Exercise 5.1.18

Answers

1.
If B is invertible, we have B1 exists and det (B)0. Now we know that
det (A + cB) = det (B)det (B1A + cI),

a nonzero polynomial of c. It has only finite zeroes, so we can always find some c sucht that the determinant is nonzero.

2.
Since we know that
det (1 1 c 1 + c ) = 1,

take A = (11 0 1 ) and B = (00 1 1 ).

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2011-06-27 00:00
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