1.3: 9,13,21; 1.4: 19,23,33,39
1.5: 5,13,19; 1.6: 3,7,31
DUE: Wed., Feb. 7, Beginning of Class
1.3:9)
x_1 [ 2 ] + x_2 [-1 ] + x_3 [ 5 ] = [ 3 ]
[ 1 ] [-8 ] [ 2 ] [ 5 ]
[ 0 ] [ 4 ] [-4 ] [ 5 ]
1.3:13) Row reduce the augmented matrix to an echelon form; if consistent,
then b is a linear combination of the columns of A:
[ 1 0 2 -5 ] [ 1 0 2 -5 ] [ 1 0 2 -5 ]
[ -2 5 0 11 ] ~ [ 0 5 4 1 ] ~ [ 0 5 4 1 ]
[ 2 5 8 -7 ] [ 0 5 4 3 ] [ 0 0 0 2 ]
The last row says "0 = 2", a contradiction, so the system is inconsistent
and hence b is not a linear combination of columns of A.
1.3:21) The system Ax=b is represented by the augmented matrix
[ 1 -5 3 ] [ 1 -5 3 ] [ 1 -5 3 ]
[ 3 -8 -5 ] ~ [ 0 7 -14 ] ~ [ 0 1 -2 ]
[ -1 2 h ] [ 0 -3 3+h] [ 0 0 h-3]
Finding h such that b is in span{a_1, a_2} is equivalent to finding h such
that the system is consistent. We need h - 3 = 0; i.e. h = 3.
1.4:19)
[-3 1 b_1 ] 2 row1 + row 2: [-3 1 b_1 ]
[ 6 -2 b_2 ] [ 0 0 2b_1 + b_2 ]
Hence the system is consistent only when 2 b_1 + b_2 = 0, that is,
b_2 = -2 b_1. The vector b must satisfy
b = [ b_1 ] = [ b_1 ] = b_1 [ 1 ] (b_1 free)
[ b_2 ] [-2b_1 ] [-2 ]
This says b must be a multiple of the vector [ 1 ] for the system to
be consistent. [-2 ]
Equivalently, the set { b: Ax=b is consistent} is span { [ 1 ] }.
[-2 ]
1.4:23) By Thm. 4, an equivalent question is: does matrix A have a pivot
position in every row?
Row reduce matrix A to an echelon form:
[ 1 3 -4 ]
[ 3 2 -6 ]
[-5 -1 8 ]
-3 row1 + row2; and 5 row1 + row3
[ 1 3 -4 ]
[ 0 -7 6 ]
[ 0 14 -12]
2 row2 + row3
[ 1 3 -4 ]
[ 0 -7 6 ]
[ 0 0 0 ]
Row3 (all zeros) does not have a pivot postion, so the columns of A do not
span R^3.
1.4:33)
A (u + v) = [ 5 1 -3 ] [ -1 ] = [-10 ]
[ 7 -2 1 ] [ 4 ] [-12 ]
[ 3 ]
Au + Av = [ 5 1 -3 ] [ 1 ] + [ 5 1 -3 ] [ -2 ]
[ 7 -2 1 ] [ 3 ] [ 7 -2 1 ] [ 1 ]
[ 3 ] [ 0 ]
= [ -1 ] + [ -9 ] = [-10 ]
[ 4 ] + [-16 ] [-12 ]
1.4:39) Row reduce to determine if the columns of the matrix span R^4:
(follow left to right)
[ 7 2 -5 8 ] row1+row4-->row4 [ 7 2 -5 8 ]
[ -5 -3 4 -4 ] 6 row2 [-30 -18 24 -54 ] row2 + row3
[ 6 10 -2 7 ] 5 row3 [ 30 50 -10 35 ] --> row3
[ -7 9 2 15 ] [ 0 11 -3 23 ]
[ 7 2 -5 8 ] [ 7 2 -5 8 ]
[-30 -18 24 -54 ] 7 row2; then [ 0 -66 18 -138] 6 row4 + row2
[ 0 32 14 -19 ] 30row1 + row2 [ 0 32 14 -19 ]
[ 0 11 -3 23 ] [ 0 11 -3 23 ]
[ 7 2 -5 8 ] No further row reduction is necessary.
