My Math Homework

Linear Algebra and Its Applications, Exercise 2.2.9

Posted on August 29, 2011 by hecker

Exercise 2.2.9. Consider the following $A$ and $b$ :

$A = \begin{bmatrix} 1&2&0&3 \\ 2&4&0&7 \end{bmatrix} \qquad b = \begin{bmatrix} b_1 \\ b_2 \end{bmatrix}$

For what values of $b_1$ and $b_2$ does the system $Ax = b$ have a solution? Also, determine the nullspace of $A$ and provide two examples of vectors within it, and find the general solution to $Ax = b$ .

Answer: We perform elimination by subtracting 2 times the first row from the third row:

$\begin{bmatrix} 1&2&0&3 \\ 2&4&0&7 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&2&0&3 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 - 2b_1 \end{bmatrix}$

This completes elimination and produces the system $Ux = c$ :

$\begin{bmatrix} 1&2&0&3 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 - 2b_1 \end{bmatrix}$

Since the pivots of $U$ are in columns 1 and 4, the basic variables are $u$ and $y$ and the free variables are $v$ and $w$ . There are no constraints on the values of $b_1$ and $b_2$ .

To find the general solution to $Ax = b$ we start with the system $Ax = 0$ , which is equivalent to $Ux = 0$ :

$\begin{bmatrix} 1&2&0&3 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

From the second equation we must have $y = 0$ . Substituting the value of $y$ into the first equation we have $u + 2v = 0$ or $u = -2v$ . The general solution to $Ax = 0$ can then be expressed in terms of the free variables $v$ and $w$ as follows:

$x_{homogeneous} = v \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + w \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}$

By setting $v = 1$ and $w = 0$ we obtain one vector $(-2, 1, 0, 0)$ in the nullspace of $A$ and by setting $v = 0$ and $w = 1$ we obtain another vector $(0, 0, 1, 0)$ also in the nullspace.

We can obtain a particular solution to $Ax = b$ by going back to the system $Ux = c$ derived above:

$\begin{bmatrix} 1&2&0&3 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 - 2b_1 \end{bmatrix}$

From the second equation we have $y = b_2 - 2b_1$ . Substituting for $y$ into the first equation we have $u + 2v + 3(b_2 - 2b_1) = u + 2v +3b_2 - 6b_1 = b_1$ or $u = -2v + 7b_1 - 3b_2$ . Setting the free variables $v$ and $w$ to zero gives the particular solution

$x_{particular} = \begin{bmatrix} 7b_1 - 3b_2 \\ 0 \\ 0 \\ -2b_1 + b_2 \end{bmatrix}$

The general solution to $Ax = b$ is then

$x = x_{particular} + x_{homogeneous} = \begin{bmatrix} 7b_1 - 3b_2 \\ 0 \\ 0 \\ -2b_1 + b_2 \end{bmatrix} + v \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + w \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}$

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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Linear Algebra and Its Applications, Exercise 2.2.8

Posted on August 28, 2011 by hecker

Exercise 2.2.8. Consider the following system of linear equations:

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcl}u&+&v&+&2w&=&2 \\ 2u&+&3v&-&w&=&5 \\ 3u&+&4v&+&w&=&c \end{array}$

For what value of $c$ does this system have a solution?

Answer: We use elimination to attempt to solve the system, starting by multiplying the first equation by 2 and subtracting it from the second, and multiplying the first equation by 3 and subtracting it from the third:

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcl}u&+&v&+&2w&=&2 \\ 2u&+&3v&-&w&=&5 \\ 3u&+&4v&+&w&=&c \end{array} \Rightarrow \begin{array}{rcrcrcl}u&+&v&+&2w&=&2 \\ &&v&-&5w&=&1 \\ &&v&-&5w&=&c-6 \end{array}$

and then substract 1 times the second equation from the third:

$\begin{array}{rcrcrcl}u&+&v&+&2w&=&2 \\ &&v&-&5w&=&1 \\ &&v&-&5w&=&c-6 \end{array} \Rightarrow \begin{array}{rcrcrcl}u&+&v&+&2w&=&2 \\ &&v&-&5w&=&1 \\ &&&&0&=&c-7 \end{array}$

In order for this system to have a solution we must have $c - 7 = 0$ or $c = 7$ .

