Matematické Fórum

stuart clark · 15. 03. 2019 07:40

Number of solution of matrix equation $\displaystyle A^2=\begin{pmatrix} 1 & 1\\ 2& 3 \end{pmatrix}$

rughar · 15. 03. 2019 10:36 — Editoval rughar (15. 03. 2019 10:36)

stuart clark · 15. 03. 2019 10:50

yes please explain me. Thanks

rughar · 15. 03. 2019 11:15

Just solve system of 4 equations. There is not much to explain.

My first step:

$a^{2} + bc = 1 \\ b(a+d) = 1 \\ c(a+d) = 2 \\ bc + d^2= 3$

My second step:

$a^{2} + bc = 1 \\ bc + d^2= 3 \\ \Rightarrow \\ (d-a)(d+a) = 2$

My third step:

$(d-a)(d+a) = 2 \\ b(a+d) = 1 \\ c(a+d) = 2 \\ \Rightarrow \\ 2b = c = d-a$

And then you have two quadratic equations.

vanok · 15. 03. 2019 11:24

Hi ↑ stuart clark:,
You can also use that your matrix is diagonalizable

stuart clark · 15. 03. 2019 11:59

Thanks ↑ rughar:

↑ vanok: can you please explain me Thanks

vanok · 15. 03. 2019 12:46

Eigenvalue of $A^2$ are $2+\sqrt 3$ and $2-\sqrt 3$ .
Also $A^2$ is diagonalizable to
$D^2=\begin{pmatrix} 2+\sqrt 3 & 0\\ 0& 2-\sqrt 3 \end{pmatrix}$ .
And $D=\begin{pmatrix} a_1 \sqrt{2+\sqrt 3} & 0\\ 0&a_ 2\sqrt{2-\sqrt 3} \end{pmatrix}$ .
$a_1,a_2 \in \{1,-1\}$ .
Remark.
$A=PDP^{-1}$ with $P=\begin{pmatrix} \frac 12(-1-\sqrt 3) &\frac 12(-1+\sqrt 3) \\ 1&1 \end{pmatrix}$