Exercises Chapter IV.
Page 154
1. Compute the determinants of each of the following matrices:
−6 0
a. det
= −12
1 2
−2 1
b. det
= −4 + 4 = 0
−4 2




1 2 3
1
2
3
−2
−10
c. det  4 6 2 = det 0 −2 −10 = det
= −8 + 60 = 52
6
4
−1 4 1
0
6
4

−1
 1

d. det 
 0
2
2
0
3
7
6
2
9
5


4
1

0
8
 = − det 
0
6
6
0
−2
2
3
11
−6
8
9
17



−4
1
4
6

12
 = −6 det  1 3 2 

6 
11 17 14
14




1 4 6
1 4
6
4  = 336
= −6(−1)2 det 0 1 4  = −6 det 0 1
0 27 52
0 0 −56
2. Compute the
1
a. det
2
1
b. det
3
determinants of each of the following matrices:
3
= 4 − 6 = −2
4
2
= 4 − 6 = −2
4


1 0 −2
4 6




4 6
0
c. det
= −2 det
+ 0 + 0 = −2(−2) = 4
1 1
1 1 0


1 4 1
4
1
d. det  0 6 1 = −2 det
=4
6 1
−2 0 0
Page 159
1. Compute the adjoint matrix of each of the following matrices and verify that A adj(A) =
det(A)In :
3 0
2 0 3 0
6 0
2 0
a. A =
, adjA =
, A(adjA) =
=
= 6I2 = det(A)I2
0 3
0 2
0 3 0 2
0 6
a 0
b 0
ab 0
b, A =
, adjA =
, A(adjA) =
= abI2 = det(A)I2
0 b
0 a
0 ab
6 1
4 −1
21 0
c. A =
, adjA =
, A(adjA) =
= det(A)I2
3 4
−3
6
0 21
−2 −4
0 0
2
4
d. A =
, adjA =
. A(adjA) =
= det(A)I2 .
−1 −2
1
2
0 0
2. Compute the adjoint matrix of each of the following matrices and
det(A)In :





−2 3
4
−2 6 2
4 0
2, adjA =  0 4 4 , A(adjA) = 0 4
a. A =  0 1
0 0 −2
0 0 −2
0 0
verify that A adj(A) =

0
0 = (det A)I3 .
4




2 6 0
−24
0
48
0 −16,
b. A =  4 0 8, adjA = −16
−2 3 0
12 −18 −24


−144
0
0
0  = (det A)I3 .
A(adjA) =  0 −144
0
0 −144






2 −1 −2
−4 8 4
0 0 0
0 −4, adjA = −6 12 6, A(adjA) = 0 0 0 = det(A)I3 .
c. A = 1
0 −1
6
−1 2 1
0 0 0
Page 162
1. Use Cramer’s rule to solve each of the following systems of equations:
a. 2x1 + 2x2 = 7
8x1 + x2 = −2
x1
7
det
−2
=
2
det
8
2
1
11
=− ,
14
2
1
2
7
det
8 −2
30
60
=
x2 =
=
14
7
2 2
det
8 1
b. −8x1 + 6x2 = 4
3x1 + 2x2 = 6
x1
4
det
6
=
−8
det
3
6
2
14
−28
=
= ,
−34
17
6
2
−8
det
3
x2 =
−8
det
3
4
6
30
−60
=
=
−34
17
6
2
3. Use Cramer’s rule to solve the following system:
2x1 − 6x2 + x3 = 2
x2 + x3 = 1
x1 − x2 − x3 = 0
Let A be the coefficient matrix of this system, then det(A) = −7.


2 −6
1
1
1
det 1
0 −1 −1
−7
=
=1
x1 =
−7
−7


2 2
1
1
det 0 1
1 0 −1
−1
1
x2 =
=
=
−7
−7
7


2 −6 2
1 1
det 0
1 −1 0
−6
6
x3 =
=
=
−7
−7
7
6. Solve the following system by using Cramer’s rule.
2x1 −
x3 = 1
2x1 + 4x2 − x3 = 0
x1 − 8x2 − 3x3 = −2
Let A be the coefficient matrix of this system, then det(A) = −20.


