Exercise 3.4.41

Answers

I = [ 1 1 1 ] [ 1 1 1 ]+ [ 1 1 1 ]+ [ 1 1 1 ] [ 1 1 1 ].

The six P ’s are dependent. Those five are independent: The 4th has P11 = 1 and cannot be a combination of the others. Then the 2nd cannot be (from P32 = 1 ) and also 5th (P32 = 1). Continuing, a nonzero combination of all five could not be zero.

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2022-01-23 16:11
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