More fun in 2016, Part 19
How many real 3 × 3 matrices A are there such that A 2 0 1 6 = − I 3 , where I 3 represents the 3 × 3 identity matrix.
Enter 666 if you come to the conclusion that infinitely many such matrices A exist.
Hint: Use determinants.
The answer is 0.
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1 solution
I must confess that I don't get it. Why must A 2 = − I hold? Where are you using the fact that we are dealing with 3 x 3 matrices?
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Note that for matrices the association rule is valid, so:
A 2 0 1 6 = A 1 0 0 8 ∗ A 1 0 0 8 : = B ∗ B = − I 3
Now it must be fulfilled: d e t ( B ∗ B ) = d e t ( B ) ∗ d e t ( B ) = [ d e t ( B ) ] 2 = d e t ( − I 3 ) = − 1
det(B) is complex in contradiction that A is real! Therefore no matrix A is existent.
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Must be fulfilled: A 2 = − I = > A 3 = − A ; A 4 = I ; A 5 = A ; A 6 = − I ; . . . We get a sequence where -I is located at places A n with n=2 + 4k (kЭN). Since 2016 is divisible by 4 the requirement can not be satified.