mathML : using mathML

1. $A=\underset{}{\left(\begin{array}{cc}3& 1\\ 1& 2\end{array}\right)}$

   Eigenvectors for the matrix $A$ :

   • $\mathit{v}=\underset{}{\left(\begin{array}{c}-0.618\\ 1\end{array}\right)}$ , eigenvalue ${\lambda }_1=1.38$
   • $\mathit{v}=\underset{}{\left(\begin{array}{c}1.62\\ 1\end{array}\right)}$ , eigenvalue ${\lambda }_2=3.62$
   Details

   1. From the definition of the eigenvector $\mathit{v}$ corresponding to the eigenvalue $\lambda$ we have $$A\mathit{v}=\lambda \mathit{v}$$
      Then: $$A\mathit{v}-\lambda \mathit{v}=\left(A-\lambda \mathrm{<span\; class="dotted-underline"\; title="Identity\; Matrix"\; aria-label="Identity\; Matrix"\; data-i="1">I</span>}\right)\cdot \mathit{v}=0$$
      Equation has a nonzero solution if and only if $$\det \left(A-\lambda \mathrm{<span\; class="dotted-underline"\; title="Identity\; Matrix"\; aria-label="Identity\; Matrix"\; data-i="1">I</span>}\right)=0$$
      $\det \left(A-\lambda \mathrm{<span\; class="dotted-underline"\; title="Identity\; Matrix"\; aria-label="Identity\; Matrix"\; data-i="1">I</span>}\right)=\underset{}{\left|\begin{array}{cc}3-\lambda & 1\\ 1& 2-\lambda \end{array}\right|}={\lambda }^2-5\lambda +5\underset{\text{<a href="https://en.wikipedia.org/wiki/Factorization\#Using\_formulas\_for\_polynomial\_roots" title="Factorization: Using formulas for polynomial roots">(?)</a>}}{=}\left(\lambda +-1.38\right)\cdot \left(\lambda -3.62\right)=0$
      Details
      1. ${\lambda }_1=1.38$
      2. ${\lambda }_2=3.62$
   2. For every $\lambda$ we find its own vectors:
      1. ${\lambda }_1=1.38$
         $A-{\lambda }_1\mathrm{<span\; class="dotted-underline"\; title="Identity\; Matrix"\; aria-label="Identity\; Matrix"\; data-i="1">I</span>}=\underset{}{\left(\begin{array}{cc}1.62& 1\\ 1& 0.618\end{array}\right)}$
         $A\mathit{v}=\lambda \mathit{v}$ *
         $\left(A-\lambda \mathrm{<span\; class="dotted-underline"\; title="Identity\; Matrix"\; aria-label="Identity\; Matrix"\; data-i="1">I</span>}\right)\cdot \mathit{v}=0$
         So we have a homogeneous system of linear equations, we solve it by Gaussian Elimination:
         $\underset{}{\left(\begin{array}{ccc}1.62& 1& 0\\ 1& 0.618& 0\end{array}\right)}$

         $\times \left(0.618\right)$

         $ ~ R 1 ∕ ( 1.62 ) → R 1 ? Divide row 1 by 1.62 : a 1 , 1 = 1.62 ⋅ ( 1 1.62 ) = 1 a 1 , 2 = 1 ⋅ ( 1 1.62 ) = 0.618 a 1 , 3 = 0 ⋅ ( 1 1.62 ) = 0 $ $\underset{}{\left(\begin{array}{ccc}1& 0.618& 0\\ 1& 0.618& 0\end{array}\right)}$

         $\times \left(-1\right)$

         $ ~ R 2 − 1 ⋅ R 1 → R 2 ? Subtract 1 × row 1 from row 2 : a 2 , 1 = 1 − 1 1 ⋅ 1 = 0 a 2 , 2 = 0.618 − 1 1 ⋅ ( 0.618 ) = 0 a 2 , 3 = 0 − 1 1 ⋅ 0 = 0 $ $\underset{}{\left(\begin{array}{ccc}1& 0.618& 0\\ 0& 0& 0\end{array}\right)}$

         $\underset{}{\{\begin{array}{rrrr}{x}_1& +0.618\cdot x_2& =& 0\end{array}}$

         (1)

