anonymous
  • anonymous
Find the products AB and BA to determine whether B is the multiplicative inverse of A.?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1360900484563:dw|
anonymous
  • anonymous
I don't know why we have to find the product of AB and then BA. My work is use Gauss-Jordan method to figure out inverse of A and then compare to B. and |dw:1360901929122:dw| is inverse of A, it is not a multiply matrix of B .
anonymous
  • anonymous
What do you mean?

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BAdhi
  • BAdhi
This uses the definition of the multiplicative inverse. ie if and only if, $$AB=BA=I$$ A is the multiplicative inverse of B So find the values of AB and BA and see whether they are equal to identity matrix
anonymous
  • anonymous
Im lost
phi
  • phi
Find the products AB and BA They want you to practice multiplying 2 matrices. can you multiply A*B ?
anonymous
  • anonymous
The bottom column of B is suppose to be 0 -1 1, my mistake. But I multiplied a * b & I got, [1 0 0] [0 1 0] [0 0 1]
phi
  • phi
OK, the matrix with 1's on the diagonal is the identity matrix (called I (eye)) a * I will give you a also, if you know a * b = I then you know b is the *inverse* of a you also know b*a= I (but I think they want you to multiply it out and see that it is) and we could just as well say a is the *inverse* of b \[ A^{-1} A = I \] (People use capital letters for matrices (bold face if you can). the use lower case, bold letters for vectors)
anonymous
  • anonymous
Wouldnt it be B = A^1?
phi
  • phi
you mean "wouldn't it be \[ B = A^{-1} \] the -1 is not an exponent, but means *inverse* You can say that. But the inverse of the inverse \[ (A^{-1})^{-1} = A\] if we take the inverse of both sides \[ B^{-1} = (A^{-1})^{-1} = A \] or \[ A = B^{-1} \] which says A is the inverse of B (or vice versa)
anonymous
  • anonymous
Ah okay, what would it mean if they "=" had a slash through it?
phi
  • phi
not equal
anonymous
  • anonymous
Got it, thank you for the help.!

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