Matrices Maths Question

Maths Matrices Question?

It is a matrices question with matrices A and B where the third part asks you to calculate AB, AB^2 and AB^3 as multiples of A, and to make a conjecture about AB^n. I calculated AB=-A, AB^2 = A and AB^3 = -A. I cannot work out how to put this into AB^n = something. With what seems to be odd powers creating -A and even powers being +A.

Matrices Question: Math?

A system is said to be inconsistent if it has no solution.
Now, I assume you have studied the Gaussian Elimination method.When using Gaussian elimination on a system, if you have zeros in an entire row to the left of the bar and do not have a zero to the right of the bar on that row, you know that you have an inconsistent system.

Note: A system is inconsistent if ANY row of the matrix has zeros left of the bar and a non-zero number right of the bar.

Math question - matrices?

Let's take care of the first part first:

The first equality can be neatly rewritten using the "coefficient matrix" and matrix multiplication.

2x - 6y= -6
3x + 2y = 13

finding the coefficient matrix is simple: just take the "numbers" in the left-hand side of each equation to get your coefficient matrix A:
A =
[2, -6]
[3, 2]

The reason this is useful is that we can now write the equality using matrix multiplication, which means that we can solve it using matrix inversion. Our equality is now:

[2, -6] * [x] = [-6]
[3, 2] [y] = [13]

By naming each of these matrices (for convenience) we can just rewrite that as

A*X = B

By solving for the matrix X, we can now find x and y. In order to solve, we need to calculate the inverse of A, and multiply it with both sides, so that

X = inv(A)*B

so now what is the inverse of A? There are several methods of finding it, but the answer you get should be

[2/22, 6/22]
[-3/22, 2/22]

After multiplying inv(A) and B, you should get X =
[3]
[2]

That is, x=3 and y=2

Part 2: So this is essentially the matrix version of "multiplication addition." You start with an "augmented matrix" like this (taking all the coefficients):
[1, -1, 1, ... 1]
[1, 1, 1, .... 5]
1, 1, -1, .... -5]

If you enter the matrix above at this website
http://www.math.odu.edu/~bogacki/cgi-bin...

it'll show the steps taken to reduce the matrix to "reduced row-echelon form" (choose m=3, n=4).

you'll end up with this matrix:
[1, 0, 0, ... -2]
[0, 1, 0, ... 2]
[0, 0, 1, .... 5]

which means, translating it back into an equation by using the coefficients,
x=-2, y=2, z=5

Hope that's all you needed. Good luck!

Matrices QUESTION MATH HELP?

Ok, so in order for the answer to have an infinite amount of solutions, there must be a free variable in the matrix. This means that the last row of the matrix, after row reductions, must be filled with zeroes. The overall system has to be consistent, thus there must also be a zero in the third row of the augmented column. Assuming I did the row reduction correctly, k=5.

2 -1 -9 7
1 2 3 1 -R3+R1
2 1 -3 k

2 -1 -9 7
1 2 3 1 -1/2R3
0 -2 -6 -k+7

2 -1 -9 7
1 2 3 1 -2R2+R1
0 1 3 (-k+7)/2

2 -1 -9 7
0 -5 -15 5 -1/5R2
0 1 3 (-k+7)/2

2 -1 -9 7
0 1 3 -1 -R3+R2
0 1 3 (-k+7)/2

2 -1 -9 7
0 1 3 -1
0 0 0 (-(-k+7)/2)+1

In other words, (-(-k+7)/2)+1=0
(k-7)/2=-1
k-7=-2
k=7-2
k=5

Maths help! Matrices! Tough question!?

Set up a matrice.
I can't do the bars so I'll just roughly explain it
So for example, the first row would be
1 2 1 0.
That's the same as x+2y+z=0. You'll have to put the answers in a separate matrix, though.
Do the same for the other equations
2 3 1 2
And
2 2 1 3

Hopefully that helped a little.

Math Help!!? Matrices question? SOS?

Okay, let's represent this in matrix form by AX=b, where A=[1, m; m-1, 2], X=[x;y], and b=[2;m]. (This notation is adapated MATLAB notation; I hope it's clear. It should be obvious from the original problem, anyway.)

Then the only time the system can have no solutions OR infinite solutions is when A is singular (has zero determinant). Thus, the system has a unique solution UNLESS

1·2 - m(m-1) = 0, or in other words,
m² -m -2 = 0. This gives
(m-2)(m+1) = 0, so
m = 2 or m = -1.

Now, we just check both values of m.
When m=2, you have the system
x + 2y = 2
x + 2y = 2.
Obviously this has infinitely many solutions.

On the other hand, when m = -1, you have
x - y = 2
-2x +2y = -1.
Multiplying the first equation by -2, you get the system
-2x + 2y = -4
-2x + 2y = -1.
Clearly this system has no solutions.

Again, whenever m is not -1 or 2, you have exactly one solution.

I'll leave the geometric interpretation up to you. :-)
§

Maths - Matrices?

Matrices-calc project please please help?
any form of help is appreciated!
note: i dont know how to make matrices on computer so each of these is a 2 by 2 matrix...just in one set of brackets. numbers before comma are top row, after comma are bottom row

1. Consider matrix M= [2 0 , 0 2] calculate M^n for n= 2, 3, 4, 5, 10, 20, 50. Describe in workds any pattern observed. use pattern to find general expression for matrix M^n in terms of n

2. Consider matrices P=[3 1, 1 3] and S= [4 2, 2 4]
P^2=[3 1, 1 3]^2 =[10 6, 6 10] = 2[5 3, 3 5] ; S^2= same type of thing as P
Calculate P^n and S^n for other values of n and describe any observed patterns

3. Consider [k+1 k-1, k-1 k+1]
steps 1 and 2 contain examples of these matrices for k=1, 2, and 3. consider other values of k and describe any patterns you observe. generalize these results in terms of k and n

4. use technology to investigate what happens when further values of k & n. state scope of limitations of k and n

Math matrix question ?

a) for which "a"-values can the matrix have an inverse?

For A to have an inverse, its determinant must be > 0.

A =
[ a... 3]
[1.a+2]

Determinant of A = |A| = a(a+2) - (1)(3) = a^2 + 2a - 3 > 0
or a^2 + 2a > 3

i.e. a ≠ 1 or a ≠ 3 <-- which are the roots of the characteristic equation.

if we pick a = 2; |A| = 4 + 4 - 3 = 5

b) determine the inverse of matrix A for the a-values for which it exists?

A =
[2 ... 3]
[1 ... 4]

|A| = 8 - 3 = 5

A^-1 =
[4 ...-1]T
[-3... 2]
------------
....|A|

A^-1 =
[4 ...-3]
[-1... 2]
------------
....5

A^-1 =
[0.8 ...-0.6]
[-0.2... 0.4]

c) If a = 2 , solve for
AX=
(0, -7,1)
( -2,7,2)

I'm not sure where this is going - if X is a vector, then multiplying A by X should give a vector (2 x 1 matrix ) of R2. Why is the RHS a 2 x 3 matrix?