Page 34 - A Course in Linear Algebra with Applications
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18 Chapter One: Matrix Algebra
1 3 \ fa b\ _ (1 0
3 9 \c d ~ [ 0 1
which leads to a set of linear equations with no solutions,
a + 3c = 1
b + 3d = 0
3a + 9c = 0
3b + 9d = 1
Indeed the first and third equations clearly contradict each
other. Hence the matrix is not invertible.
Example 1.2.10
Show that the matrix
A- r ~ 2
is invertible and find an inverse for it.
Suppose that B = I , I is an inverse of A. Write out
the product AB and set it equal to I 2, just as in the previous
example. This time we get a set of linear equations that has
a solution,
Indeed there is a unique solution a = 1, b = 2, c = 0, d = 1.
Thus the matrix
