Two Simple Linear Algebra Problems To Study for A Test From Best answer on the web
solutions, as well as steps to solve these 2 relatively simple
problems, to study for a test. Thus, i need each step so that I can
follow what is happening. I need the solutions to these problems no
later than 12:01pm September 18th, Eastern Time.
Question #1
Find all values of a for which the resulting linear systemhas (a) no
solution, (b) a unique solution, and (c) infinitely many solutions.
x + y = 3
x + ((a^2)-8)y = a
Question 2
Find a 3x1 matrix x with the entries not all zero such that Ax=1x.
Where:
A= 1 2 -1
1 0 1
4 -4 5
I guess I get some kind of sick happiness out of answering other people's math questions! I was originally only going to give the basics, but I kind of got carried away :) Besides the question was asked with good intentions (studying for a test rather than "can you do my homework") so how could I say no? :)))
Anyway, people keep asking $30 economics 101 questions, so it kind of evens out after a while. And it's another good way to procrastinate for a 30 minutes and feel like you accomplished something :)
calebu2-ga
My aim in this answer is to do two things :
1) To give you the working out/answers for the two questions you list. 2) To give you a general background to the questions so that you understand why the answers are what they are (that way you are prepared for a generic test question).
I have to guess at your level of understanding - I may shoot too low, may aim too high, but hopefully I'll be clear enough that it doesn't matter.
Question 1
----------
Background
One of the best ways to think about sets of linear equations in two variables (x, y) is as sets of lines. For example the equation
ax + by = c
is an equation with slope -a/b and y-intercept c/b. The solution to a set of linear equations is simply the point at which these equations intersect. This is where our ability to picture straight lines helps us understand the different possibilities for two equations (two straight lines).
If the lines have differing slopes, then there must be a point somewhere at which the lines intersect. Because the lines are straight there is at most one (a unique) solution. An example of this would be
x + y = 2 (1)
x + 2y = 4 (2)
Subtracting (1) from (2) gives y = 2 and substituting into either equation gives x = 0 - so x=0, y=2 is a unique solution.
If the lines have the same slope then there are two possibilities. One is that they are the same line - ie. two different linear equations represent the same line. In which case there is an infinite number of solutions (every point on the line is a solution) - for example the equations
x + y = 2 (1)
2x + 2y = 4 (2)
Any solution to (1) is a solution to (2). eg. x=0, y=2 satistfies both. As does x=1, y=1. The easiest way to notice this circumstance is if the ratio of the coefficients a, b and c remains constant (in this case 1:1:2) - if a is doubled, b is doubled and c is doubled.
The other possibility for lines with the same slope is that they are parallel but not the same. In which case the lines never intersect and there is no solution. For example
x + y = 2 (1)
x + y = 3 (2)
Any solution to (1) does not sastify (2). eg. x = 0, y = 2 does not satisfy (2). The easiest way to notice this possibility is when, after scaling one equation so that the a and b coefficients are identical, the coefficient for c differs between equations.
With this in mind we can solve question 1 :
1x + 1y = 3 (1)
1x + ((a^2)-8)y = a (2)
Part (b) - Unique solution. This is the easiest set of values to find. The slope of equation (1) is 1/1 = 1. The slope of equation (2) is 1/((a^2)-8). As long as these two equations have different slopes there will be a unique solution.
In other words as long as 1/((a^2)-8) does not equal 1, there is a unique solution. Solving the inequality formally (using != for not equals) :
1/((a^2)-8) != 1
(a^2)-8 != 1
a^2 != 9
a != +/-3.
So as long as a is not -3 or +3, there is a unique solution.
Given this part, we now focus on parts (a) and (c). For either of these situations, we require that the slopes are the same, ie. a = -3 or +3.
If substituting a into the equation makes the coefficients of (2) a multiple of (1) then the equations coincide and we have infinitely many solutions. Otherwise the equations have no solution.
Because we have only two possible values of a that can meet the requirements of parts (a) or (c), we can just investigate each one in turn and see which description (infinitely many solutions or no solutions) best describes the situation.
When a = -3, equation 2 becomes :
x + y = -3
Clearly, any values that satistfy (1) cannot satisfy (2), so there are no solutions.
When a = +3, equation 2 becomes :
x + y = 3
This is the same as (1), therefore there are infinitely many solutions.
So summarizing:
(a) No solutions : a = -3
(b) unique solution : a != -3 and a != +3
(c) Infinitely many solutions : a = +3
Question 2
----------
Background
We're now being a little bit more formal with how we approach linear equations - using matrices to represent the possible solutions. The same idea for the first equation applies here, it's just a little tougher to visualise in 3 dimensions (instead of lines, we have planes and the solutions are given by where the planes intersect). It's also a little harder to eyeball the equations like we did in the first part and find a solution.
Solution
We can write the question of finding the solution to Ax = Ix out full as :
(1 2 -1)(x_1) (1 0 0)(x_1)
(1 0 1)(x_2) = (0 1 0)(x_2)
(4 -4 5)(x_3) (0 0 1)(x_3)
We can do addition and subtraction of matrices in very much the same way we handle normal numbers, so our equation can be rewritten as (A - I)x = 0, where (A - I) is the matrix :
(0 2 -1)
(1 -1 1)
(4 -4 4)
(all I did was subtract element by element)
So in full, our problem becomes
(0 2 -1)(x_1) (0)
(1 -1 1)(x_2) = (0)
(4 -4 4)(x_3) (0)
We now have 3 equations that we need to solve :
2*x_2 -1*x_3 = 0 (1)
1*x_1 -1*x_2 +1*x_3 = 0 (2)
4*x_1 -4*x_2 +4*x_3 = 0 (3)
First notice that the coefficients for equations 2 and 3 are in the same ratio. Therefore if the matrix x satisfies equation (2), then it also satisfies (3).
(We now know at this point that there is not a single solution to the problem, because in effect we have 2 equations (1) and (2/3) and 3 unknowns (x_1, x_2, x_3) - the question asks you for "a" matrix - meaning that you might get away with a single numerical solution, but I'm going to be technically correct and give a general solution first).
Lets try and get a solution involving just x_2 by combining equations (1) and (2).
(1) + (2) gives : x_1 + x_2 = 0 (4)
So let x_1 = -x_2.
Substituting into (2) gives -2*x_2 + x_3 = 0.
Hence x_3 = 2*x_2
If we let k = x_2, we can write :
x_1 = -k
x_2 = k
x_3 = 2k
This is the generic matrix that solves our question. To get a specific solution, we can substitute in a number such as k = 1 and get the matrix :
(-1)
x = ( 1)
( 2)
Check this works :
Ax : 1*-1 + 2*1 -1*2 = -1
1*-1 + 0*1 +1*2 = 1
4*-1 - 4*1 +5*2 = 2
Of course any value of k gives a valid solution.
Good luck on the test - feel free to ask for clarification if you need any additional help with the listed questions before your deadline.
calebu2-ga
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July 30th, 2010 edit