I'm familiar with Gaussian Elimination from doing it using row operations to solve systems of linear equations in the past, but I've started reading a formal textbook on it, namely "Linear Algebra and its Applications" by Gilbert Strang but I'm having a bit of trouble understanding some details he was mentioning about pivot elements and singular cases. I think by explaining what I think I know about what I'm doing can in turn clue others in as to where I'm getting confused.
Basically, as far as I'm concerned, when we have a system of linear equations, like, off the top of my head and just for illustration,
$$x+y+z = 5$$
$$3x+2y-5z=20$$
$$-4x-2y+z=3$$ the vector-space interpretation is that if $x, y$ and $z$ are variables in 3 different equations with varying coefficients and numerical solutions, it can be interpreted in the column picture as $x$, $y$, and $z$ being scalars needed to produce a vector that is formed by the constants on the $RHS$ of the 3 equations in column form. So it's the required scalars for each $x, y$ and $z$ needed to create the $RHS$ vector through vector addition. Hopefully this wasn't a complete word salad.
In a row picture interpretation, it's the $x$, $y$, and $z$ value or values that specify the intersection of the three planes given by the 3 equations.
Now, in elimination, there is something I'm being told about called pivot elements, and from what I gather, they're basically when, at least when you're using Gaussian Elimination with matrices, the entry in a matrix you plan on combining with a newly operated on row to sum to 0, so that it leads you to isolating one variable to one solution, and then allows you to use back-substitution to solve the system of linear equations. For instance, if I, in the system above, subtracted 3 times equation 1 from equation 2, $3$ would be a pivot element as it causes $x=0$ in a row.
However, Strang notes that "we divide by them" in reference to pivot elements, so I think my idea of what a pivot element therefore must be false.
Proceeding, he mentions that in a singular case, a $0$ occurs in a pivot position, and elimination then must stop.
I'm not exactly sure what that means, but he gave an example of a system of linear equations that couldn't be solved by elimination:
I myself can't use elimination on this from my own experience because I can't get zeroes on the bottom left entry, middle left entry and bottom middle entry — I end up with two zeroes in the middle left and middle middle (for lack of a better description) entry, as well as two zeros in the bottom left and bottom middle entries. This kind of screws things up when I solve it, so I can see why it's unsolvable but not what he means by "$0$ being in a pivot position" and since we apparently divide by pivot elements this causes a problem, since we cannot divide by zero.
All of this aside, I have two main questions:
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What of my pivot element interpretation is wrong, and how then is it that we divide them?
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What does he mean by "$0$ being in a pivot position" causes a singular case in, if possible, the context of how I found eliminating that example of a system of linear equations he provided impossible?
I apologize if this is a lot to answer, but I felt the questions weren't too long-winded to address to necessitate splitting each question into its own thread.
Best Answer
• So basically, when you obtain a Triangular system you have to do a process of back substitution to obtain the value of the variables.
Now, if the Pivot element is $0$ then the solution would be indeterminable. Hence, Pivot elements can't be $0$ by definition.
• While manipulating the original system into a triangular system, we take the pivot element of the $1^{st}$ equation to make the $a_{n1} = 0$ and so on and so forth. So, while you are unable to obtain a non-zero pivot element to make $a_{n2} = 0$ for the rest of the equations, you can't go further with Gaussian elimination.