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Unformatted text preview: 3k+1 = 3 · 3k , the induction hypothesis gives
3k+1 = 3 · 3k > 3(100k ) = 300k . We are done if we can show 300k > 100(k + 1) for k ≥ 6.
Solving 300k > 100(k + 1) we get k > 1 . Since k ≥ 6, we know this is true. Putting all of
this together, we have 3k+1 = 3 · 3k > 3(100k ) = 300k > 100(k + 1), and hence P (k + 1) is
true. By induction, 3n > 100n for all n ≥ 6.
4. To prove this determinant property, we use induction on n, where we take P (n) to be that
the property we wish to prove is true for all n × n matrices. For the base case, we note that if
A is a 1 × 1 matrix, then A = [a] so A = [ca]. By deﬁnition, det(A) = a and det(A ) = ca so
we have det(A ) = c det(A) as required. Now suppose that the property we wish to prove is
true for all k × k matrices. Let A be a (k + 1) × (k + 1) matrix. We have two cases, depending
on whether or not the row R being replaced is the ﬁrst row of A.
Case 1: The row R being replaced is the ﬁrst row of A. By deﬁnit...
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