2.7:5,7; 3.1:9,10,24,38,41; 3.2:6,7,22,34
DUE: Friday Oct. 13, Beginning of Class
2.7:5) Solve x = Cx + d for the production level x.
(I-C)x = d. Find inverse of (I-C)
I-C = [ 1 -.5 ]
[ -.6 .8 ]
(I-C)^-1 = 1/(.8 - .3) [ .8 .5 ] = [ 1.6 1 ]
[ .6 1 ] [ 1.2 2 ]
Now x = ((I-C)^-1)d.
x = [ 1.6 1 ] [ 50 ] = [ 1.6(50) + 30 ] = [ 110 ]
[ 1.2 2 ] [ 30 ] [ 1.2(50) + 60 ] [ 120 ]
2.7:7) a) The final demand vector is d = [ 1 ]
[ 0 ]
Thus x = [ 1.6 1 ] [ 1 ] = [ 1.6 ]
[ 1.2 2 ] [ 0 ] [ 1.2 ]
b) The final demand vector is [ 51 ]
[ 30 ]
Thus x = [ 1.6 1 ] [ 51 ] = [ 111.6 ]
[ 1.2 2 ] [ 30 ] [ 121.2 ]
c) The production level for final demand [ 51 ] equals the sum of the
[ 30 ]
production level necessary for final demand [ 50 ] and that for [ 1 ], i.e.
[ 30 ] [ 0 ]
[ 111.6 ] = [ 110 ] + [ 1.6 ]
[ 121.2 ] [ 120 ] [ 1.2 ]
3.1:9) Using cofactor expansion along 3rd row:
det [ 6 0 0 5 ] = 2 det [ 0 0 5 ] = 2(5) det [ 7 2 ] = 2(5)(7-6) = 10
[ 1 7 2 -5 ] [ 7 2 -5 ] [ 3 1 ]
[ 2 0 0 0 ] [ 3 1 8 ]
[ 8 3 1 8 ]
3.1:10) Use expansion along row 2:
det [ 1 -2 5 2 ] = 3(-1)det [ 1 -2 2 ] = -3(5 det [-2 2 ] + 4 det [1 -2])
[ 0 0 3 0 ] [ 2 -6 5 ] [-6 5 ] [2 -6]
[ 2 -6 -7 5 ] [ 5 0 4 ]
[ 5 0 4 4 ]
= -3{ 5(-10+12) + 4 (-6+4) } = -3(2) = -6
3.1:24) Rows 1 and 2 are switched to arrive at the second matrix. Thus the
determinant of second matrix is (-1) multiplied by determinant of first.
Verify directly by Cofactor expansion for each matrix.
Expand along row 1:
det [ a b c ] = a(12-10) - b(18-12) + c(15-12) = 2a - 6b + 3c
[ 3 2 2 ]
[ 6 5 6 ]
Expand along row 2:
det [ 3 2 2 ] = -a(12-10) + b(18-12) - c(15-12) = - 2a + 6b - 3c
[ a b c ]
[ 6 5 6 ]
3.1:38)
det kA = det [ ka kb ] = (k^2)ad - (k^2)bc = (k^2)(ad-bc) = (k^2)det A
[ kc kd ]
3.1:41) u = [ 3 ] , v = [ 1 ] , u+v = [ 4 ]
[ 0 ] [ 2 ] [ 2 ]
The area of parallelogram is (base)(height) = 3(2) = 6.
det [ u v ] = det [ 3 1 ] = 6, equal to the area above.
[ 0 2 ]
If u = [ 3 ], v = [ x ], the area equals 3(2) = 6 = det [ 3 x ]
[ 0 ] [ 2 ] [ 0 2 ]
It is the same for any x since the height of the parallelogram remains the same
3.2:6) Row reduce to an echelon form
det[ 1 5 -3 ] = det[ 1 5 -3 ] = det[ 1 5 -3 ] = -det[ 1 5 -3 ]
[ 3 -3 3 ] [ 0 -18 12 ] [ 0 0 6 ] [ 0 3 -1 ]
[ 2 13 -7 ] [ 0 3 -1 ] [ 0 3 -1 ] [ 0 0 6 ]
= -(1)(3)(6) = -18
3.2:7) Row reduce to an echelon form
detA = det[ 1 3 0 2 ] = det[ 1 3 0 2 ]
[-2 -5 7 4 ] [ 0 1 7 8 ]
[ 3 5 2 1 ] [ 0 -4 2 -5 ]
[ 1 -1 2 -3 ] [ 0 -4 2 -5 ]
Rows 3 and 4 are equal (so row vectors linearly dependent) so detA = 0.
3.2:22) Expanding down 1st column:
detA = det[ 5 0 -1 ] = 5(-9+10) - 1(0+5) = 5 - 5 = 0
[ 1 -3 -2 ]
[ 0 5 3 ]
Thus A is not invertible.
3.2:34) By multiplicative property for determinants (see Thm. 6),
det(PAP^-1) = (detP)(detA)(detP^-1) = detA, since
(detP)(detP^-1) = det(PP^-1) = detI = 1.