Expert: Scott A Wilson - 5/5/2009

Question
QUESTION: 1.find the determinant of the following 5x5 matrix

4 1 8 2 5
1 0 0 3 1
4 2 3 1 3
6 4 4 1 0
7 3 1 2 2

i have tried using the row reducing method such that i could get a triangular matrix so that i could simply add multiply the major row but i got zero

2.solve the following systems of equations using the row echelon method
W+2X-Y+4Z=7
W+3Y+2Z=10
2W+6X+Y-5Z=8
3W-3X+2Y+Z=4
PLEASE SHOW ALL THE WORKING THANK YOU VERY MUCH."

3)find which of the matrices are  diagnosable.for those  that are, find P and D such that P^-1(inverse)AP=D(D is a diagonal matrix)
a)1 1 -2
 4 0  4

b)1  2  3
 0 -1 -2
 0  0  2

ANSWER: Sorry it took so long - I just got back from thanksgiving a few hundred miles away.

1) To find the determinant, put it in triangular form by adding rows to other rows.  Each row must be left as it is.

4 1 8 2 5
1 0 0 3 1
4 2 3 1 3
6 4 4 1 0
7 3 1 2 2

The first step is to multiply the first row by -1/4 and add to the second row.  Multiply the first row by -1 and add to the third row.  Multiply the first row by -6/4 and add to the fourth row.  Multiply the first row by -7/4 and add to the last row.  This will make a 4 at the top of the first column with 0's below.

AFter this, I would multiply the second row by -4 and multiply that row by -1/4 and add to the third, multiply it by 10 and add it to the third, and multiply it by 5 and add to the last.  Note the first row can be left as is.

Repeat this, putting the matrix in upper triangular form and never dividing the row in question by anything.  The determinant will be the product of the diagonal elements.


2) Take  the matrix
1 2 -1 4 7
1 0 3 2 10
2 6 1 -5 8
3 -3 2 1 4
and start by zeroing the first column below the 1 on top.

Multiply it by -1, -2, and -3 and add to the lower rows, giving
1 2 -1 4 7
0 -2 4 -2 3
0 2 3 -13 -6
0 -9 5 -11 -17.

Do the same to column 2 by multiplying by 1 and -9/2 and adding to the third and fourth rows.  Since we are not worried about the determinant, divide the second row by -2.

1 2 -1 4 7
0 1 -2 1 -3/2
0 0 7 -15 -3
0 0 -13 -2 -61/2.

If you are having trouble seeing these in columns, they can be rewritten in Excel.

Do the same with the next column, getting a zero in the bottom row.  
Now all that has to be done is to divide the fourth row and get a solution and then back substitute.

Once an answer is gotten, remember each of the columns 1, 2, 3, and 4 are W, X, Y, and Z, respectively.


3) a) I don't see how it can be done since the matrix is not square.

b) Line it up the matrix D next to and identiy matrix.  The first column on the matrix is already set as 1, 0, 0.  Adjust the matrix by the second column, making that element on the diagonal a 1.

1   2   3   1   0   0
0   -1   -2   0   1   0
0   2   0   0   1   0,

1   0   -1   1   2   0
0   1   2   0   -1   0
0   0   2   0   0   1.

Adjust the matrix to make the 2 in the third row a 1 and zeroing the entries above by adding the appropriate multiple of the last row.

1   0   0   1   2   0.5
0   1   0   0   -1   -1
0   0   1   0   0   0.5.

The inverse, D^-1, is the the matrix on the right,
1   2   0.5
0   -1   -1
0   0   0.5.

The result is the original matrix is D and it's in the last display, known as D^-1.  What you have is D^-1 D = I, but I don't recall how to find a matrix P such that P^-1 A P = D.


