anonymous
  • anonymous
linear algebra help
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@amistre64
phi
  • phi
to show 1) A+B=B+A let A be (a11 a12 ; a21 a22) (that is supposed to be a 2 x 2 matrix) and B be (b11 b12; b21 b22) now add the two matrices (you add corresponding elements) you get A+B= (a11 + b11 a12+b12; a21+b21 a22+b22) each entry in the the matrix can be "swapped" because addition is commutative so we can say = (b11+a11 b12+a12; b21+a21 b22+a22) and that is the same as B+A
anonymous
  • anonymous
ok so wat about the 2nd property how would i do that?

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phi
  • phi
let A be (a11 a12 ; a21 a22) what would you add to A to get 0's in every location?
anonymous
  • anonymous
(-a11 -a12 -a21 -a22)?
phi
  • phi
yes, and that works for all possible a's so A^(-1) in this problem would be -A or A with each entry negated
anonymous
  • anonymous
ok and tthe third property?
phi
  • phi
if you multiply a matrix by a scalar , you multiply each entry in A by that scalar in other words b A = [ b*a11 b*a12; b*a21 b*a22] and of course, vice versa: [ b*a11 b*a12; b*a21 b*a22] = b*A
phi
  • phi
so show (ab) which is a scalar, times A gives a*b*a11 a*b*a12; a*b*a21 a*b*a22 now do b*A and then multiply that result by a you should get the same as the line up above, showing (ab)*A = a*(b*A) (btw the scalar a is not the same as the entries a11, etc in the matrix)
phi
  • phi
I am naming the entries in matrix A \(a_{11}\) etc (I have to call them something) but your problem says show (ab)A and their a is just some number. if you like, change the name of the numbers in A to \(c_{11}, c_{12}\) , etc
phi
  • phi
the idea is (ab) means multiply a*b to get a number and then multiply each entry in matrix A by that number in other words we would say (ab) is (a*b) and multiply each entry by (a*b) then drop the parens, so each entry would be \( a\cdot b \cdot c_{11} \) etc
anonymous
  • anonymous
ok i understand and when u do it for b*a u shd get the same right ?
phi
  • phi
on the other hand (bA) means first multiply b times each entry in A. You should get \(b\cdot c_{11}\) etc then multiply that new matrix by a
anonymous
  • anonymous
ok i have 1 more question
phi
  • phi
They are asking if you can add any two members of S_2, and the result is also in S_2 or if we multiply by a scalar, the result is in S_2 when they say y=-x are positive, I think that means x is negative and the members in S_2 are (-1,1), (-2,2) , (-3,3), (-1.111, 1.111), etc
phi
  • phi
I can see a problem if we multiply by the scalar -1 for example -1 * (-1,1) becomes (1,-1) and y is -1 and not a positive number. In other words when we multiply by -1, we get a result that does not meet the definition of S_2. That means we do not have a subspace
anonymous
  • anonymous
o ok so its not a subspace of V
phi
  • phi
yes, it is not a subspace of V

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