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

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Linear Algebra and Its Applications, Exercise 2.2.7

Posted on August 27, 2011 by hecker

Exercise 2.2.7. Consider the following system of linear equations:

$\begin{bmatrix} 1&0 \\ 0&1 \\ 2&3 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}$

Find the values of $b$ for which the system has a solution. What is the rank? How many basic variables and free variables are there?

Answer: We perform elimination by first subtracting 2 times the first row from the third row:

$\begin{bmatrix} 1&0 \\ 0&1 \\ 2&3 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&0 \\ 0&1 \\ 0&3 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 - 2b_1 \end{bmatrix}$

and then subtracting 3 times the second row from the third:

$\begin{bmatrix} 1&0 \\ 0&1 \\ 0&3 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 - 2b_1 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&0 \\ 0&1 \\ 0&0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 - 2b_1 -3b_2 \end{bmatrix}$

This completes elimination and produces the system $Ux = c$ .

From the third equation we must have $b_3 -2b_1 -3b_2 = 0$ in order for the system to have a solution. The matrix $U$ has pivots in columns 1 and 2, so that the basic variables are $u$ and $v$ ; there are no free variables. The rank is 2 (the number of basic variables).

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

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Linear Algebra and Its Applications, Exercise 2.2.6

Posted on August 26, 2011 by hecker

Exercise 2.2.6. Consider the following system of linear equations:

$\begin{bmatrix} 1&2&2 \\ 2&4&5 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} 1 \\ 4 \end{bmatrix}$

Find the general solution expressed as the sum of a particular solution to $Ax = b$ and the solution to $Ax = 0$ .

Answer: We perform elimination by subtracting 2 times the first row from the second row:

$\begin{bmatrix} 1&2&2 \\ 2&4&5 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&2&2 \\ 0&0&1 \end{bmatrix}$

This completes elimination. The matrix $U$ has pivots in columns 1 and 3, so that the basic variables are $u$ and $w$ and the free variable is $v$ .

We can replace the system $Ax = 0$ with the new system $Ux = 0$ :

$\begin{bmatrix} 1&2&2 \\ 0&0&1 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

From the second equation we have $w = 0$ . Substituting into the first equation gives us $u +2v + 2w = u + 2v = 0$ or $u = -2v$ . The general solution to $Ax = 0$ is thus

$x = \begin{bmatrix} -2v \\ v \\ 0 \end{bmatrix} = v \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix}$

We now consider the inhomogeneous system $Ax = b$ or

$\begin{bmatrix} 1&2&2 \\ 2&4&5 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} 1 \\ 4 \end{bmatrix}$

The elimination sequence from above produces the system $Ux = c$ or

$\begin{bmatrix} 1&2&2 \\ 0&0&1 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$

From the second equation we have $w = 2$ . Substituting into the first equation and setting the free variable $v = 0$ produces $u + 2v + 2w = u + 0 + 4 = 1$ or $u = -3$ . The particular solution is thus $x = (-3, 0, 2)$ .

We can combine the particular solution to this system with the general solution to $Ax = 0$ to produce the general solution $x$ for the system $Ax = b$ :

$x = \begin{bmatrix} -3 \\ 0 \\ 2 \end{bmatrix} + v \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix}$

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

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Linear Algebra and Its Applications, Exercise 2.2.5

Posted on August 25, 2011 by hecker

Exercise 2.2.5. Consider the system of linear equations represented by the following matrix:

$A = \begin{bmatrix} 0&0 \\ 1&2 \\ 4&8 \\ 0&0 \end{bmatrix}$

(This is the transpose of the matrix from exercise 2.2.4.) Find the echelon matrix $U$ , a set of basic variables, a set of free variables, and the general solution to $Ax = 0$ . Then use elimination to find when the system $Ax = b$ has a solution, and express that solution as the sum of a particular solution and the general solution to $Ax = 0$ . Finally, find the rank of $A$ .

