1 1 0 0 2 9 1 0 0 9 0 0 1 9 9 5Expanding with respect to the fourth column gives
1 1 0 5 det 2 9 1 0 9 0and 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 8Expanding with respect to the first column gives
2 1 0 1 3 det 0 2 0 2 5 3 9 9 7 4 0 1Now use the third column to get
2 1 1 3(9) det 0 2 2 7 4 1We can factor 2 out of the second row, which gives
2 1 1 54 det 0 1 1 7 4 1Subtracting the second row from the first gives
2 0 0 54 det 0 1 1 7 4 1and expanding with respect to the first row gives
108 det 1 1 = 108(1-4) = -324 4 132(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 fsince
3 1 1 5 = 1 + 2 2 , 7 1 3and 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_2is 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 29and 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.