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I was attempting this question.

Is $w=(-12,20)$ a linear combination of $v_1 =(-1,2)$ and $v_2=(4,-6)$?

I understand that you have do the following: $$w = c_1v_1 +c_2v_2 \\ (-12,20)= c_1 (-1,2) + c_2 (4,-6)$$ and put it in a system like:

\begin{cases}-c_1 + 4c_2 = -12 \\ 2c_1- 6c_2 = 20 \end{cases} and to make sure the system has one solution and is consistent and finally to verify the system has a unique solution, the augmented matrix must be used to find $c_1 \text{and} c_2$, however, this is where I'm stuck on, I have used the augmented matrix and have done the row operations but can someone help me extracting the values from the augmented matrix The row operations i did was :

-1 2 |1 0
4 -6 |0 1
R1=3R1+R2
1 0 | 3 1
4 -6| 0 1
R2=4R1-R2
1 0 | 3 1
0 -6| 12 3
R2=R2/-6
1 0 | 3 1
0 1 | -2 -1/6

Thank You in advance!

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"I have used the augmented matrix and have done the row operations" - please show what you did on your paper... –  J. M. Apr 15 '12 at 17:12
    
@J.M. ive added it, but the format comes out quite weird –  Xabi Apr 15 '12 at 17:22
    
@Fatz: I fixed that part of it. For future reference, use a double space at the end of a line for a line break. –  El'endia Starman Apr 15 '12 at 17:41
    
@El'endiaStarman okay thanks :) –  Xabi Apr 15 '12 at 18:18
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2 Answers 2

up vote 2 down vote accepted

When writing the system $$\tag{1} \eqalign{ \color{maroon}{-1}\cdot c_1+\color{darkgreen}4c_2&=\color{darkblue}{-12}\cr \color{maroon}2c_1\color{darkgreen}{-6}c_2&=\color{darkblue}{20} } $$ as an augmented matrix: $$\tag{2} \left[ \matrix{\color{maroon}{ -1}&\color{darkgreen}4 \cr\color{maroon} 2&\color{darkgreen}{-6}} \biggl| \matrix{\color{darkblue}{-12}\cr \color{darkblue}{20}}\right] $$

You're essentially just writing the original system down, but you are not writing the variables or the equality sign. In $(2)$, the red terms are the coefficients of the variable $c_1$ in the system of equations $(1)$. So column one of $(2)$ is the ''$c_1$-column". Similarly, column two of $(2)$ is the ''$c_2$-column" and the third column corresponds to the constants on the right hand side of the equalities of the system of equations $(1)$.

Performing a row operation on the matrix $(2)$ corresponds to performing the same operation on the system.

A reduced form of the augmented matrix $(2)$ is

$$\tag{3} \left[ \matrix{\color{maroon}{ -1}&\color{darkgreen}4 \cr\color{maroon} 0&\color{darkgreen}{2}} \biggl| \matrix{\color{darkblue}{-12}\cr \color{darkblue}{-4}}\right] $$ This was obtained by adding twice row one to row two. (Note that this corresponds to adding twice equation one to equation two in the system of equations $(1)$.)

Translating the matrix $(3)$ back to the corresponding system of equations:

Row one of $(3)$ gives $\color{maroon}{-1}c_1+\color{darkgreen}4c_2=\color{darkblue}{-12}$

Row two of $(3)$ gives $\color{maroon} 0c_1+\color{darkgreen}2c_2=\color{darkblue}{-4}.$


You could reduce $(3)$ further to

$$\tag{4} \left[ \matrix{\color{maroon}{ 1}&\color{darkgreen}0 \cr\color{maroon} 0&\color{darkgreen}{1}} \biggl| \matrix{\color{darkblue}{ 4}\cr \color{darkblue}{-2}}\right] $$ And the corresponding system would be $\eqalign{\color{maroon}1c_1 +\color{darkgreen}0 c_2&=\color{darkblue}4\cr \color{maroon}0c_1+\color{darkgreen}1c_2&=\color{darkblue}{-2} }$, or simply $c_1=4$, $c_2=-2$.


Just keep in the back of your mind that writing

$$ \left[ \matrix{\color{maroon}{ a}&\color{darkgreen}b \cr\color{maroon} c&\color{darkgreen}{d}} \biggl| \matrix{\color{darkblue}{ e}\cr \color{darkblue}{f}}\right] $$

is shorthand notation for writing $$ \eqalign{\color{maroon}a\,c_1 +\color{darkgreen}b\, c_2&=\color{darkblue}e\cr \color{maroon}c\,c_1+\color{darkgreen}d\,c_2&=\color{darkblue}{f} }. $$

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Thank you sooo much! a really clear and concise explanation :) –  Xabi Apr 15 '12 at 17:41
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Notice that your system of linear equations in matrix equation $AX=B$ would look like:

$$\begin{bmatrix} -1 & 4 \\2 & -6 \end{bmatrix} \begin{bmatrix} c_1 \\c_2 \end{bmatrix}=\begin{bmatrix}-12\\20 \end{bmatrix}$$ The augmented matrix $[A,B]$ should look like:

\begin{bmatrix} -1 & 4 &-12\\2& -6 &20 \end{bmatrix}

Notice that the elementary row transformation $R_2 \leftrightarrow R_2+2R_1$ takes the matrix to Row-Echelon form:

$$\sim \begin{bmatrix} -1 &4 &-12\\0 &2 &-4 \end{bmatrix}$$

Now, notice that, your matrix equation is equivalent to $$\begin{bmatrix} -1 & 4 \\0 & 2 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} -12\\-4 \end{bmatrix}$$

Can you take it from here?

So, $c_2=-2$ and hence $c_1=4$. This means that $(-12,20)$ is a linear combination of $(-1,2)$ and $(4,-6)$. $\blacksquare$

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