I'm asking this for clarification. A matrix/system of equations is singular is there are infinite solutions, but iff there is a unique solution then its non-singular?
I haven't learned how to take a determinant yet. However, my professor went over how to determine if it is singular or non-singular without needing to take it. I've looked in my notes, but haven't been able to find it.
Any clarity would be much appreciated.
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$\begingroup$Suppose the linear system we have is
\begin{equation} A x = b \end{equation}
where $\newcommand{\reals}{{\mathbf R}}A\in\reals^{n\times n}$ and $x, b \in\reals^n$.
You need to be a bit more precise to be correct to relate the number (or existence) of solutions to the singularity of $A$.
The following statements are correct:
- A linear system has a unique solution if and only if the matrix is non-singular.
- A linear system has either no solution or infinite number of solutions if and only if the matrix is singular.
- A linear system has a solution if and only if $b$ is in the range of $A$.
Now by definition,
- The matrix is non-singular if and only if the determinant is nonzero.
However, like your professor mentioned, you do not need to evaluate the determinant to see whether a matrix is singular or not (though most such methods evaluates the determinant as by-product).
For example, you can use Gaussian elimination to tell whether a matrix is singular. This has the following advantages.
- The time complexity of Gaussian elimination is $O(n^3)$ (whereas brute-force evaluation of determinant by the original definition takes $O(n!)$).
- Gaussian elimination evaluates the determinant as by-product (i.e., with no additional cost).
Hope this helps you!
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