# If A=[(-1,2), (3,1)], how do you find F(A) where f(x)=x^2-2x+3?

Mar 7, 2017

$2 \left[\begin{matrix}6 & - 2 \\ - 3 & 4\end{matrix}\right]$

See below.

#### Explanation:

Write it as:

$f \left(A\right) = \left[\begin{matrix}- 1 & 2 \\ 3 & 1\end{matrix}\right] \cdot \left[\begin{matrix}- 1 & 2 \\ 3 & 1\end{matrix}\right] - 2 \left[\begin{matrix}- 1 & 2 \\ 3 & 1\end{matrix}\right] + 3 I$

Where $I$ is the identity matrix: $\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]$

And then process the algebra.

If you need an answer check, I get:

$f \left(A\right) = 2 \left[\begin{matrix}6 & - 2 \\ - 3 & 4\end{matrix}\right]$

Mar 8, 2017

$f \left(A\right) = {A}^{2} - 2 A + 3 I = \left(\begin{matrix}12 & - 4 \\ - 6 & 8\end{matrix}\right)$

#### Explanation:

Given:

$A = \left(\begin{matrix}- 1 & 2 \\ 3 & 1\end{matrix}\right)$

Then:

${A}^{2} = \left(\begin{matrix}- 1 & 2 \\ 3 & 1\end{matrix}\right) \left(\begin{matrix}- 1 & 2 \\ 3 & 1\end{matrix}\right) = \left(\begin{matrix}7 & 0 \\ 0 & 7\end{matrix}\right) = 7 I$

So:

${A}^{2} - 2 A + 3 I = 7 I - 2 A + 3 I$

$\textcolor{w h i t e}{{A}^{2} - 2 A + 3 I} = 10 I - 2 A$

$\textcolor{w h i t e}{{A}^{2} - 2 A + 3 I} = \left(\begin{matrix}10 & 0 \\ 0 & 10\end{matrix}\right) - \left(\begin{matrix}- 2 & 4 \\ 6 & 2\end{matrix}\right)$

$\textcolor{w h i t e}{{A}^{2} - 2 A + 3 I} = \left(\begin{matrix}12 & - 4 \\ - 6 & 8\end{matrix}\right)$

$\textcolor{w h i t e}{}$
Footnote

Note that $A$ is a "square root of $7$". That is, it is a root of the polynomial:

${x}^{2} - 7 = 0$

As a result, we find that the set of matrices of the form $p I + q A$ where $p$ and $q$ are rational is a field under matrix addition and multiplication. This field is essentially the same as (i.e. isomorphic to) the set of numbers of the form $p + q \sqrt{7}$ for rational multipliers $p , q$.

Note that this matrix $A$ is not the only possible way to represent $\sqrt{7}$ with a rational matrix. For example, you could use $\left(\begin{matrix}0 & 7 \\ 1 & 0\end{matrix}\right)$.