Answer: We must do a row exchange to exchange the first and second rows. This is equivalent to multiplying by a permutation matrix $P$ as follows:

$PA = \begin{bmatrix} 0&1&0&0 \\ 1&0&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} 0&0 \\ 1&2 \\ 4&8 \\ 0&0 \end{bmatrix} = \begin{bmatrix} 1&2 \\ 0&0 \\ 4&8 \\ 0&0 \end{bmatrix}$

We then perform elimination on $PA$ by subtracting 4 times the first row from the third row (i.e., using the multiplier $l_{31} = 4$ ):

$\begin{bmatrix} 1&2 \\ 0&0 \\ 4&8 \\ 0&0 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&2 \\ 0&0 \\ 0&0 \\ 0&0 \end{bmatrix}$

This completes elimination, and leaves us with the echelon matrix

$U = \begin{bmatrix} 1&2 \\ 0&0 \\ 0&0 \\ 0&0 \end{bmatrix}$

and the factorization $PA = LU$ :

$\begin{bmatrix} 0&1&0&0 \\ 1&0&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} 0&0 \\ 1&2 \\ 4&8 \\ 0&0 \end{bmatrix} = \begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 4&0&1&0 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} 1&2 \\ 0&0 \\ 0&0 \\ 0&0 \end{bmatrix}$

We now solve for

$Ax = \begin{bmatrix} 0&0 \\ 1&2 \\ 4&8 \\ 0&0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = 0$

which is equivalent to

$Ux = \begin{bmatrix} 1&2 \\ 0&0 \\ 0&0 \\ 0&0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = 0$

The pivot in $U$ is in column 1, so the (only) basic variable is $u$ and the free variable is $v$ .

From the first row of $U$ we have $u + 2v = 0$ and thus $u = -2v$ . We then have

$x = \begin{bmatrix} -2v \\ v \end{bmatrix} = v \begin{bmatrix} -2 \\ 1 \end{bmatrix}$

In considering the inhomogeneous system $Ax = b$ or

$\begin{bmatrix} 0&0 \\ 1&2 \\ 4&8 \\ 0&0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \\ b_4 \end{bmatrix}$

the elimination sequence above (including the initial row exchange) would produce the system $Ux = c$ or

$\begin{bmatrix} 1&2 \\ 0&0 \\ 0&0 \\ 0&0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} b_2 \\ b_1 \\ b_3 - 4b_2 \\ b_4 \end{bmatrix} \Rightarrow \begin{bmatrix} u+2v \\ 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} b_2 \\ b_1 \\ b_3 - 4b_2 \\ b_4 \end{bmatrix}$

From the second and fourth equations we must have $b_1 = b_4 = 0$ . From the third equation we must have $b_3 - 4b_2 = 0$ or $b_3 = 4b_2$ . Thus for $Ax = b$ to have a solution the vector $b$ must have the form $(0, b_2, 4b_2, 0)$ .

Taking $b = (0, b_2, 4b_2, 0)$ produces the system

$\begin{bmatrix} 0&0 \\ 1&2 \\ 4&8 \\ 0&0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} 0 \\ b_2 \\ 4b_2 \\ 0 \end{bmatrix}$

which after row exchange and elimination becomes

$\begin{bmatrix} 1&2 \\ 0&0 \\ 0&0 \\ 0&0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} b_2 \\ 0 \\ 0 \\ 0 \end{bmatrix}$

The first equation gives $u + 2v = b_2$ or $u = b_2 - 2v$ . Setting the free variable $v$ to zero produces the particular solution $x = (u, v) = (b_2, 0)$ to the system

$\begin{bmatrix} 0&0 \\ 1&2 \\ 4&8 \\ 0&0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} 0 \\ b_2 \\ 4b_2 \\ 0 \end{bmatrix}$

We can combine the particular solution to this system with the solution to $Ax = 0$ to produce the general solution $x$ for the system $Ax = b$

$x = \begin{bmatrix} b_2 \\ 0 \end{bmatrix} + v \begin{bmatrix} -2 \\ 1 \end{bmatrix}$

We can check this solution as follows:

