You are not logged in.
Hello there. I find myself stuck with a matrix exercise:
Consider the matrices
A= ( k 3 0
8 2 -1
0 k 1)
and
B= (1 4 2
2 0 -2)
k is a parameter, element of IR
b) For which value(s) of k, if any, does the matrix equation AM=A+M have a unique solution for M?
c) let k=5. Find all solutions M, if any, of the matrix equation AM=B, respectively MA=B
For a I already found a solution so I won't bother you with it.
For these two others I am lacking an idea how to even start this and google doesn't help because it is to specific I guess.
So, any advice, hint, idea and ideally showing of steps will be greatly appreciated.
Thank you,
Nadine
Offline
Hi;
Did you copy B right?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
I did. Yes. And I figured it might be a hint to the solution maybe not existing with the different shape?
Offline
What can you multiply a 3 x 3 matrix with to get a 2 x 3?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
A 2 x 3?
Offline
Hi;
m x n . n x p = m x p
A 3 x 3 multiplied by a 2 x 3 is not legal.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
Hi NadineSch,
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
Offline
Hi bobbym,
Did you notice "if any" in the problem statement. that fixes it. AM=B is not possible but MA=B is possible.
Hi NadineSch,
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
Offline
Hi;
Please check your answer, there is a sign error, should be -6.
I had not looked at MA = B, I was trying to get her to understand that AM = B was not possible. From the OP's comment in post #5 I thought she might be a bit confused about that.
Anyway once she understands that, MA = B can be solved without resorting to linear algebra. You need to solve the simultaneous set,
5 a + 8 b = 1,
3 a + 2 b + 5 c = 4
-b + c = 2
5 d + 8 e = 2
3 d + 2 e + 5 f = 0
-e + f = -2
This is just tedious algebraic substitutions and yields the answer.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
Hi there. Yes, I needed about two pages as well for the algebra. But, the good and most important news: I found the same solution!!! Jippieh!!! Thank you so much!
Offline
That is good.
Please memorize this formula (m x n) . (n x p) = (m x p)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
For everyone that helped: I have passed my exam!! Thank you very much. I only found it polite to come back to you with this.
Offline
Wunderbar!
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline