Bildy,
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
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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
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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 |
