1.3.11 Determine if b is a linear combination

[karush]

Determine if $b$ is a linear combination of $a_1,a_2$ and $a_3$
$$a_1\left[
\begin{array}{r}
1\\-2\\0 \end{array}\right],
a_2\left[
\begin{array}{r}
0\\1\\2
\end{array}\right],
a_3\left[
\begin{array}{r}
5\\-6\\8
\end{array}\right],
b=\left[
\begin{array}{r}
2\\-1\\6
\end{array}\right]$$
(rref) augmented matrix is
$$\left[
\begin{array}{ccc|c}
1 & 0 & 5 & 2 \\
0 & 1 & 4 & 3 \\
0 & 0 & 0 & 0
\end{array} \right]$$
from observation this is not a combination ok well I thot if the bottom row is all 0's then you have 2 equations and 3 answers so notalso, I thot (rref) was all just 1's or 0's
I did it on eMh

 

[Opalg]

Determine if $b$ is a linear combination of $a_1,a_2$ and $a_3$
$$a_1\left[
\begin{array}{r}
1\\-2\\0 \end{array}\right],
a_2\left[
\begin{array}{r}
0\\1\\2
\end{array}\right],
a_3\left[
\begin{array}{r}
5\\-6\\8
\end{array}\right],
b=\left[
\begin{array}{r}
2\\-1\\6
\end{array}\right]$$
(rref) augmented matrix is
$$\left[
\begin{array}{ccc|c}
1 & 0 & 5 & 2 \\
0 & 1 & 4 & 3 \\
0 & 0 & 0 & 0
\end{array} \right]$$
from observation this is not a combination ok well I thot if the bottom row is all 0's then you have 2 equations and 3 answers so notalso, I thot (rref) was all just 1's or 0's
I did it on eMh

You want to know whether $b$ is a linear combination of $a_1$, $a_2$ and $a_3$. So you want to know whether the equation $xa_1 + ya_2 + za_3 = b$ can be solved for $x$, $y$ and $z$. The rref matrix tells you that this is equivalent to solving the equations $$x + 5z = 2,$$ $$y+4z=3$$ (two equations for three unknowns, so there should be plenty of solutions). Try taking it from there.

 

[HallsofIvy]

It is remarkable how many people on this and other boards post problem of the form "show that 'a' is an 'X'" without knowing the definition of 'X'! I would think that, in that situation, the first thing one would do is look up the definition of 'X'.

 

[karush]

It is remarkable how many people on this and other boards post problem of the form "show that 'a' is an 'X'" without knowing the definition of 'X'! I would think that, in that situation, the first thing one would do is look up the definition of 'X'.

They are both coefficients

 

[HallsofIvy]

They are both coefficients

I have no idea what this is supposed to mean or what "both" refers to here. The "X" in my previous post was the term "linear combination". I was concerned with whether or not you know the definition of "linear combination".

A vector "u" is a "linear combination" of vectors $$v_1$$, $$v_2$$, …, $$v_n$$ if and only if there exist scalars (numbers) $$C_1$$, $$C_2$$, … , $$C_n$$ such that $$C_1v_1+ C_2v_2+ \cdot\cdot\cdot+ C_nv_n= u$$.

Here the question is whether or not $$b= \begin{bmatrix}2 \\ -1 \\ 6 \end{bmatrix}$$ is a linear combination of vectors [tex]a_1=

\begin{bmatrix}1 \\ -2 \\ 0 \end{bmatrix}​

[/tex]

, [tex]a_2=

\begin{bmatrix}0 \\ 1 \\ 2 \end{bmatrix}​

[/tex]

, and [tex]a_3=

\begin{bmatrix}5 \\ -6 \\ 8 \end{bmatrix}​

[/tex]

.

That is, do there exist numbers, $$C_1$$, $$C_2$$, and $$C_3$$ such that [tex]C_1

\begin{bmatrix}1 \\ -2 \\ 0 \end{bmatrix}+ C_2

\begin{bmatrix}0 \\ 1 \\ 2 \end{bmatrix}+ C_3

\begin{bmatrix}5 \\ -6 \\ 8 \end{bmatrix}= \begin{bmatrix}C_1+ 5C_2 \\ -2C_1+ C_2- 6C_3 \\ 2C_2+ 8C_3\end{bmatrix}=

\begin{bmatrix}2 \\ -1 \\ 6 \end{bmatrix}​

​

​

​

[/tex]

.

That is the same as the three equations $$C_1+ 5C_2= 2$$, $$-2C_1+ C_2- 6C_3= -1$$, and $$2C_2+ 8C_3= 6$$.

Yes, you can use the "augmented matrix" to solve those equations but jumping directly to that matrix makes me wonder if you understand where the matrix comes from rather than just using a memorized formula. From the first equation we get $$C_1= 2- 5C_2$$. From the third equation we get $$C_3= \frac{3}{4}- \frac{1}{4}C_2$$. Replacing $$C_1$$ and $$C_3$$ in the second equation with those, we get $$-2(2- 5C_2)+ C_2- 6\left(\frac{3}{4}- \frac{1}{4}C_2\right)= -4+ 10C_2+ C_2- \frac{9}{2}+ \frac{3}{2}C_2= \frac{25}{2}C_2- \frac{17}{2}= -1$$. $$\frac{25}{2}C_2= \frac{15}{2}$$ so $$C_2= \frac{3}{5}$$. Then $$C_1= 2- 5C_2= 2- 3= -1$$ and $$C_3= \frac{3}{4}- \frac{1}{4}C_2= \frac{3}{4}- \frac{3}{20}= \frac{15- 3}{20}= \frac{3}{5}$$.

The problem only asked whether b could be written as a linear combination of the other vectors but showing that linear combination is the best way to show it exists.
​

​

​

​

​

​

​