         • Find the variable $x_1$ from the equation $1$ of the system (1) :
           $x_1=-0.618\cdot x_2$

         Answer:

         • $x_1=-0.618\cdot x_2$
         • $x_2=x_2$

         General Solution : $X=\underset{}{\left(\begin{array}{c}-0.618\cdot x_2\\ x_2\end{array}\right)}$

         The solution set: $\{x_2\cdot \underset{}{\left(\begin{array}{c}-0.618\\ 1\end{array}\right)}\}$
         Let $x_2=1$ , ${\mathit{v}}_1=\underset{}{\left(\begin{array}{c}-0.618\\ 1\end{array}\right)}$
      2. ${\lambda }_2=3.62$
         $A-{\lambda }_2\mathrm{<span\; class="dotted-underline"\; title="Identity\; Matrix"\; aria-label="Identity\; Matrix"\; data-i="1">I</span>}=\underset{}{\left(\begin{array}{cc}-0.618& 1\\ 1& -1.62\end{array}\right)}$
         $A\mathit{v}=\lambda \mathit{v}$ *
         $\left(A-\lambda \mathrm{<span\; class="dotted-underline"\; title="Identity\; Matrix"\; aria-label="Identity\; Matrix"\; data-i="1">I</span>}\right)\cdot \mathit{v}=0$
         So we have a homogeneous system of linear equations, we solve it by Gaussian Elimination:
         $\underset{}{\left(\begin{array}{ccc}-0.618& 1& 0\\ 1& -1.62& 0\end{array}\right)}$

         $\times \left(-1.62\right)$

         $ ~ R 1 ∕ ( − 0.618 ) → R 1 ? Divide row 1 by − 0.618 : a 1 , 1 = − 0.618 ⋅ ( 1 − 0.618 ) = 1 a 1 , 2 = 1 ⋅ ( 1 − 0.618 ) = − 1.62 a 1 , 3 = 0 ⋅ ( 1 − 0.618 ) = 0 $ $\underset{}{\left(\begin{array}{ccc}1& -1.62& 0\\ 1& -1.62& 0\end{array}\right)}$

         $\times \left(-1\right)$

         $ ~ R 2 − 1 ⋅ R 1 → R 2 ? Subtract 1 × row 1 from row 2 : a 2 , 1 = 1 − 1 1 ⋅ 1 = 0 a 2 , 2 = − 1.62 − 1 1 ⋅ ( − 1.62 ) = 0 a 2 , 3 = 0 − 1 1 ⋅ 0 = 0 $ $\underset{}{\left(\begin{array}{ccc}1& -1.62& 0\\ 0& 0& 0\end{array}\right)}$

         $\underset{}{\{\begin{array}{rrrr}{x}_1& -1.62\cdot x_2& =& 0\end{array}}$

         (1)

         • Find the variable $x_1$ from the equation $1$ of the system (1) :
           $x_1=1.62\cdot x_2$

         Answer:

         • $x_1=1.62\cdot x_2$
         • $x_2=x_2$

         General Solution : $X=\underset{}{\left(\begin{array}{c}1.62\cdot x_2\\ x_2\end{array}\right)}$

         The solution set: $\{x_2\cdot \underset{}{\left(\begin{array}{c}1.62\\ 1\end{array}\right)}\}$
         Let $x_2=1$ , ${\mathit{v}}_2=\underset{}{\left(\begin{array}{c}1.62\\ 1\end{array}\right)}$
2. $A=\underset{}{\left(\begin{array}{cc}3& 1\\ 1& 2\end{array}\right)}$

   Eigenvectors for the matrix $A$ :

   • $\mathit{v}=\underset{}{\left(\begin{array}{c}\frac{-\sqrt{5}+1}{2}\\ 1\end{array}\right)}$ , eigenvalue ${\lambda }_1=\frac{-\sqrt{5}+5}{2}$
   • $\mathit{v}=\underset{}{\left(\begin{array}{c}\frac{\sqrt{5}+1}{2}\\ 1\end{array}\right)}$ , eigenvalue ${\lambda }_2=\frac{\sqrt{5}+5}{2}$
   Details

