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Unformatted text preview: trix and then guide you through some
exercises using a 3 × 3 matrix.
Consider the matrix 6 15
14 35 C= from Exercise 1 above. We know that det(C ) = 0 which means that CX = 02×2 does not
have a unique solution. So there is a nonzero matrix Y such that CY = 02×2 . In fact, every
matrix of the form
is a solution to CX = 02×2 , so there are inﬁnitely many matrices such that CX = 02×2 . But
consider the matrix
It is NOT a solution to CX = 02×2 , but rather,
CX41 = 3
7 6 15
14 35 = 123
7 = 41 In fact, if Z is of the form
CZ = 6 15
14 35 3
7t t = 3
7t 41t = 41 3
7t t = 41Z for all t. The big question is “How did we know to use 41?”
We need a number λ such that CX = λX has nonzero solutions. We have demonstrated that
λ = 0 and λ = 41 both worked. Are there others? If we look at the matrix equation more
closely, what we really wanted was a nonzero solution to (C − λI2 )X = 02×2 which we know
exists if and only if the determinant of C...
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