$Ax = \begin{bmatrix} 0&0 \\ 1&2 \\ 4&8 \\ 0&0 \end{bmatrix} \left( \begin{bmatrix} b_2 \\ 0 \end{bmatrix} + v \begin{bmatrix} -2 \\ 1 \end{bmatrix} \right) = \begin{bmatrix} 0&0 \\ 1&2 \\ 4&8 \\ 0&0 \end{bmatrix} \begin{bmatrix} b_2 - 2v \\ v \end{bmatrix}$

$= \begin{bmatrix} 0 \\ (b_2 - 2v) + 2v \\ 4(b_2 - 2v) + 8v \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ b_2 - 2v + 2v \\ 4b_2 - 8v + 8v \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ b_2 \\ 4b_2 \\ 0 \end{bmatrix}$

The rank of $A$ is 1, the number of basic variables (or pivots).

UPDATE: Corrected a typo found by Argiris.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

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Linear Algebra and Its Applications, Exercise 2.2.4

Posted on August 24, 2011 by hecker

Exercise 2.2.4. Consider the system of linear equations represented by the following matrix:

$A = \begin{bmatrix} 0&1&4&0 \\ 0&2&8&0 \end{bmatrix}$

Find the echelon matrix $U$ , a set of basic variables, a set of free variables, and the general solution to $Ax = 0$ . Then use elimination to find when the system $Ax = b$ has a solution, and express that solution as the sum of a particular solution and the general solution to $Ax = 0$ . Finally, find the rank of $A$ .

Answer: We perform elimination on $A$ by subtracting 2 times the first row from the second row (i.e., using the multiplier $l_{21} = 2$ ):

$\begin{bmatrix} 0&1&4&0 \\ 0&2&8&0 \end{bmatrix} \Rightarrow \begin{bmatrix} 0&1&4&0 \\ 0&0&0&0 \end{bmatrix}$

This completes elimination, and leaves us with the echelon matrix

$U = \begin{bmatrix} 0&1&4&0 \\ 0&0&0&0 \end{bmatrix}$

and the factorization $A = LU$ :

$\begin{bmatrix} 0&1&4&0 \\ 0&2&8&0 \end{bmatrix} = \begin{bmatrix} 1&0 \\ 2&1 \end{bmatrix} \begin{bmatrix} 0&1&4&0 \\ 0&0&0&0 \end{bmatrix}$

We now solve for

$Ax = \begin{bmatrix} 0&1&4&0 \\ 0&2&8&0 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = 0$

which is equivalent to

$Ux = \begin{bmatrix} 0&1&4&0 \\ 0&0&0&0 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = 0$

The pivot in $U$ is in column 2, so the (only) basic variable is $v$ and the free variables are $u$ , $w$ and $y$ .

From the first row of $U$ we have $v + 4w = 0$ and thus $v = -4w$ . We then have

$x = \begin{bmatrix} u \\ -4w \\ w \\ y \end{bmatrix} = u \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + w \begin{bmatrix} 0 \\ -4 \\ 1 \\ 0 \end{bmatrix} + y \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$

In considering the inhomogeneous system $Ax = b$ or

$\begin{bmatrix} 0&1&4&0 \\ 0&2&8&0 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \end{bmatrix}$

the elimination sequence above would produce the system $Ux = c$ or

$\begin{bmatrix} 0&1&4&0 \\ 0&0&0&0 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 - 2b_1 \end{bmatrix}$

From the second equation we must have $b_2 - 2b_1 = 0$ or $b_2 = 2b_1$ . Thus for $Ax = b$ to have a solution the vector $b$ must lie on the line passing through the origin and the point $(1, 2)$ so that $b = (b_1, 2b_1)$ .

