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.