Question

Does anyone know of a relationship between combining (adding/subtracting) equations to solve them and linear combinations?

Answer 1

i would assume that combining equations, both in equation and matrix form, is essentially taking linear combinations, but is there a more in depth relationship?

Answer 2

Yes that is technically correct. This seems to be on the topic of linear algebra, so i'll give it my best shot. A matrix is a represntation of a linear system of equations. For example:
\[\left(\begin{matrix}x _{1} + x _{2} +x_3= 2 \\ x _{1} + x_2 + 2x_3 = 4\\x_2 + 2x_3=4\end{matrix}\right)\] is a system of linear equations, which we can represent in matrix form by using the coefficents of the variables.
\[\left[\begin{matrix}1 & 1 & 1 & 2 \\ 1 & 1 & 2 & 4\\ 0 & 1 & 2& 4\end{matrix}\right]\]
In either case, we will use linear combinations of rows or individual equations, in equation and matrix form respectively, to solve for the variables. In linear algebra to solve the system of equations we would want to put it into what is called Row-Reduced Echlon Form (RREF). After which it would look like this:
\[\left[\begin{matrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 0 & 0 & 1& 2\end{matrix}\right]\]
Which means x1 = 0, x2 = 0, and x3 = 2. If you were to plug it into the original set of equations you would see it works. How we get to the RREF of the matrix is by preforming a set of what are called elementary row operations on the rows. 1. You can exchange any row with another. 2. Multiply a row by a nonzero-constant. 3. Add a multiple of a row to another. In this case take row 1, multiply it by -1 and to the second row. Add -2 times the second row to the third. Add -1 times the second row to the first. Add -1 times the third row to the first. These set of instructions if applied to the sets of linear equations would also yield the answer to the variables.
All in all, the matrix form is a convenient way to perform linear combinations on sets of equations. In addition, as you delve deeper into linear algebra it makes your work much neater and easier to work with while performing various things on the matricies.

Answer 3

i had figured so much, yet neither professor nor google specified that this could be considered a type of reasoning behind linear combinations

Answer 4

Indeed. If your interested further in that sort of relationship you should find an article or book on linear algebra. The reason we make the matrixs in such a way is because we are using the the matrix equation Ax = b. Where A is the coefficent matrix caring the coefficient of the variables. x is a matrix of the variables. And b is the solution matrix per se.
\[A = \left[\begin{matrix}1 & 1 & 1\\ 1 & 1 & 2\\ 0&1&2\end{matrix}\right],x = \left[\begin{matrix}x _{1} \\ x _{2}\\ x _{3}\end{matrix}\right], b=\left[\begin{matrix}2 \\ 4\\ 4\end{matrix}\right]\]
If we look at just the first row. According to matrix multiplication:
\[A \times x = 1*x _{1} + 1*x _{2} + 1*x _{3} = x _{1} + x _{2} + x _{3} \]
With Ax = b, we can say \[x _{1} + x _{2} + x _{3} = 2\]
Which is equal to the first linear equation in the set. Represented in matrix form, you can also solve the set using methods other than by putting it into RREF. With the matrix equation Ax = b and us trying to solve for x, we can isolate x by multiplying both sides by the inverse of A known as A^-1. Without delving to deeply into everything, we can do this because A, in this case, is invertible and thus when matrix A is multiplied by its inverse the result is the identity matrix that looks like:
\[I =\left[\begin{matrix}1 & 0 & 0 \\ 0& 1 & 0\\ 0 & 0 & 1\end{matrix}\right]\]
Which when multiplied by any matrix (x) will equal x.
Ex: A*I = A, B*I = B, C*I = C, and so on.
Any invertible matrix multiplied by its inverse will yield the identity matrix I.
Ex: A * A^-1 = I, B * B^-1 = I, and so on.

This multiplying Ax=b by A^-1, we get (A^-1)*A*x = (A^-1)b, and with (A^-1)*A = I, we get I*x=(A^-1)b. I*x = x so we get x = (A^-1)b.

We find A^-1 by adjoining it to an identity matrix and performing Gauss-Jordan elimination to put it into RREF, however, we apply the elementary row operations that we are performing on the matrix A, also to the identity matrix I. Once matrix A is in RREF, matrix I will be the inverse of A. Then multiplying this by matrix b will directly give the values of x_1, x_2, and x_3. Don't mean to overload you. I just find linear algebra a fun subject.