Taking $b = (b_1, 2b_2)$ produces the system

$\begin{bmatrix} 0&1&4&0 \\ 0&2&8&0 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = \begin{bmatrix} b_1 \\ 2b_1 \end{bmatrix}$

which after elimination becomes

$\begin{bmatrix} 0&1&4&0 \\ 0&0&0&0 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = \begin{bmatrix} b_1 \\ 0 \end{bmatrix}$

The first equation gives $v + 4w = b_1$ or $v = b_1 - 4w$ . Setting the free variables $u$ , $w$ , and $y$ all to zero produces the particular solution $x = (0, b_1, 0, 0)$ to the system

$\begin{bmatrix} 0&1&4&0 \\ 0&2&8&0 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = \begin{bmatrix} b_1 \\ 2b_1 \end{bmatrix}$

We can combine the particular solution to this system with the solution to $Ax = 0$ to produce the general solution for the system

$x = \begin{bmatrix} 0 \\ b_1 \\ 0 \\ 0 \end{bmatrix} + u \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + w \begin{bmatrix} 0 \\ -4 \\ 1 \\ 0 \end{bmatrix} + y \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$

We can check this solution as follows:

$Ax = \begin{bmatrix} 0&1&4&0 \\ 0&2&8&0 \end{bmatrix} \left( \begin{bmatrix} 0 \\ b_1 \\ 0 \\ 0 \end{bmatrix} + u \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + w \begin{bmatrix} 0 \\ -4 \\ 1 \\ 0 \end{bmatrix} + y \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \right)$

$= \begin{bmatrix} 0&1&4&0 \\ 0&2&8&0 \end{bmatrix} \begin{bmatrix} u \\ b_1 - 4w \\ w \\ y \end{bmatrix} = \begin{bmatrix} (b_1-4w) + 4w \\ 2(b_1-4w) + 8w \end{bmatrix}$

$= \begin{bmatrix} b_1 - 4w + 4w \\ 2b_1 - 8w + 8w \end{bmatrix} = \begin{bmatrix} b_1 \\ 2b_1 \end{bmatrix}$

The rank of $A$ is 1, the number of basic variables (or pivots).

UPDATE: I expanded the answer to conform to the presentation in the answer to exercise 2.2.5.

UPDATE 2: I corrected the value for $U$ used in various places. Thanks go to Zoi for catching this error!

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

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Linear Algebra and Its Applications, Exercise 2.2.3

Posted on August 23, 2011 by hecker

Exercise 2.2.3. Consider the system of linear equations represented by the following matrix:

$A = \begin{bmatrix} 1&2&0&1 \\ 0&1&1&0 \\ 1&2&0&1 \end{bmatrix}$

Find the factorization $A = LU$ , a set of basic variables, a set of free variables, a general solution to $Ax = 0$ (expressed as a linear combination as in equation (1) on page 73), and the rank of $A$ .

Answer: We perform elimination on $A$ by subtracting the first row from the third (i.e., using the multiplier $l_{31} = 1$ ):

$\begin{bmatrix} 1&2&0&1 \\ 0&1&1&0 \\ 1&2&0&1 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&2&0&1 \\ 0&1&1&0 \\ 0&0&0&0 \end{bmatrix}$

This completes elimination, and leaves us with the factorization

$U = \begin{bmatrix} 1&2&0&1 \\ 0&1&1&0 \\ 0&0&0&0 \end{bmatrix} \qquad L = \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 1&0&1 \end{bmatrix}$

so that

$LU = \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 1&0&1 \end{bmatrix} \begin{bmatrix} 1&2&0&1 \\ 0&1&1&0 \\ 0&0&0&0 \end{bmatrix} = \begin{bmatrix} 1&2&0&1 \\ 0&1&1&0 \\ 1&2&0&1 \end{bmatrix} = A$

We now solve for

$Ax = \begin{bmatrix} 1&2&0&1 \\ 0&1&1&0 \\ 1&2&0&1 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = 0$

The pivots in $U$ are in columns 1 and 2, so the basic variables are $u$ and $v$ and the free variables are $w$ and $y$ . The rank of $A$ is 2, the number of basic variables (or pivots).

