2.6.1 Lab 4¶

Part I. Linear Independence¶

Determine if the sets of vectors are linearly independent or linearly dependent. Explain your reasoning for each set. If they are linearly dependent, give the dependence relation.

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 $\vec y$.

First:

\[\begin{split}\left\{ \vec v_1 =\left[\begin{array}{r}-1\\0\\0\\0\\\end{array}\right],\vec v_2=\left[\begin{array}{r}-2\\15\\-6\\3\\\end{array}\right],\vec v_3=\left[\begin{array}{r}4\\9\\-3\\5\\\end{array}\right],\vec v_4=\left[\begin{array}{r}0\\10\\-4\\2\\\end{array}\right] \right\}\end{split}\]

v1 = [-1 ; 0 ; 0 ; 0 ] ;
v2 = [-2 ; 15 ; -6 ; 3 ] ;
v3 = [4 ; 9 ; -3 ; 5 ] ;
v4 = [0 ; 10 ; -4 ; 2 ] ;

Second: $$\left\{ \vec v_5 =\left[\begin{array}{r}5\\0\\0\\0\\\end{array}\right],\vec v_6=\left[\begin{array}{r}-1\\2\\-2\\0\\\end{array}\right],\vec v_7=\left[\begin{array}{r}2\\-2\\7\\15\\\end{array}\right],\vec v_8=\left[\begin{array}{r}1\\-3\\5\\8\\\end{array}\right] \right\}$$

v5 = [5 ; 0 ; 0 ; 0 ] ;
v6 = [-1 ; 2 ; -2 ; 0 ] ;
v7 = [2 ; -2 ; 7 ; 15 ] ;
v8 = [1 ; -3 ; 5 ; 8 ] ;

First:

\[\begin{split}\left\{ \vec x_1 =\left[\begin{array}{r}0\\9\\-1\\1\\\end{array}\right],\vec x_2=\left[\begin{array}{r}4\\-11\\1\\-1\\\end{array}\right],\vec x_3=\left[\begin{array}{r}7\\-17\\2\\-2\\\end{array}\right],\vec x_4=\left[\begin{array}{r}3\\12\\-1\\1\\\end{array}\right] \right\}\end{split}\]

x1 = [0 ; 9 ; -1 ; 1 ] ;
x2 = [4 ; -11 ; 1 ; -1 ] ;
x3 = [7 ; -17 ; 2 ; -2 ] ;
x4 = [3 ; 12 ; -1 ; 1 ] ;

Second:

\[\begin{split}\left\{ \vec x_5 =\left[\begin{array}{r}-1\\-2\\-1\\-2\\\end{array}\right],\vec x_6=\left[\begin{array}{r}-1\\5\\-2\\2\\\end{array}\right],\vec x_7=\left[\begin{array}{r}-1\\-12\\0\\-9\\\end{array}\right],\vec x_8=\left[\begin{array}{r}-1\\-15\\0\\-11\\\end{array}\right] \right\}\end{split}\]

x5 = [-1 ; -2 ; -1 ; -2 ] ;
x6 = [-1 ; 5 ; -2 ; 2 ] ;
x7 = [-1 ; -12 ; 0 ; -9 ] ;
x8 = [-1 ; -15 ; 0 ; -11 ] ;

First:

\[\begin{split}\left\{ \vec y_1 =\left[\begin{array}{r}-1\\3\\1\\0\\\end{array}\right],\vec y_2=\left[\begin{array}{r}49\\-122\\-49\\5\\\end{array}\right],\vec y_3=\left[\begin{array}{r}34\\-82\\-34\\4\\\end{array}\right],\vec y_4=\left[\begin{array}{r}-44\\247\\84\\-1\\\end{array}\right] \right\}\end{split}\]

y1 = [-1 ; 3 ; 1 ; 0 ] ;
y2 = [49 ; -122 ; -49 ; 5 ] ;
y3 = [34 ; -82 ; -34 ; 4 ] ;
y4 = [-44 ; 247 ; 84 ; -1 ] ;

Second:

\[\begin{split}\left\{ \vec y_5 =\left[\begin{array}{r}12\\0\\0\\12\\\end{array}\right],\vec y_6=\left[\begin{array}{r}0\\4\\0\\0\\\end{array}\right],\vec v_3=\left[\begin{array}{r}19\\-3\\1\\19\\\end{array}\right],\vec v_4=\left[\begin{array}{r}76\\-6\\5\\77\\\end{array}\right] \right\}\end{split}\]

y5 = [12 ; 0 ; 0 ; 12 ] ;
y6 = [0 ; 4 ; 0 ; 0 ] ;
y7 = [19 ; -3 ; 1 ; 19 ] ;
y8 = [76 ; -6 ; 5 ; 77 ] ;

Part II. Null Space in Vector Form¶

Find the null space for the given matrix, and write the solution set in vector form.

If your first name begins with:

A through D, do A.
D through J, do D.
J through Z, do J.

\[\begin{split}A = \left[\begin{array}{rrrr}1&2&1&1\\-2&1&1&-3\\1&-3&-1&3\\\end{array}\right]\end{split}\]

A = [1 2 1 1 ; -2 1 1 -3 ; 1 -3 -1 3 ] ;

\[\begin{split}D = \left[\begin{array}{rrrr}2&6&-4&11\\0&-8&5&-14\\0&12&-7&19\\\end{array}\right]\end{split}\]

D = [2 6 -4 11 ; 0 -8 5 -14 ; 0 12 -7 19 ] ;

\[\begin{split}J = \left[\begin{array}{rrrr}1&8&5&4\\2&91&40&38\\-2&-16&-9&-10\\\end{array}\right]\end{split}\]

J = [1 8 5 4 ; 2 91 40 38 ; -2 -16 -9 -10 ] ;

Part III: Column Space¶

The column space of matrix $A$ is the span of its columns. Find the column space for the following matrices.

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.

\[\begin{split}B = \left[\begin{array}{rrrr}2&2&5&1\\-2&-2&-5&-2\\0&0&0&-1\\2&2&5&1\\\end{array}\right]\end{split}\]

B = [2 2 5 1 ; -2 -2 -5 -2 ; 0 0 0 -1 ; 2 2 5 1 ] ;

\[\begin{split}H = \left[\begin{array}{rrrr}0&-2&-1&5\\0&-2&-1&5\\5&1&-2&0\\0&1&1&-1\\\end{array}\right]\end{split}\]

H = [0 -2 -1 5 ; 0 -2 -1 5 ; 5 1 -2 0 ; 0 1 1 -1 ] ;

\[\begin{split}R = \left[\begin{array}{rrrrr}6&6&-12&37&-8\\-4&6&-2&-7&3\\2&-2&0&5&-2\\\end{array}\right]\end{split}\]

R = [-3 2 2 0 ; -1 -1 1 4 ; -3 -3 3 17 ; 0 0 0 -1 ] ;

Linear Algebra with MATLAB

2.6.1 Lab 4¶

Part I. Linear Independence¶

Part II. Null Space in Vector Form¶

Part III: Column Space¶