[Math] Can you solve a quadratic equation using matrices

matricesquadratics

I was wondering whether there are any alternatives or more efficient methods to finding a solution to a quadratic equation other than simply trial and error or by using the quadratic formula.

I was once told that it could be very easily done using matrices. How would this work?

Additionally, are there any other "better" alternatives?

I would really appreciate if you were to give me examples and explain how to solve them with the alternate method.

Thank you 🙂

Best Answer

You can solve the quadratic equation $x^2 + ax+b=0$ by computing the eigenvalues of the companion matrix $$ \pmatrix{ 0 & - b \\ 1 & -a}. $$ Simple and efficient.

Of course not! It requires to solve exact the same equation most of the time. The characteristic polynimial of the matrix is $x^2 + ax+b$.

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[Math] Generating Different types of Matrices in Matlab

I think that you're looking for the gallery function. It has over 60 types of test matrices. For the particular case of diagonally dominant matrices, you might look at the 'dorr' option:

A = full(gallery('dorr',5,0.01))

You can find the code for most of these matrix generation functions in the matlabroot/toolbox/matlab/elmat/private/ directory, where matlabroot is the root directory of your Matlab installation.

If you need random diagonally dominant matrices, then you might look at the answers to this StackOverflow question. I believe that this is equivalent Matlab code to the accepted answer (you'll have to check if the resultant matrices are indeed diagonally dominant):

rng(1)
n = 5;
mmax = 100;
mat1 = 2*mmax*rand(n)-mmax;
mat = mat1+diag(sum(abs(mat1),2).*sign(diag(mat1)))

[Math] How to derive the quadratic formula using calculus

If f is a polynomial of degree $n$ then for all $x, y$ we have $$f(x)=\sum_{j=0}^n(x-y)^jf^{(j)}(y)/j!$$ where $f^{(0)}=f$ and $f^{(j)}$ is the $j$th derivative of $f$ when $j>0.$... And with the usual convention that $0^0=1 $ (i.e. the term $(x-y)^j$ for $j=0$, when $x=y$).

When $f(x)=Ax^2 +Bx+C$ with $A \ne 0,$ then $f'(x)=2Ax+B$ is equal to $0$ when $x=x_0=-B/2A.$ For all $x$ we have $$f(x)=f(x_0)+(x-x_0)f'(x_0)+(x-x_0)^2f''(x_0)/2!.$$ But $f'(x_0)=0$ and $f''(x_0)=2A,$ so for all $x$ we have $$f(x)=f(x_0)+(x-x_0)^2\cdot A.$$ This "completes the square" for us.

Best Answer

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[Math] Generating Different types of Matrices in Matlab

[Math] How to derive the quadratic formula using calculus

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