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P. 341

Practical Questions





                 1.   Find the Transpose of the following matrix.

                                                                2    –5     3
                 Ans.      2    –1                             –1     0    2/5
                          –5     0
                          3      2/5

                 2.   If A = B, then find the value of x, y and z.

                          x + 2     2              6   2
                      A =                    B =
                          z – 5    xy             10   8
                 Ans.  As        A=B, so  x+2 = 6
                            x = 6 – 2 = 4
                            xy = 8

                     (Putting x=4), 4y=8     y = 8/4 =2
                     z – 5 = 10    z = 10+5 = 15
                     Hence, x=4, y=2 and z=15
                                       1    3
                               –1
                 3.   Find A + A  if A =   2  4
                          1    3
                 Ans.  A =  2  4
                                  1     3
                     |A|          2     4   = (1 × 4) – (3 × 2) = 4 – 6 = –2    So |A| = –2

                           1    4   –3    –2   –3/2
                       –1
                     A  =              =
                           2  –2     1     1    –1/2
                               1    3    –2    –3/2    1–2    3+3/2      –1     9/2
                     A + A  =  2    4   +    1    –1/2       2+1  4–1/2          3   7/2
                          –1

                 4.   If A= {1,2,3,4,5} and B = {2,4,6} then using set theory, find A-B.
                 Ans.  A – B is denoted by the set containing elements that are in A but not in B. Hence, A – B = {1,3,5}

                 5.   Give a = (3,1, –5) and b = (0, –2, 4) find the cross product a × b.

                 Ans.  ā = (3, 1, –5) and b = (0, –2, 4)
                              i    j      k

                     a  × b =   3    1   –5
                              0   –2      4

                           1   –5        3    –5        3    1
                     = i  –2     4      –j  0      4  +k  0  –2

                     = ((1 × 4) – (–5 × –2)) i – ((3 × 4) – (–5 × 0)) j + ((3 × –2) – (1 × 0)) k
                     = (4 – 10)i – 12j – 6k

                     = 6i – 12j – 6k


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