   1. From the definition of the eigenvector $\mathit{v}$ corresponding to the eigenvalue $\lambda$ we have $$A\mathit{v}=\lambda \mathit{v}$$
      Then: $$A\mathit{v}-\lambda \mathit{v}=\left(A-\lambda \mathrm{<span\; class="dotted-underline"\; title="Identity\; Matrix"\; aria-label="Identity\; Matrix"\; data-i="1">I</span>}\right)\cdot \mathit{v}=0$$
      Equation has a nonzero solution if and only if $$\det \left(A-\lambda \mathrm{<span\; class="dotted-underline"\; title="Identity\; Matrix"\; aria-label="Identity\; Matrix"\; data-i="1">I</span>}\right)=0$$
      $\det \left(A-\lambda \mathrm{<span\; class="dotted-underline"\; title="Identity\; Matrix"\; aria-label="Identity\; Matrix"\; data-i="1">I</span>}\right)=\underset{}{\left|\begin{array}{cc}3-\lambda & 1\\ 1& 2-\lambda \end{array}\right|}={\lambda }^2-5\lambda +5\underset{\text{<a href="https://en.wikipedia.org/wiki/Factorization\#Using\_formulas\_for\_polynomial\_roots" title="Factorization: Using formulas for polynomial roots">(?)</a>}}{=}\left(\lambda +\frac{\sqrt{5}-5}{2}\right)\cdot \left(\lambda -\frac{\sqrt{5}+5}{2}\right)=0$
      Details
      1. ${\lambda }_1=\frac{-\sqrt{5}+5}{2}$
      2. ${\lambda }_2=\frac{\sqrt{5}+5}{2}$
   2. For every $\lambda$ we find its own vectors:
      1. ${\lambda }_1=\frac{-\sqrt{5}+5}{2}$
         $A-{\lambda }_1\mathrm{<span\; class="dotted-underline"\; title="Identity\; Matrix"\; aria-label="Identity\; Matrix"\; data-i="1">I</span>}=\underset{}{\left(\begin{array}{cc}\frac{\sqrt{5}+1}{2}& 1\\ 1& \frac{\sqrt{5}-1}{2}\end{array}\right)}$
         $A\mathit{v}=\lambda \mathit{v}$ *
         $\left(A-\lambda \mathrm{<span\; class="dotted-underline"\; title="Identity\; Matrix"\; aria-label="Identity\; Matrix"\; data-i="1">I</span>}\right)\cdot \mathit{v}=0$
         So we have a homogeneous system of linear equations, we solve it by Gaussian Elimination:
         $\underset{}{\left(\begin{array}{ccc}\frac{\sqrt{5}+1}{2}& 1& 0\\ 1& \frac{\sqrt{5}-1}{2}& 0\end{array}\right)}$

         $\times \left(\frac{\sqrt{5}-1}{2}\right)$

         $ ~ R 1 ∕ ( 5 + 1 2 ) → R 1 ? Divide row 1 by 5 + 1 2 : a 1 , 1 = 5 + 1 2 ⋅ ( 1 5 + 1 2 ) = 1 a 1 , 2 = 1 ⋅ ( 1 5 + 1 2 ) = 5 − 1 2 a 1 , 3 = 0 ⋅ ( 1 5 + 1 2 ) = 0 $ $\underset{}{\left(\begin{array}{ccc}1& \frac{\sqrt{5}-1}{2}& 0\\ 1& \frac{\sqrt{5}-1}{2}& 0\end{array}\right)}$

         $\times \left(-1\right)$

         $ ~ R 2 − 1 ⋅ R 1 → R 2 ? Subtract 1 × row 1 from row 2 : a 2 , 1 = 1 − 1 1 ⋅ 1 = 0 a 2 , 2 = 5 − 1 2 − 1 1 ⋅ ( 5 − 1 2 ) = 0 a 2 , 3 = 0 − 1 1 ⋅ 0 = 0 $ $\underset{}{\left(\begin{array}{ccc}1& \frac{\sqrt{5}-1}{2}& 0\\ 0& 0& 0\end{array}\right)}$

         $\underset{}{\{\begin{array}{rrrr}{x}_1& +\frac{\sqrt{5}-1}{2}\cdot x_2& =& 0\end{array}}$

         (1)