---------- FOLLOW-UP ----------

QUESTION: dear sir, thanks for the previous help, i would like to get your opinion on this no.
with part one i listed the possible outcomes that one can get from the two dice,but am confused where to put the the case where   when you throw the dice and you get ((2,2)or(4,4)do you take one of them and consider it as maximum or i simply ignore it.
when i ignore it i get 0.5 as the answer
As for part two i wanted to find a way of using combinations but i can not figure it out  


1A pair of fair dice is tossed and the ‘output’ is the maximum number that comes on top.  What is the probability that
a.The output is even
b.The output is prime
2Assuming the dice pair in (1) above was tossed 30 times, what is the probability that the output of 4 is got 16 or 17 times?


Answer
1. When the dice are the same, both of them have the maximum number.
The only way a maximum of 1 is rolled is if a 1-1 is tossed,
so the chances are only 1/36.

A maximum of 2 could be the max if a 1-2, 2-2, or 2-1 is rolled,
so the chances are 3/36.

A maximum of 3 could be rolled if a 1-3, 2-3, 3-3, 3-2, or 3-1 is rolled, so the chances of a 3 being the max is 5/36.

So far we have
1 1/36
2 3/36 and
3 5/36.

That looks like a pattern - 1, 3, 5 - so the next ones should be 7, 9, and 11.  If this is thought about for a few minutes, the chance of a certain number n being the max is 2(n-1) + 1.  That is, there are n-1 possibilities for the other die.  It is times to since either one could be the max.  A one is added in case a pair is tossed.

So overall, the table looks like this:
1  1/36
2  3/36
3  5/36
4  7/36
5  9/36
6  11/36

Not only that, but the sum of odd numbers is always the square of the number of odd numbers if they are taken in order.  For example,
1² = 1, and that's the first one.  We can also see that
1+3 = 4 = 2²
1+3+5= 9 = 3²
1+3+5+7 = 16 = 4²
etc.

Maybe that wasn't part of the answer, but it was trivia to know.

I believe the best way to do it is to say what was just said.  
The number of ways of getting, for example, of rolling both dice numbered 1 to 3 with 6 sided dice is 3²/6² = 9/36.  To find 3 as a maximum, subtract off the number of ways of rolling 2 dice and only getting 1 or 2 on each of them.  That would be 2².  So for a 3 max, take 9-4 = 5.

To get the chance of an even, add up the chances of a 2, 4, or 6.
That would be 3/36 + 7/36 + 11/36.

The chance of it being prime is gotten by knowing which ones are prime.  They are 2, 3, and 5.  The numbers 4 and 6 are not since they are 2*2 and 2*3.  The number 1 is not since it only has 1 factor.  All primes have two distinct factors - 1 and themselves.
THe chances of a 2, 3, or 5 is found by adding together the chances for each.  That would be 3/36 + 5/36 + 9/36.


2. The chances of rolling a number k times out of n rolls is given by
p^k)(1-p)^(n-k) n!/k!(n-k)!  = C(n,k).  

For example, rolling 3 2's in 5 rolls.
The chance of a 3 on each die is 1/6.  
The chance each not a 3 on each die is then 1 - 1/6 = 5/6.
So in order to get 3 2's in 5 rolls, the chances would be
(1/6)^2 * (5/6)^3 * C(5,2) = (1/36)(125/216)(120/6*2) =
1250/7776 = 0.016075103.  That's not very much.

To get 16 or 17 rolls of a 4 on 30 dice would also be small.
It would be (1/6)^16 * (5/6)^14 * C(30,16)  +
(1/6)^17 * (5/6)^13 * C(30,17).

Since it is known that C(30,16) = 30!/16!14!, this can be seen to be
30*29*28*27*26*25*24*20*19*18*17 all over
14*13*12*11*10*9*8*7*6*5*4*3*2*1.
This may be large, but (5/6)^14 is small and 1/6^16 is even smaller.
For C(30,16), I get 145,422,675.  
For (5/6)^14, I get 0.077886566
For 1/6^16, I get 3.5447E-13.
Multiplying all of these together gives 4.0149E-06.
As expected, the chance of getting 16 4's is not very much.

Now just think - the odds of winning the lottery are even smaller...