311.1.3.12 linear combination

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  • Thread starter karush
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  • #1
karush
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$\tiny{311.1.3.12}$
Determine if $b$ is a linear combination of $a_1,a_2$ and $a_3$
$ a_1\left[\begin{array}{r} 1\\0\\1 \end{array}\right],
a_2\left[\begin{array}{r} -2\\3\\-2 \end{array}\right],
a_3\left[\begin{array}{r} -6\\7\\5 \end{array}\right],
b=\left[\begin{array}{r} -7\\13\\4 \end{array}\right]$

ok I don't think this is too difficult to do.
but these matrix problems are very error prone
so thot I would just do a step at a time here
from the example I looked at this is the same thing as
$\left[\begin{array}{lll}a_1&+(-2a_2)&+(-6a_3)\\
&+3a_2 &+7a_3\\
a_1&+(-2a_2)&+5a_3) \end{array}\right]
=\left[\begin{array}{r} -7\\13\\4 \end{array}\right]$
I left all the + signs in since I think this is what a combination is, so then
$\left[\begin{array}{rrr|r}1&-2&-6&-7\\
0&3&7&13\\
1&-2&5&4 \end{array}\right]$
by RREF I got $a_1=3,\quad a_2=2\quad a_3=1$
 
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Answers and Replies

  • #3
HOI
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I don't see any matrix!

These vectors will be independent if and only if whenever
\(\displaystyle A\begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}+ B\begin{bmatrix}-2 \\ 3 \\ -2\end{bmatrix}+ C\begin{bmatrix}-6 \\ 7 \\ 5\end{bmatrix}+ D\begin{bmatrix}-7 \\ 13 \\ 4 \end{bmatrix}= 0\) we must have A= B= C= D= 0.

\(\displaystyle \begin{bmatrix}A- 2B- 6C- 7D \\ 3B+ 7C+ 13D \\ A- 2B+ 5C+ 4D\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}\).

So we need to solve A- 2B- 6C- 7D= 0., 3B+ 7C+ 13D= 0, and A- 2B+ 5C+ 4D= 0.
If A= B= C= D= 0 is the only solution the vectors are independent. If there are other solutions they are dependent.

Of course, if we were really clever we would have said, right at the start, that four vectors in three dimensional space CAN'T be independent!

No nasty matrices!
 
  • #4
karush
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well that's good to know😎
 
  • #5
karush
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wanted to add this true or false statement and justification:

Each matrix is row equivalent to one to one and only one reduced echelon matrix
ok I don't think this is true because some matrix are not one to one

The row reduction algorithm applies only to augmented matrices for a linear system
no really sure isn't an augmented matrix Ax=b in one place
 
  • #6
HOI
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I have no idea what you mean by a matrix being "one to one".

In any case, this question has nothing to do with "one to one". Also row reduction can be applied to any matrix. The matrix doesn't have to be "augmented" and row reduction is not only used to solve equations.

This question is only asking if a row reduction is "unique"- if a given matrix has only one row reduced form..
 
  • #7
karush
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I thot one to one meant like 3 variables 3rows 3cols in Ax=b
 
  • #8
HOI
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No, "one to one" means that each value of the independent variable gives one unique value of the dependent value. It does NOT necessarily have anything to do with matrices or linear transformations.
 

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