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Math 105
Semester Review - Solutions
7. Solve the following systems of linear equations using (i) Cramer’s rule (ii) the inverse.
3x
− y + = 1
a)
5x
+ 3z
= 14
x + 3y
− z = 4
i – Cramer’s Rule
3 −1 0
0 3 5 3
det ( A) = 5 0 3 = 3 −( − 1)
0 =−27 −8 =− 35
3 −1 1 −1
1 3 −1
1 −1 0
0 3 14 3
det ( A()
1 ) = 14 0 3 = −( − 1)
+ 0 =−9 −26 =− 35
3 −1 4 −1
4 3 −1
3 1 0
14 3 5 3
det ( A ( 2)
) = 5 14 3 = 3 − + 0 =− 78 + 8 =− 70
4 −1 1 −1
1 4 −1
3 −1 1
0 14 5 14 5 0
det ( A) = 5 0 14 = 3 −( − 1)
+ =− 126 + 6 + 15 =− 105
3 4 1 4 1 3
1 3 4
x
z
det ( A()
1 ) −35 1
( A( ))
= = =
y
det ( A)
− 35
( A)
det ( A( 3)
) −105
= = = 3 Solution: ( 1, 2, 3 )
det ( A)
− 35
ii - The inverse.
⎡−9 8 15 ⎤
cof ( A)
=
⎢
1 3 10
⎥
⎢
− − −
⎥
⎢⎣−3 −9 5 ⎥⎦
9 1 3
⎡
35 35 35
⎤
−1 1
−8 3 9
A = adj( A)
=
⎢ ⎥
35 35 35
det ( A)
⎢ ⎥
⎢
−3 2 −1
⎣ ⎥
7 7 7 ⎦
9 1 3
⎡35 35 35
⎤⎡ 1⎤ ⎡1⎤
−1 −8 3 9
X = A b= ⎢ ⎥⎢
35 35 35
14
⎥ ⎢
2
⎥
⎢ ⎥⎢ ⎥
=
⎢ ⎥
⎢−
3 2 −1
⎣ ⎥⎢
7 7 7 ⎦⎣ 4⎥⎦ ⎢⎣3⎥⎦
det 2 −70
= = = 2
det −35
⎡−9 −−1
−3⎤
adj( A)
=
⎢
8 3 9
⎥
⎢
− −
⎥
⎢⎣
15 −10 5 ⎥⎦
Winter 2006 Martin Huard 6