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I'm taking a linear algebra class now, and I was introduced to vector equations. Consider the system
$$ \left\{ \begin{array}{rcl} x-4y &=& 8\\ 2x+3y &=& 6\\ \end{array} \right.$$

I want to understand why I can factor out the x and y variables to create the vector equation $$x\begin{bmatrix} 1 \\ 2 \end{bmatrix}+y\begin{bmatrix} -4 \\ 3 \end{bmatrix}=\begin{bmatrix} 8 \\ 6 \end{bmatrix}$$

are $x$ and $y$ scalars here? Is the column vector $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$ the same as the vector $\left< 1, 2\right>$? And lastly, how does this notation help me solve for solutions?

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    $\begingroup$ "I want to understand why I can..." Because the two ways of expressing the system of equations are equivalent. The one implies the other and vice versa. The two sides of the equation in the vector format are equal if and only if the first coordinate in the left hand side is equal to the first coordinate of the right hand side (which is equivalent to the first equation in the first presentation)... similarly for the second. "Are $x$ and $y$ scalars here?" Scalar unknowns, yes... you might also choose to call them variables. $\endgroup$
    – JMoravitz
    Aug 24, 2021 at 17:32
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    $\begingroup$ "Is the column vector $\begin{bmatrix}1\\2\end{bmatrix}$ the same as the vector $\langle 1,2\rangle$?" Strictly? No, however they can be used interchangably. Situations where one is used can be rewritten and possibly rearranged so as to use the other form instead... so they can be thought of as effectively being the same thing... but when it comes to using multiple different shapes simultaneously in the same equation we will be stricter as to which shape can be used when. For instance $[1~2] \begin{bmatrix}3\\4\end{bmatrix}$ is not the same as $\begin{bmatrix}1\\2\end{bmatrix}[3~4]$ $\endgroup$
    – JMoravitz
    Aug 24, 2021 at 17:34
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    $\begingroup$ "How does this notation help me solve for solutions" Taking it a step further, you have $\begin{bmatrix}1&-4\\2&3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}8\\6\end{bmatrix}$ and you can now use matrix algebra to continue... finding that $\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}1&-4\\2&3\end{bmatrix}^{-1}\begin{bmatrix}8\\6\end{bmatrix}$. Solving $Av=b$ for $v$ given an $A$ and $b$ is one of the first uses of Linear Algebra and is useful for many fields. $\endgroup$
    – JMoravitz
    Aug 24, 2021 at 17:36
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    $\begingroup$ @JMoravitz I'm not very familiar with the policies of SE but is there any specific reason you wrote all of this in comments? I mean you answered each query of OP perfectly but you could've written it as answer, so why comment? $\endgroup$ Aug 24, 2021 at 18:03

3 Answers 3

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Yes $x,y$ are scalars. Yes the column vectors are the same as you mentioned. This notation is classical in linear algebra because you can go ahead and write the above as $$\begin{bmatrix} 1 & -4 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8 \\ 6 \end{bmatrix}$$ where now all you have to do is "invert" the matrix to obtain $x,y$.

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The fact is that any linear system in cartesian form can also be expressed in matrix form $Au=b$ and the product $Au$ can also be viewed as the linear combination of the columns of matrix $A$ by the components of the vector $u$.

$$\overbrace{\boxed{\left\{ \begin{array}{rcl} a_{11}x+a_{12}y &=& b_1\\ a_{21}x+a_{22}y &=& b_2\\ \end{array} \right.}}^{\text{cartesian form}} \,\equiv\, \overbrace{\boxed{\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \end{bmatrix}}}^{\text{matrix form}} \,\equiv\, \overbrace{\boxed{x\begin{bmatrix} a_{11} \\ a_{21} \end{bmatrix}+y\begin{bmatrix} a_{12} \\ a_{22} \end{bmatrix}=\begin{bmatrix} b_1 \\ b_2 \end{bmatrix}}}^{\text{vector form}}$$

In some case this interpretation (i.e. solution as a combination of matrix column vectors) can be very useful. In this particular case it seems not much enlightening to find the solution.

As reference I suggest to take a look to the first lesson of the famous Linear Algebra course by Prof. Gilbert Strang.

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I agree with the other answers that the matrix form is the more standard way of viewing the system of linear equations.

But the vector way of viewing things can be thought of as in a 2-d diagram, as we are dealing with two-dimensional vectors.

Clearly $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} -4 \\ 3 \end{bmatrix}$ are linearly independent, as one is not a scalar multiple of the other.

Therefore, the equation:

$$x\begin{bmatrix} 1 \\ 2 \end{bmatrix}+y\begin{bmatrix} -4 \\ 3 \end{bmatrix}=\begin{bmatrix} 8 \\ 6 \end{bmatrix}$$

can be thought of as, "What unique linear combination of the $2-$d vectors $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} -4 \\ 3 \end{bmatrix}$ is equal to $\begin{bmatrix} 8 \\ 6 \end{bmatrix},$ and personally I kind of find this uniqueness kind of "interesting" or "nice", I suppose.

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  • $\begingroup$ I guess this doesn't really answer the question, but I think my answer is still very relevant, and the uniqueness and the diagram aspect suggest some kind of general motivation for applications. $\endgroup$ Aug 24, 2021 at 17:48
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    $\begingroup$ I fully agree, this way to interpret linear system is really enlightening for many applications. $\endgroup$
    – user
    Aug 24, 2021 at 17:54
  • $\begingroup$ Agree too, and also makes it clear what least squares approximation means for finite dimensional vectors. $\endgroup$
    – Paul
    Aug 24, 2021 at 17:59

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