1
0 −1
4 −1
det  0
−2 −8 −3
−28
7
=
=
x1 =
−20
−20
5


2
1 −1

0 −1
det 2
1 −2 −3
5
1
x2 =
=
=−
−20
−20
4


2
0
1

4
0
det 2
1 −8 −2
−36
9
x3 =
=
=
−20
−20
5
Page 168
1. Calculate the area of the triangles whose vertices are:
a. (0,0), (1,6), (−2, 3)
1 1 −2 15
Area = det
.
=
6
3 2
2
b. (8,17), (9,2), (4,6)
For this problem pick one of the vertices from which vectors, which terminate at the
other two vertices, start.
1 4
5 71
.
Area = det
=
11 −4 2
2
Note: The area of the triangle is half that of the parallelogram formed by the given
vectors.
2.
Calculate the volume of the tetrahedron whose vertices are:
a. (0,0,0), (1, −1, 2), (−3, 6, 7), (1,1,1)


1
−3
1
6 1 = 29 .
Area = det −1
2
7 1
b. (1,1,1), (−1, −1, −1), (0,4,8), (−3, 0, 2)


−2
−1
−4
3 −1 = 8 .
Area = det −2
−2
7
1
5. Show that the straight line passing through the two points (a1 , a2 ) and (b1 , b2 ) has equation


x1 x2 1
det a1 a2 1 = 0
b1 b2 1
The equation is linear in the variables x1 and x2 , so it is an equation for a straight line.
Moreover, since a determinant with two identical rows equals zero, the points (a1 , a2 )
and (b1 , b2 ) are easily seen to satisfy this equation.
6. Let L, a linear transformation from R2 to R2 , have matrix representation
1
2
A=
−1 −2
Let P be the parallelogram generated by the vectors (−1, 2) and (1,3). Sketch P and
L(P ) and compute their areas. Then verify (4.10).
−1
3
1
7
L
=
L
=
2
−3
3
−7
L maps the parallelogram onto a straight line segment, whose area is zero. Since the
determinant of A is equal to zero, we have verified the equation
Area of L(P ) = |det(A)| Area of P.
y
y
(0, 5)
x
(1, 3)
P
L(P ) lies on line:
y = −x
(−1, 2)
x
(10, −10)
Page 169


−2 x
1
1. Find all values of x for which det(A) = 0.
3. Let A =  x 1
2 3 −1



−2 x
1
−2
x
1 = det  x
1
det(A) = det  x 1
2 3 −1
0 3+x
−2 1
= −(3 + x) det
x 1

1
1
0
= −(3 + x)(−2 − x) = (x + 3)(x + 2) .
The values of x are: −3 and − 2.
4. Find all values of λ for which the following system of equations has a nontrivial solution:
2x1 − 7x2 = λx1
−4x1 + x2 = λx2
The system will have a non-trivial solution if and only if the determinant of the coefficient
matrix equals zero.
2 − λ −7
det
= (2 − λ)(1 − λ) − 28 = λ2 − 3λ − 26
−4 1 − λ
Thus, λ =
√
3± 113
.
2
a(t) b(t)
5. Let f (t) = det
, which a(t), b(t), c(t), and d(t) are differential functions of t.
c(t) d(t)
0 0
a b
a b
0
a. Show that f (t) = det
+ det 0 0 .
c d
c d
d
d
a b
0
f (t) =
det
= (ad − bc) = a0 d + ad0 − b0 c − bc0
c d
dt
dt
0 0
a b
a b
0
0
0
= (a d − b c) + (ad − bc ) = det
+ det 0 0 .
c d
c d
b. Derive a similar formula for the determinant of a 3 × 3 matrix of functions.
The formula, which is given below, is proved the same way
fact, there is a similar formula for n × n matrices.


 0 0 0 

a b c
a b c
a b
d










d e f
d e f
d0 e0
det
= det
+det
dt
g h i
g h i
g h
as for the 2 × 2 case. In



c
a b c
f 0 +det  d e f 
i
g 0 h0 i0