y&&&+&w&=&4\\ x&-&y&+&z&+&w&=&1\end{array}     (f)  x+z+w=42x+y−w=23x+y+z=7\begin{array}{rcrcrcrcr}x&&&+&z&+&w&=&4\\ 2x&+&y&&&-&w&=&2\\ 3x&+&y&+&z&&&=&7\end{array}

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Exercise 3

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

  • (a)  2x+y−z=14x−y=3\begin{array}{rcrcrcr}2x&+&y&-&z&=&1\\ 4x&-&y&&&=&3\end{array}     (b)  x−z=1y+2z−w=3x+2y+3z−w=7\begin{array}{rcrcrcrcr}x&&&-&z&&&=&1\\ &&y&+&2z&-&w&=&3\\ x&+&2y&+&3z&-&w&=&7\end{array}     (c)  x−y+z=0y+w=03x−2y+3z+w=0−y−w=0\begin{array}{rcrcrcrcr}x&-&y&+&z&&&=&0\\ &&y&&&+&w&=&0\\ 3x&-&2y&+&3z&+&w&=&0\\ &&-y&&&-&w&=&0\end{array}     (d)  a+2b+3c+d−e=13a−b+c+d+e=3\begin{array}{rcrcrcrcrcr}a&+&2b&+&3c&+&d&-&e&=&1\\ 3a&-&b&+&c&+&d&+&e&=&3\end{array}

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Exercise 4

For the system

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

  • (a)  (0−350)\begin{pmatrix}0\\ -3\\ 5\\ 0\end{pmatrix}     (b)  (2110)\begin{pmatrix}2\\ 1\\ 1\\ 0\end{pmatrix}     (c)  (−1−48−1)\begin{pmatrix}-1\\ -4\\ 8\\ -1\end{pmatrix}

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Exercise 5

Lemma 3.7 says that we can use any particular solution for p⃗\vec{p}. Find, if possible, a general solution to this system

that uses the given vector as its particular solution.

  • (a)  (0004)\begin{pmatrix}0\\ 0\\ 0\\ 4\end{pmatrix}     (b)  (−51−710)\begin{pmatrix}-5\\ 1\\ -7\\ 10\end{pmatrix}     (c)  (2−111)\begin{pmatrix}2\\ -1\\ 1\\ 1\end{pmatrix}

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Exercise 6

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

  • (a)  (134−12)\begin{pmatrix}1&3\\ 4&-12\end{pmatrix}     (b)  (13412)\begin{pmatrix}1&3\\ 4&12\end{pmatrix}

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Exercise 7

Singular or nonsingular?

  • (a)  (1213)\begin{pmatrix}1&2\\ 1&3\end{pmatrix}     (b)  (12−3−6)\begin{pmatrix}1&2\\ -3&-6\end{pmatrix}     (c)  (121131)\begin{pmatrix}1&2&1\\ 1&3&1\end{pmatrix}     (d)  (121113347)\begin{pmatrix}1&2&1\\ 1&1&3\\ 3&4&7\end{pmatrix}     (e)  (221105−114)\begin{pmatrix}2&2&1\\ 1&0&5\\ -1&1&4\end{pmatrix}

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Exercise 8

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

  • (23),\begin{pmatrix}2\\ 3\end{pmatrix}, {(14),(15)}\{\hskip 1.00006pt\begin{pmatrix}1\\ 4\end{pmatrix},\begin{pmatrix}1\\ 5\end{pmatrix}\hskip 1.00006pt\}

  • (−101),\begin{pmatrix}-1\\ 0\\ 1\end{pmatrix}, {(210),(101)}\{\hskip 1.00006pt\begin{pmatrix}2\\ 1\\ 0\end{pmatrix},\begin{pmatrix}1\\ 0\\ 1\end{pmatrix}\hskip 1.00006pt\}

  • (130),\begin{pmatrix}1\\ 3\\ 0\end{pmatrix}, {(104),(215),(330),(421)}\{\hskip 1.00006pt\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}\hskip 1.00006pt\}

  • (1011),\begin{pmatrix}1\\ 0\\ 1\\ 1\end{pmatrix}, {(2101),(3002)}\{\hskip 1.00006pt\begin{pmatrix}2\\ 1\\ 0\\ 1\end{pmatrix},\begin{pmatrix}3\\ 0\\ 0\\ 2\end{pmatrix}\hskip 1.00006pt\}

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Exercise 9

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

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Exercise 10

In the proof of Lemma 3.6, what happens if there are no non-0=00=0 equations?

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Exercise 11

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

  • (a)  s⃗+t⃗\vec{s}+\vec{t}     (b)  3s⃗3\vec{s}     (c)  ks⃗+mt⃗k\vec{s}+m\vec{t} for k,m∈Rk,m\in\mathbb{R}

What’s wrong with this argument: “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.”?

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Exercise 12

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?

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Adapted from Linear Algebra © 2016 Jim Hefferon. Licensed under CC BY-SA 3.0 US. Source: Linear Algebra. Imported at revision df2262e.