Math2270, Spring 2007: Homework 11 Solutions
Dr. David M. Goulet
April 3, 2007
1. §6.1.
(a) # 16.
1 2 3 4 k 5 = −2 4 5
6 8
6 7 8 +k 1 3
6 8
− 7 1 3
4 5
= −10k + 45
(1)
So The matrix is invertible iff k 6= 9/2.
(b) # 44. If A is an n × n matrix, and k is an arbitrary constant, what is the
relationship between det(A) and det(kA)?
When we form the matrix kA, each entry of A is multiplied by k. This
means that each row is multiplied by k. Because there are n rows, the
determinant changes by a factor of k n . So det(kA) = k n det(A).
2. §6.2.


v~1
 v~2 

(a) # 14. Given det 
 v~3  = 8. By elementary row operations we have
v~4



v~1
 v~2 + 9v~4 

 = det 
det 



v~3
v~4
1

v~1
v~2 
=8
v~3 
v~4
(2)
(b) # 38. If det(A) = 3 for some n × n matrix, what is det(AT A)? From class
we know
det(AT A) = det(AT )det(A) = det(A)det(A) = 9
(3)
(c) # 50.
1 . . . 1 2 . . . 2 3 . . . 3 = ..
.. .
. 1 2 3 ... n 1
1
1
..
.
1
2
2
..
.
1 . . . 1 1 . . . 1 1 . . . 1 = 1
..
.. .
. 0 0 0 ... 1 1
0
0
..
.
1
1
0
..
.
(4)
This reduction was found by subtracting each row from the subsequent
row, starting from the bottom. The final result is 1 because, as we showed
in class, the determinant of a triangular matrix is the product of diagonal
elements.
3. §6.3.
(a) # 12. Consider those 4 × 4 matrices whose entries are all 1, −1, or 0.
What is the maximal value of the determinant of a matrix of this type?
Give an example whose determinant has this maximal value.
We know that the absolute value of the determinant gives the volume
of a parallelpiped. If we can only use the numbers
√ 1, −1, and 0, then the
longest any edge of the parallelpiped can be is 1 + 1 + 1 + 1 = 2. Also,
the volume will be largest if the sides of the parallelpiped are orthogonal,
because this will form a 4D cube. In this case, the volume of the hypercube would be 24 = 16. So we wish to find a matrix with each column
of length 2 and with the columns orthogonal to one another. Let’s see if
this is possible. Consider the matrix


−1
1
1 −1
 −1 −1
1
1 

A=
(5)
 −1
1 −1
1 
−1 −1 −1 −1
Notice that each column has length 2 and that the columns are orthogonal.
Also notice that det(A) = 16, the predicted maximum value.
2
(b) # 22. Us Cramer’s rule.
1
3
x = 3
4
3
4
y = 3
4
−10
=
= −2
5
1 3 5
= =1
5
7 11 7
11
7
11
(6)
(7)
4. §Chapter 6 Review.
(a) # 6. The equation det(−A) = det(A) holds for all 6 × 6 matrices.
True. By problem 44 in §6.1, det(−A) = (−1)6 det(A) = det(A).
(b) # 20. There exists a non-zero 4×4 matrix, A, such that det(A) = det(4A).
True. Let A be any non-zero 4 × 4 matrix with det(A) = 0. Then,
by problem 44 in §6.1, det(4A) = 44 det(A) = 0.
(c) # 28. There exists invertible 2 × 2 matrices A and B such that det(A +
B) = det(A) + det(B).
True. For example.
0 1
2 0
+
1 0 = −1
0 1 1 1 = −1
2 1 (8)
(9)
(d) # 32. If A is any skew-symmetric 4 × 4 matrix, then det(A) = 0.
False. For example.
0
0 0
0
0 1
0 −1 0
−1
0 0
3
1
0
0
0
=1
(10)
(e) # 40. There exists invertible 3 × 3 matrices A and S such that S T AS =
−A.
False. det(S T AS) = det(S T )det(A)det(S) = (det(S))2 det(A) while det(−A) =
−det(A) by problem 44 in §6.1. These can only be equal if (det(S))2 = −1,
which is impossible because det(S) is a real number.
1 1 1 1
1
1 1 2 3 4
5
6 1 3 6 10 15 21 = 1.
5. With as little work as possible, show that 1 4 10 20 35 56 1 5 15 35 70 126 1 6 21 46 126 252 1 1 1 1
1
1
0 1 2 3
4
5 0 2 5 9 14 20 . Now
Subtract the first row from each of the others 0 3 9 19 34 55 0 4 14 34 69 125 0 5 20 45 125 251 subtract twice the second row from the third, 3 times the second row from the
1 1 1 1
1
1 0 1 2 3
4
5 0 0 1 3
6 10 fourth, etc. . Continue this process of Gaussian
0 0 3 10 22 40 0 0 6 22 53 105 0 0 10 30 105 226 1 1 1 1 1 1 0 1 2 3 4 5 0 0 1 3 6 10 .
elimination so as to create an upper triangular matrix 0 0 0 1 4 10 0 0 0 0 1 5 0 0 0 0 0 1 So the determinant is 1.
4