[ 0 0 0 0 ] Row 2 all zeros (no pivot position)
[ 0 32 14 -19 ] implies that the columns of original matrix
[ 0 11 -3 23 ] do not span R^4 (Theorem 4).
1.5:5)
[ 1 -3 -2 0 ] [ 1 -3 -2 0 ]
[ 0 1 -1 0 ] ~ [ 0 1 -1 0 ] ~
[ -2 3 7 0 ] [ 0 -3 3 0 ]
[ 1 -3 -2 0 ] [ 1 0 -5 0 ]
[ 0 1 -1 0 ] ~ [ 0 1 -1 0 ]
[ 0 0 0 0 ] [ 0 0 0 0 ]
Hence x_3 is free variable, so the homogeneous system has the vector
solution:
x = [ x_1 ] = [ 5x_3 ] = x_3 [ 5 ]
[ x_2 ] [ x_3 ] [ 1 ]
[ x_3 ] [ x_3 ] [ 1 ]
x_3 is a (free) parameter, so this is the parametric form for the
solution set: a line through the origin.
1.5:13) Get augmented matrix in reduced echelon form
[ 1 -3 -2 -5 ] [ 1 -3 -2 -5 ]
[ 0 1 -1 4 ] ~ [ 0 1 -1 4 ] ~
[ -2 3 7 -2 ] [ 0 -3 3 -12 ]
[ 1 -3 -2 -5 ] [ 1 0 -5 7 ]
[ 0 1 -1 4 ] ~ [ 0 1 -1 4 ] ~
[ 0 0 0 0 ] [ 0 0 0 0 ]
Hence x_3 is free and the solution is
x = [ x_1 ] = [7 + 5x_3 ] = [ 7 ] + x_3 [ 5 ]
[ x_2 ] [4 + x_3 ] [ 4 ] [ 1 ]
[ x_3 ] [ x_3 ] [ 0 ] [ 1 ]
This is a line parallel to the solution of the homogeneous system of
Exercise 1.5:5, shifted to pass through the point ( 7, 4, 0 ) rather than
the origin. Note the form for the solution set is a particular solution
plus the solution set of the associated homogeneous system.
1.5:19) The direction of the line is the vector q - p ( or you could use
p - q ), where
q - p = [ 0 ] - [-1 ] = [ 1 ] = v
[ 7 ] [ 4 ] [ 3 ]
So x = p + tv is our parametric vector equation for the line,
where t is a parameter (any real #)
x = [ x_1 ] = [-1 ] + t [ 1 ]
[ x_2 ] [ 4 ] [ 3 ]
1.6:3) Linearly dependent by inspection, since one is a multiple of the other
[ -2 ] = -2 [ 1 ]
[ 10 ] [ -5 ]
1.6:7) Yes, linearly dependent since there are more column vectors (4) than
entries in a column vector (3). See Theorem 8, p.62.
Alternatively, row reduction will show there is a free variable (or
use previous homework 1.2:31 characterizing an 'underdetermined' system) for
the homogeneous system Ax=0. Each non-zero value of the free variable
determines a non-zero solution ( x_1, x_2, x_3, x_4 ) of
x_1 [ 1 ] + x_2 [ 3 ] + x_3 [-2 ] + x_4 [ 0 ] = [ 0 ]
[ 3 ] [10 ] [-7 ] [ 1 ] [ 0 ]
[-5 ] [-5 ] [ 3 ] [ 7 ] [ 0 ]
showing that the column vectors are linearly dependent by definition.
1.6:31) True.
v_3 is a linear combination of the others since
v_3 = 2 v_1 + v_2 + 0 v_4 ; now utilize Theorem 7, p.61.