Math 2270 Spring 2004
Homework 6 Solutions
Section 2.3) 1, 4, 33, 34
(1) Here we can use Fact 2.3.6 which was derived in class. For the given matrix we have: a =
2, b = 3, c = 5 and d = 8. The determinant is det(A) = ad − bc = 2(8) − 3(5) = 16 − 15 = 1.
Therefore,
"
#
8
−3
A−1 =
−5 2
(4) Here we must use Fact 2.3.5.






1 0 −1 | 3 −2 0
1 2 1 | 1 0 0
1 2 1 | 1 0 0





1 1 | −1 1 0  →  0 1 1 | −1 1 0 

 1 3 2 | 0 1 0 → 0
0 0 2 | −3 2 1
0 −2 0 | −1 0 1
1 0 1 | 0 0 1




→



1 0 −1 | 3
−2 0
0 1
1 | −1
1
0
0 0
1 | − 32
1
1
2








→






1 0 0 |
0 1 0 |
3
2
1
2
−1
1
2
0 0 1 | − 32
0
− 21
1
2
1








(33) Using Fact 2.3.6 with d = −a, b = c and det(A) = −a2 − b2 , we have
A−1 =
−1
a2 +b2
"
−a −b
−b a
#
=
1
a2 +b2
"
a b
b −a
#
This inverse exists as long as a2 + b2 6= 0, or as long as a and b are not both zero. The inverse
is equal to the original matrix A when a2 + b2 = 1.
(34) Here we must use Fact 2.3.5.




1 0 0 | 1/a 0
0
a 0 0 | 1 0 0



1/b 0 

 0 b 0 | 0 1 0 → 0 1 0 | 0
0 0 1 | 0
0 1/c
0 0 c | 0 0 1
A is invertible if a, b and c are non-zero. In general, a diagonal matrix is invertible as long
as there are no zeros on the diagonal.
1