Initial polynom of f (x) (its coefficients) The argument is a square matrix with elements
 Polynomial Variable x = Result of calculations

Let's consider in this material one of labor-consuming tasks in the higher mathematics which sounds so: The polynomial is given a task to find to what

$f(x)=a_0x^{n}+a_1x^{n-1}+a_2x^{n-2}+.....+a_{n-1}x+a_n$

if the argument is a square matrix, that is  $x=\begin{pmatrix}x_{00} & x_{10} &...&x_{i0}\\ x_{01} & x_{11}&...&x_{i1}\\ ... & ...&...&... \\ x_0j} & x_{1j}&...&x_{ij}\end{pmatrix}$

And if the principle of calculation is clear, especially if you in perfection understood how to multiply matrixes, then direct calculation, for me personally it is considered a routine which it is whenever possible necessary to avoid.

At once it would be desirable to tell where this calculator is useful. For teachers, teachers, for creators of textbooks, for those who need to create original tasks of this subject.

Also it is useful for students or graduate students who write papers, course, diplomas. For all others, it is an easy way to check a mistake in the set example, to solve, without long intermediate calculations, an objective.

When the calculator was written, it turned out that the websites which were posveshchenna to this subject contained mistakes in intermediate calculations and as as result were incorrect.

This calculator, I hope is saved from mistakes and you will be able safely to solve any examples.

As well as the vast majority of calculators on this website, values of both polynom coefficients, and matrix elements, can be complex values.

It for the end of 2017, more you will not find anywhere, apart from of course the special created mathematical programs.

Let's start examples?

To find value of a polynom  $f(x)=2x^2-3x+4$  from a matrix $x=\begin{pmatrix}-1 &2 &0\\2 & 1&-3\\{0} & -1&2\end{pmatrix}$

 Polynom $f(x) = (2)*x^{2}+(-3)*x+(4)$ Variable x = $x = \begin{pmatrix}-1 & 2 & 0 \\ 2 & 1 & -3 \\ 0 & -1 & 2 \\ \end{pmatrix}$ Result of calculations $f(x) = \begin{pmatrix}17 & -6 & -12 \\ -6 & 17 & -9 \\ -4 & -3 & 12 \\ \end{pmatrix}$

One more example

What the polynom is equal to  $f(x)=ix^5+(2-i)x^2-11x$  if $x=\begin{pmatrix}2 &2-i &3\\-7 & 0&i\\1+i &2&0\end{pmatrix}$

 Polynom $f(x) = (i)*x^{5}+(2-i)*x^{2}+(-11)*x$ Variable x = $x = \begin{pmatrix}2 & 2-i & 3 \\ -7 & 0 & i \\ 1+i & 2 & 0 \\ \end{pmatrix}$ Result of calculations $f(x) = \begin{pmatrix}2774-2058i & -1092-741i & -66-1293i \\ -1336+2039i & 1937+1391i & 995+2236i \\ 1300+279i & 389+117i & 543+401i \\ \end{pmatrix}$

To find value of a polynomial  $f(x)=x^4-x-1$ from a complex matrix

$x=\begin{pmatrix}-1 & 0 & 0 & -1 & 0 & 0 \\ 0 & -1 & 1 & 0 & -1 & 1 \\ -1 & 0 & i & i & i & -1 \\ -1 & -1 & i & -1 & -1 & i \\ 0 & 1 & -1 & 0 & -1 & i \\ 1 & 0 & 0 & 0 & i & 0 \\ \end{pmatrix}$

 Polynom $f(x) = x^{4}+(-1)*x+(-1)$ Variable x = $x = \begin{pmatrix}-1 & 0 & 0 & -1 & 0 & 0 \\ 0 & -1 & 1 & 0 & -1 & 1 \\ -1 & 0 & i & i & i & -1 \\ -1 & -1 & i & -1 & -1 & i \\ 0 & 1 & -1 & 0 & -1 & i \\ 1 & 0 & 0 & 0 & i & 0 \\ \end{pmatrix}$ Result of calculations $f(x) = \begin{pmatrix}7+2i & 3 & -2-2i & 6+i & 1+2i & -3-7i \\ 4+7i & -3-5i & 6+i & 5 & 3-2i & 11 \\ 3-5i & 3+i & -7+i & -2-2i & -4+9i & -4+i \\ 7+6i & 5-i & -1 & 5+4i & 0+6i & 1-2i \\ -2-5i & 3+4i & -8-5i & -2-i & -9-9i & -8+i \\ -5-i & -2+i & -2+3i & -5 & -8+4i & 0+2i \\ \end{pmatrix}$

Successful calculations!

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