From the second row of $A$ we have $v + w = 0$ and thus $v = -w$ . From the first row of $A$ we have $u + 2v + y = 0$ and can substitute for $v$ to obtain $u - 2w + y = 0$ or $u = 2w - y$ . We then have

$x = \begin{bmatrix} 2w - y \\ -w \\ w \\ y \end{bmatrix} = w \begin{bmatrix} 2 \\ -1 \\ 1 \\ 0 \end{bmatrix} + y \begin{bmatrix} -1 \\ 0 \\ 0 \\ 1 \end{bmatrix}$

UPDATE: Corrected the final equation (had $w$ multiplying the second vector instead of $y$ ).

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

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Linear Algebra and Its Applications, Exercise 2.2.2

Posted on August 22, 2011 by hecker

Exercise 2.2.2. Consider a system of linear equations that has more unknowns than there are equations and that has no solution. Provide the smallest example of such a system you can think of.

Answer: Systems of one linear equation and two unknowns (i.e., of the form $ax + by = c$ ) always have solutions. To find a system with no solution we therefore have to look at systems of two equations and three unknowns. For example, the system

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcr}u&+&v&+&w&=&0 \\ u&+&v&+&w&=&1 \end{array}$

has no solution: Gaussian elimination (i.e., by subtracting the first equation from the second) would produce the contradiction 0 = 1.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

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Linear Algebra and Its Applications, Exercise 2.2.1

Posted on August 21, 2011 by hecker

Exercise 2.2.1. Consider the set of 2 by 3 echelon matrices. How many patterns are there for such matrices?

Answer: Suppose the first row of a 2 by 3 echelon matrix has a pivot in column 1. The second row could then have a pivot in column 2 or in column 3, or could have no pivot at all. This produces three possible patterns:

$\begin{bmatrix} *&*&* \\ 0&*&* \end{bmatrix} \qquad \begin{bmatrix} *&*&* \\ 0&0&* \end{bmatrix} \qquad \begin{bmatrix} *&*&* \\ 0&0&0 \end{bmatrix}$

If the first row has no pivot in column 1 but a pivot in column 2, then the second row could have a pivot in column 3 or no pivots at all:

$\begin{bmatrix} 0&*&* \\ 0&0&* \end{bmatrix} \qquad \begin{bmatrix} 0&*&* \\ 0&0&0 \end{bmatrix}$

If the pivot for the first row is in column 3, then the second row can have no pivots:

$\begin{bmatrix} 0&0&* \\ 0&0&0 \end{bmatrix}$

Finally, there could be no pivots in the first row, in which case the second row would not have pivots either; in other words, this corresponds to the zero matrix:

$\begin{bmatrix} 0&0&0 \\ 0&0&0 \end{bmatrix}$

There are thus seven possible patterns in all.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

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Linear Algebra and Its Applications, Exercise 2.1.9

Posted on August 20, 2011 by hecker

Exercise 2.1.9. Consider the set of all nonsingular 2 by 2 matrices. Is it a vector space? How about the set of singular 2 by 2 matrices?

Answer: The set of all nonsingular 2 by 2 matrices is not closed under addition; for example, the matrices

$I = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} \qquad P = \begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix}$

are both nonsingular (and in fact are their own inverses), but the sum

$I + P = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} + \begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix} = \begin{bmatrix} 1&1 \\ 1&1 \end{bmatrix}$

is singular. Since the set of nonsingular 2 by 2 matrices is not closed under addition it is not a vector space.

Next, consider the two singular matrices

$A = \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} \qquad B = \begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix}$

Their sum is nonsingular:

$A + B = \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} + \begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix} = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} = I$

So the set of singular 2 by 2 matrices is also not closed under addition, and thus is also not a vector space.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

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My Math Homework

Linear Algebra and Its Applications, Exercise 2.2.9

Linear Algebra and Its Applications, Exercise 2.2.8

Linear Algebra and Its Applications, Exercise 2.2.7

Linear Algebra and Its Applications, Exercise 2.2.6

Linear Algebra and Its Applications, Exercise 2.2.5

Linear Algebra and Its Applications, Exercise 2.2.4

Linear Algebra and Its Applications, Exercise 2.2.3

Linear Algebra and Its Applications, Exercise 2.2.2

Linear Algebra and Its Applications, Exercise 2.2.1

Linear Algebra and Its Applications, Exercise 2.1.9

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