## COMMENTS ON HOMEWORK 9

**
2. The problem is to find the determinant of **
1 1 0 0
2 9 1 0
0 9 0 0
1 9 9 5

Expanding with respect to the fourth column gives
1 1 0
5 det 2 9 1
0 9 0

and expanding with respect to the third row gives
5(-9) det 1 0
2 1

= (-45)(1)(1) (since the matrix is now lower triangular)
= -45.
4. Here the matrix is

0 2 1 0 1
0 0 2 0 2
0 5 3 9 9
0 7 4 0 1
3 9 5 4 8

Expanding with respect to the first column gives
2 1 0 1
3 det 0 2 0 2
5 3 9 9
7 4 0 1

Now use the third column to get
2 1 1
3(9) det 0 2 2
7 4 1

We can factor 2 out of the second row, which gives
2 1 1
54 det 0 1 1
7 4 1

Subtracting the second row from the first gives
2 0 0
54 det 0 1 1
7 4 1

and expanding with respect to the first row gives
108 det 1 1 = 108(1-4) = -324
4 1

32(a) By linearity in the second column
a 3 d a 1 d
det b 3 e = 3 det b 1 e = 3(7) = 21
c 3 f c 1 f

(b) Again by linearity in the second column
a 3 d a 1 d a 1 d
det b 5 e = det b 1 e + 2 det b 2 e ,
c 7 f c 1 f c 3 f

since
3 1 1
5 = 1 + 2 2 ,
7 1 3

and this is 7 + 2(11) = 29.
5.3

2. The area of the specified triangle is one half of the area of
the parallelogram determined by the specified vectors, which
is half the determinant of the matrix

3 8

7 2

that has those vectors as columns. The determinant is 6 - 56 = -50,
and so the area of the triangle is |-50/2| = 25.

6. By problems 5.3.4 and 5.3.5 (whose solutions were provided)
the volume of the tetrahedron with the given columns is 1/6
the absolute value of the determinant of the matrix they form, i.e.,

a_1 b_1 c_1
a_2 b_2 c_2
1 1 1 ,

while the area of the triangle with vertices
a_1 b_1 c_1
a_2 b_2 c_2

is half the the absolute value of that same determinant. Therefore,
the volume of the tetrahedron is 1/3 the area of the triangle.
14. By the formula developed in class, the 3-volume of the
3-parallepiped determined by the columns of A =

1 1 1

0 1 2

0 1 3

0 1 4

is the square root of det (A^T A). In this case, A^T =

1 0 0 0

1 1 1 1

1 2 3 4

and A^T A =

1 1 1
1 4 10
1 10 30

(The computation is slightly simplified by using the
fact that we know A^T A is symmetric.)
Subtracting the first row from the second and third gives

1 1 1
0 3 9
0 9 29

and subtracting 3 times the second row from the third gives
1 1 1

0 3 9

0 0 2

**
Thus, the determinant is 6, and the answer is the
square root of 6.
**