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2.4.1 Lab 3

Part I. Determine if a vector is in the span of a set of vectors

If your birthday is in:

  • January through April, do the sets of vectors \(\vec v\).

  • May through August, do the sets of vectors \(\vec x\).

  • September through December, do the sets of vectors \(\vec y\).

Note

If the columns of matrix \(A\) are the vectors \(\{\vec a_1,\vec a_2, \dots , \vec a_k \}\), then

\[A = [\vec a_1,\vec a_2, \dots , \vec a_k ]\]

The same basic notation works in MATLAB to create \(A\) from vectors a1, a2 and a3.

A = [a1, a2, a3]

Determine if the vector

\[\begin{split}\vec v = \left[\begin{array}{r}1\\4\\1\\7\\\end{array}\right]\end{split}\]

is in the \(\text{Span}\{\vec v_1,\vec v_2, \vec v_3 \}\) where

\[\begin{split}\vec v_1 =\left[\begin{array}{r}0\\0\\-1\\0\\\end{array}\right],\vec v_2=\left[\begin{array}{r}3\\6\\0\\12\\\end{array}\right],\vec v_3=\left[\begin{array}{r}1\\1\\1\\2\\\end{array}\right]\end{split}\]

Determine if the vector

\[\begin{split}\vec x = \left[\begin{array}{r}-1\\0\\-1\\-1\\\end{array}\right]\end{split}\]

is in the \(\text{Span} = \{\vec x_1,\vec x_2, \vec x_3 \}\) where

\[\begin{split}\vec x_1 =\left[\begin{array}{r}-2\\0\\0\\0\\\end{array}\right],\vec x_2=\left[\begin{array}{r}14\\5\\0\\0\\\end{array}\right],\vec x_3=\left[\begin{array}{r}-1\\-1\\1\\1\\\end{array}\right]\end{split}\]
x1 = [-2 ; 0 ; 0 ; 0 ] ;
x2 = [14 ; 5 ; 0 ; 0 ] ;
x3 = [-1 ; -1 ; 1 ; 1 ] ;
x = [-1 ; 0 ; -1 ; -1 ] ;

Determine if the vector

\[\begin{split}\vec y=\left[\begin{array}{r}2\\-3\\-1\\-1\\\end{array}\right]\end{split}\]

is in the \(\text{Span}\{\vec y_1,\vec y_2, \vec y_3 \}\) where

\[\begin{split}\vec y_1 =\left[\begin{array}{r}3\\-3\\0\\0\\\end{array}\right],\vec y_2=\left[\begin{array}{r}3\\2\\0\\0\\\end{array}\right],\vec y_3=\left[\begin{array}{r}5\\3\\1\\0\\\end{array}\right]\end{split}\]
y1 = [3 ; -3 ; 0 ; 0 ] ;
y2 = [3 ; 2 ; 0 ; 0 ] ;
y3 = [5 ; 3 ; 1 ; 0 ] ;
y = [2 ; -3 ; -1 ; -1 ] ;

Part II. Matrix-vector multiplication

Use the dot product function dot to show matrix-vector multiplication.

Hint

You can create a row vector from the row of a matrix, for example, the second row as shown below.

A = randi(5,5,4)
r2 = A(2,:)

Tip

The MATLAB function dot works equally well with row vectors and column vectors.

If your first name begins with:

  • A through D, do A.

  • D through J, do D.

  • J through Z, do J.

Multiply \(A\vec v\) where

\[\begin{split}A = \left[\begin{array}{rrrrr}7&-3&6&5&9\\7&2&5&-3&-4\\-3&7&6&4&0\\0&5&8&-3&4\\\end{array}\right], \vec v = \left[\begin{array}{r}7\\-1\\1\\-2\\8\\\end{array}\right]\end{split}\]
A = [7 -3 6 5 9 ;7 2 5 -3 -4 ;-3 7 6 4 0 ;0 5 8 -3 4 ] ;
v = [ 7 ; -1 ; 1 ; -2 ; 8 ] ;

Multiply \(D\vec x\) where

\[\begin{split}D = \left[\begin{array}{rrrrr}2&6&-2&-2&-5\\7&-3&-3&10&10\\3&5&-3&-5&-1\\-2&5&-3&10&3\\\end{array}\right], \vec x = \left[\begin{array}{r}6\\-1\\3\\-1\\7\\\end{array}\right]\end{split}\]
D = [2 6 -2 -2 -5 ;7 -3 -3 10 10 ;3 5 -3 -5 -1 ;-2 5 -3 10 3 ] ;
x = [6 ;-1 ;3 ;-1 ;7 ] ;

Multiply \(J\vec y\) where

\[\begin{split}J = \left[\begin{array}{rrrrr}7&-3&-1&2&2\\9&7&-2&10&2\\2&9&3&4&7\\8&8&1&10&6\\\end{array}\right], \vec y = \left[\begin{array}{r}7\\0\\6\\5\\-2\\\end{array}\right]\end{split}\]
J = [7 -3 -1 2 2 ;9 7 -2 10 2 ;2 9 3 4 7 ;8 8 1 10 6 ] ;
y = [7 ;0 ;6 ;5 ;-2 ] ;

Part III: Solve the homogeneous system of equations

Solve the homogenenous system of equations which are of the form \(A\vec v = \vec 0\), but use row operations instead of the rref function. Write the solution set in vector form.

If your last name begins with:

  • B through G, do B.

  • H through Q, do H.

  • R through Z, do R.

  • A pick any of the three.

\[\begin{split}B = \left[\begin{array}{rrrrr|r}7&18&39&-16&-1&0\\0&-1&-1&2&0&0\\-9&-22&-49&18&1&0\end{array}\right]\end{split}\]
B = [7 18 39 -16 -1 ;0 -1 -1 2 0 ;-9 -22 -49 18 1 ] ;
z = zeros(3,1) ;
[B,z]

ans =

     7    18    39   -16    -1     0
     0    -1    -1     2     0     0
    -9   -22   -49    18     1     0
\[\begin{split}H = \left[\begin{array}{rrrrr|r}1&-5&-2&-1&-4&0\\-5&25&7&8&20&0\\-2&10&-2&11&11&0\end{array}\right]\end{split}\]
H = [1 -5 -2 -1 -4 ;-5 25 7 8 20 ;-2 10 -2 11 11 ] ;
z = zeros(3,1) ;
[H,z]

ans =

     1    -5    -2    -1    -4     0
    -5    25     7     8    20     0
    -2    10    -2    11    11     0
\[\begin{split}R = \left[\begin{array}{rrrrr}6&6&-12&37&-8\\-4&6&-2&-7&3\\2&-2&0&5&-2\\\end{array}\right]\end{split}\]
R = [6 6 -12 37 -8 ;-4 6 -2 -7 3 ;2 -2 0 5 -2 ] ;
z = zeros(3,1) ;
[R,z]

ans =

     6     6   -12    37    -8     0
    -4     6    -2    -7     3     0
     2    -2     0     5    -2     0
2.4 Solution Sets 2.5 Linear Independence