Find vectors u,v that satisfy the equation A u = e 1 , and A v = e 2 I am really new to matrices and these concepts, so any help would be much appreciated. A = [ 1 − 3 0 1 4 − 3 0 1 5 ] We're asked to let each row = a 1 , a 2 , a 3 going down and then calculate the cross products and triple scalar products which I was able to do.I calculated the cross products of a 1 × a 3 , and then a 2 × a 3 : which are: a 1 × a 3 = [ − 15 − 5 1 ] and a 2 × a 3 = [ 23 − 5 1 ] If A u = e 1 and A v = e 2 , and I need to determine vectors u and v, could someone explain how to find u and v. I know that if I multiply A by the cross product vectors I get [38,0,0] and [0,−38,0] but this confuses me because I thought to satisfy the equation the answer should be [1,0,0] and [0,1,0]. I have also tried finding the inverse of A and multiply this by e 1 and e 2 , but the results were not even close to the result of the cross products?? I am really trying to learn, would appreciate some advice.Is there another method to solve this?

Question

Find vectors u,v that satisfy the equation   A  u  =      e    1  , and   A  v  =      e    2  I am really new to matrices and these concepts, so any help would be much appreciated.  A  =      [                            1                          −          3                          0                                      1                          4                          −          3                                      0                          1                          5                      ]  We&#039;re asked to let each row =       a    1    ,      a    2    ,      a    3   going down and then calculate the cross products and triple scalar products which I was able to do.I calculated the cross products of       a    1    ×      a    3  , and then       a    2    ×      a    3  : which are:      a    1    ×      a    3    =      [                            −          15                                      −          5                                      1                      ]  and      a    2    ×      a    3    =      [                            23                                      −          5                                      1                      ]  If   A  u  =      e    1   and   A  v  =      e    2  , and I need to determine vectors u and v, could someone explain how to find u and v. I know that if I multiply A by the cross product vectors I get [38,0,0] and [0,−38,0] but this confuses me because I thought to satisfy the equation the answer should be [1,0,0] and [0,1,0]. I have also tried finding the inverse of A and multiply this by       e    1   and       e    2  , but the results were not even close to the result of the cross products?? I am really trying to learn, would appreciate some advice.Is there another method to solve this?

Trevon Hughes · Accepted Answer

You are very close. Your cross product strategy gives you a vector that produces multiples of       e    1   and       e    2  . But remember that since A is a linear transformation,   A  (  λ  x  )  =  λ  (  A  x  ) for any scalar lambda and any vector x. So how can you choose scalars   λ  ,  μ so that   A  (  λ  u  )  =      e    1   and   A  (  μ  v  )  =      e    2  ? (Here u and v are the vectors you calculated).You were also close with the strategy of finding the inverse! Note that if the inverse has columns       x    1  ,       x    2  , and       x    3   (in that order from left to right), then  A      A          −      1        =  I  =  [      e    1    ,      e    2    ,      e    3    ]  =  A  [      x    1    ,      x    2    ,      x    3    ]  =  [  A      x    1    ,  A      x    2    ,  A      x    3    ]So in fact that first column of the inverse is the u you&#039;re looking for, and the second column of the inverse is the v you&#039;re looking for.You asked a good question, namely whether the constants   λ and   μ can be deduced without having to calculate Au and Av and then reverse-engineering them. And the answer is yes. We&#039;ll show it for   λ, but the process is identical for   μ. Plugging in our expression for u, we notice that  A  u  =  A  (      a    2    ×      a    3    )  =      [                                        a            1                    ⋅          (                      a            2                    ×                      a            3                    )                                                  a            2                    ⋅          (                      a            2                    ×                      a            3                    )                                                  a            3                    ⋅          (                      a            2                    ×                      a            3                    )                      ]    =      [                                        a            1                    ⋅          (                      a            2                    ×                      a            3                    )                                      0                                      0                      ]  the last equality following from the properties of the triple product. So we can deduce that      e    1    =      [                            1                                      0                                      0                      ]    =            (              a        2            ×              a        3            )                      a        1            ⋅      (              a        2            ×              a        3            )      And you can further note that this expression is only ill-defined (i.e. the denominator is zero) when       a    1   is a linear combination of       a    2   and       a    3  , which would mean that A is not invertible, in which case we would have to adopt a different method to solve this problem.

Injetaangetly2u · Accepted Answer

You have  A  =      (                            1                          −          3                          0                                      1                          4                          −          3                                      0                          1                          5                      )  You seek a vector  v  =      (                            x                                      y                                      z                      )  with the property that  A  v  =      e    1    =      (                            1                                      0                                      0                      )    .Let&#039;s write that out:      (                            1                          −          3                          0                                      1                          4                          −          3                                      0                          1                          5                      )        (                            x                                      y                                      z                      )    =      (                            1                                      0                                      0                      )    .Doing the matrix multiplication, we get      (                            1          x          −          3          y          +          0          z                                      1          x          +          4          y          −          3          z                                      0          x          +          1          y          +          5          z                      )    =      (                            1                                      0                                      0                      )    .Without the b rackets, that&#039;s just                    1                            =        x        −        3        y                            0                            =        x        +        4        y        −        3        z                            0                            =        y        +        5        z            The last equation lets you replace all ys with −5z, so we get                    1                            =        x        +        15        z                            0                            =        x        +        4        (        −        5        z        )        −        3        z        =        x        −        23        z                            0                            =        y        +        5        z            From the first two, we have  x  =  1  −  15  z  ;  x  =  23  zso we conclude that  23  z  =  1  −  15  z    38  z  =  1    z  =      1    38  From this, we can obtain x and y and we&#039;re done.As you take more linear algebra, you&#039;ll learn more methodical approaches to solving problems like these, but this gets you started for now

Find vectors u,v that satisfy the equation Au=e_1, and Av=e_2

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