         • Find the variable $x_1$ from the equation $1$ of the system (1) :
           $x_1=\frac{-\sqrt{5}+1}{2}\cdot x_2$

         Answer:

         • $x_1=\frac{-\sqrt{5}+1}{2}\cdot x_2$
         • $x_2=x_2$

         General Solution : $X=\underset{}{\left(\begin{array}{c}\frac{-\sqrt{5}+1}{2}\cdot x_2\\ x_2\end{array}\right)}$

         The solution set: $\{x_2\cdot \underset{}{\left(\begin{array}{c}\frac{-\sqrt{5}+1}{2}\\ 1\end{array}\right)}\}$
         Let $x_2=1$ , ${\mathit{v}}_1=\underset{}{\left(\begin{array}{c}\frac{-\sqrt{5}+1}{2}\\ 1\end{array}\right)}$
      2. ${\lambda }_2=\frac{\sqrt{5}+5}{2}$
         $A-{\lambda }_2\mathrm{<span\; class="dotted-underline"\; title="Identity\; Matrix"\; aria-label="Identity\; Matrix"\; data-i="1">I</span>}=\underset{}{\left(\begin{array}{cc}\frac{-\sqrt{5}+1}{2}& 1\\ 1& \frac{-\sqrt{5}-1}{2}\end{array}\right)}$
         $A\mathit{v}=\lambda \mathit{v}$ *
         $\left(A-\lambda \mathrm{<span\; class="dotted-underline"\; title="Identity\; Matrix"\; aria-label="Identity\; Matrix"\; data-i="1">I</span>}\right)\cdot \mathit{v}=0$
         So we have a homogeneous system of linear equations, we solve it by Gaussian Elimination:
         $\underset{}{\left(\begin{array}{ccc}\frac{-\sqrt{5}+1}{2}& 1& 0\\ 1& \frac{-\sqrt{5}-1}{2}& 0\end{array}\right)}$

         $\times \left(\frac{-\sqrt{5}-1}{2}\right)$

         $ ~ R 1 ∕ ( − 5 + 1 2 ) → R 1 ? Divide row 1 by − 5 + 1 2 : a 1 , 1 = − 5 + 1 2 ⋅ ( 1 − 5 + 1 2 ) = 1 a 1 , 2 = 1 ⋅ ( 1 − 5 + 1 2 ) = − 5 − 1 2 a 1 , 3 = 0 ⋅ ( 1 − 5 + 1 2 ) = 0 $ $\underset{}{\left(\begin{array}{ccc}1& \frac{-\sqrt{5}-1}{2}& 0\\ 1& \frac{-\sqrt{5}-1}{2}& 0\end{array}\right)}$

         $\times \left(-1\right)$

         $ ~ R 2 − 1 ⋅ R 1 → R 2 ? Subtract 1 × row 1 from row 2 : a 2 , 1 = 1 − 1 1 ⋅ 1 = 0 a 2 , 2 = − 5 − 1 2 − 1 1 ⋅ ( − 5 − 1 2 ) = 0 a 2 , 3 = 0 − 1 1 ⋅ 0 = 0 $ $\underset{}{\left(\begin{array}{ccc}1& \frac{-\sqrt{5}-1}{2}& 0\\ 0& 0& 0\end{array}\right)}$

         $\underset{}{\{\begin{array}{rrrr}{x}_1& -\frac{\sqrt{5}+1}{2}\cdot x_2& =& 0\end{array}}$

         (1)

         • Find the variable $x_1$ from the equation $1$ of the system (1) :
           $x_1=\frac{\sqrt{5}+1}{2}\cdot x_2$

         Answer:

         • $x_1=\frac{\sqrt{5}+1}{2}\cdot x_2$
         • $x_2=x_2$

         General Solution : $X=\underset{}{\left(\begin{array}{c}\frac{\sqrt{5}+1}{2}\cdot x_2\\ x_2\end{array}\right)}$

         The solution set: $\{x_2\cdot \underset{}{\left(\begin{array}{c}\frac{\sqrt{5}+1}{2}\\ 1\end{array}\right)}\}$
         Let $x_2=1$ , ${\mathit{v}}_2=\underset{}{\left(\begin{array}{c}\frac{\sqrt{5}+1}{2}\\ 1\end{array}\right)}$