MATH 240

Wed. September 11th, 2019

Homogeneous Systems of Equations (cont.)

A=(35−40−3−2406180)→(10−43001000000)A=\begin{pmatrix} 3&5&-4&0\\ -3&-2&4&0\\ 6&1&8&0 \end{pmatrix} \rightarrow \begin{pmatrix} 1&0&-\frac{4}{3}&0\\ 0&1&0&0\\ 0&0&0&0 \end{pmatrix}

The solution set of the homogeneous system of equations is equivalent to
Span{(4301)},\text{Span}\{\begin{pmatrix} \frac{4}{3}\\0\\1 \end{pmatrix}\},
i.e. it can be represented as c(4301),c\begin{pmatrix} \frac{4}{3}\\0\\1 \end{pmatrix}, where cc is a scalar.

Connecting to the Non-Homogeneous Case

If b=(7−1−4),b=\begin{pmatrix} 7\\-1\\-4 \end{pmatrix}, then find the solution to Ax=b.Ax=b.

(35−47−3−24−1618−4)→(10−43−101020000)\begin{pmatrix} 3&5&-4&7\\ -3&-2&4&-1\\ 6&1&8&-4 \end{pmatrix} \rightarrow \begin{pmatrix} 1&0&-\frac{4}{3}&-1\\ 0&1&0&2\\ 0&0&0&0 \end{pmatrix}
{x1−43x3=−1x2=2x3 is free→(−120)+x3(4301)\begin{cases} x_1-\frac{4}{3}x_3=-1\\ x_2=2\\ x_3\text{ is free} \end{cases} \rightarrow \begin{pmatrix} -1\\2\\0 \end{pmatrix} + x_3\begin{pmatrix} \frac{4}{3}\\0\\1 \end{pmatrix}

The solution to the non-homogeneous system is just the solution to the homogeneous case plus some constant vector.

Theorem: Suppose the equation Ax=bAx=b is consistent and pp is a solution. Then, the solution set of Ax=bAx=b is the set of all vectors of the form w=p+vhw=p+v_h, where vhv_h is any solution of Ax=0Ax=0.

Linear Dependence/Independence

Let S={v1,...vp}S=\{v_1,...v_p\} be an indexed set of vectors. We say that SS is linearly independent if x1v1+...+xpvp=0x_1v_1+...+x_pv_p=0 implies that x1,...xp=0x_1,...x_p=0. (i.e. the homogeneous system formed by using this set of vectors as columns only has the trivial solution.)

If there is a solution where not all of the xix_i are zero, then SS is linearly dependent.

Linearly independent vectors span the entire space (always consistent) and span it uniquely (so there’s only one possible solution for each point), which makes them useful for labelling points in space.

Ex.

v1=(123), v2=(456), v3=(210)v_1=\begin{pmatrix} 1\\2\\3 \end{pmatrix}, \ v_2=\begin{pmatrix} 4\\5\\6 \end{pmatrix}, \ v_3=\begin{pmatrix} 2\\1\\0 \end{pmatrix}
Is {v1,v2,v3}\{v_1,v_2,v_3\} linearly independent?
x1v1+x2v2+x3v3=0x_1v_1+x_2v_2+x_3v_3=0(142025103600)→(14200−3−300−6−60)→(14200−3−300000)\begin{pmatrix} 1&4&2&0\\ 2&5&1&0\\ 3&6&0&0 \end{pmatrix}\rightarrow \begin{pmatrix} 1&4&2&0\\ 0&-3&-3&0\\ 0&-6&-6&0 \end{pmatrix}\rightarrow \begin{pmatrix} 1&4&2&0\\ 0&-3&-3&0\\ 0&0&0&0 \end{pmatrix}
We can see that we have a free variable from the bottom row. Therefore {v1,v2,v3}\{v_1,v_2,v_3\} is linearly dependent.
(14200−3−300000)→(142001100000)→(10−2001100000)\begin{pmatrix} 1&4&2&0\\ 0&-3&-3&0\\ 0&0&0&0 \end{pmatrix}\rightarrow \begin{pmatrix} 1&4&2&0\\ 0&1&1&0\\ 0&0&0&0 \end{pmatrix}\rightarrow \begin{pmatrix} 1&0&-2&0\\ 0&1&1&0\\ 0&0&0&0 \end{pmatrix}{x1=2x3x2=−x3x3 is free→x3(2−11)\begin{cases} x_1=2x_3\\ x_2=-x_3\\ x_3\text{ is free} \end{cases}\rightarrow x_3\begin{pmatrix} 2\\-1\\1 \end{pmatrix}2v1−1v2+1v3=0.2v_1-1v_2+1v_3=0.