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.