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# Challenging Matrices Questions watch

1. Hi
Please could somebody help me with this matrices homework? I have done part a, and in part b I have established that the product of A and B is the identity matrix, but I don't know how to use that information to find the solutions to the simultaneous equations
Any advice would be much appreciated!
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2. You turn the simultaneous equations into matrices which is what they gave you above:
(don't know how to illustrate it on here but I'll try
3. So you've solved AB=KI where k is a scalar and I is the identity matrix, notice that

M(X,Y,Z)=(10,11,5) HENCE X,Y,Z=M^-1(10,11,5) where M is matrix B

hence to find x y and z we'd need to inverse matrix, we can find that by manipulating AB=KI, by post multiplying by B^-1 we get A=kB^-1 hence A/k=b^-1
4. (Original post by Katie21v9)
Hi
Please could somebody help me with this matrices homework? I have done part a, and in part b I have established that the product of A and B is the identity matrix, but I don't know how to use that information to find the solutions to the simultaneous equations
Any advice would be much appreciated!
You can write the simultaneous equations using matrices:

How can you solve this to find ?
5. This works because of the way that you multiply matrices.
Can you go from there?
Attached Images

6. (Original post by surina16)
This works because of the way that you multiply matrices.
Can you go from there?
OK is this along the right lines?
As AB=I, B is the inverse of A
B.(x,y,z) = (11,10,15)
B.B^-1(x,y,z) =B^-1.(11,10,15)
(x,y,z) = A. (11,10,15)
7. (Original post by Katie21v9)
OK is this along the right lines?
As AB=I, B is the inverse of A
B.(x,y,z) = (11,10,15)
B.B^-1(x,y,z) =B^-1.(11,10,15)
(x,y,z) = A. (11,10,15)
All looks good, but isn't the last value 5 and not 15?
8. (Original post by surina16)
All looks good, but isn't the last value 5 and not 15?
Oh yes, thanks for that!

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