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Preguntas 10
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 #3
avatar+18393 
0

Heureka, how did you get that ϕ^n=Fnϕ+Fn-1 ?

 

Formula:  \( \phi^2 = \phi +1\)

mathematical proof:

\(\begin{array}{|rcll|} \hline \phi = \frac{1+\sqrt{5}} {2} \\\\ \phi^2 &=& \left( \frac{1+\sqrt{5}} {2} \right)^2 \\ &=& \frac{\left( 1+\sqrt{5}\right)^2 } {4} \\ &=& \frac{ 1+2\sqrt{5}+5 } {4} \\ &=& \frac{ 6+2\sqrt{5} } {4} \\ &=& \frac{ 3+\sqrt{5} } {2} \\ &=& \frac{ 1+2+\sqrt{5} } {2} \\ &=& \frac{ 1+\sqrt{5}+2 } {2} \\ &=& \frac{ 1+\sqrt{5} } {2} +\frac{2}{2} \\ &=& \frac{ 1+\sqrt{5} } {2} + 1 \\ &=& \mathbf{ \phi + 1 } \\ \hline \end{array}\)

 

\(\small{ \begin{array}{|lclclclcl|} \hline \phi^1&&&& &=& 1\phi+0 &=& F_1\phi + F_0 \\ \phi^2&&&& &=& 1\phi+1 &=& F_2\phi + F_1 \\ \phi^3 = \phi\phi^2 = \phi(\phi+1) &=&1\phi^2 + 1\phi &=&1(\phi+1)+1\phi &=& 2\phi+1 &=& F_3\phi + F_2 \\ \phi^4 = \phi\phi^3 = \phi(2\phi+1) &=&2\phi^2 + 1\phi &=&2(\phi+1)+1\phi &=& 3\phi+2 &=& F_4\phi + F_3 \\ \phi^5 = \phi\phi^4 = \phi(3\phi+2) &=&3\phi^2 + 2\phi &=&3(\phi+1)+2\phi &=& 5\phi+3 &=& F_5\phi + F_4 \\ \phi^6 = \phi\phi^5 = \phi(5\phi+3) &=&5\phi^2 + 3\phi &=&5(\phi+1)+3\phi &=& 8\phi+5 &=& F_6\phi + F_5 \\ \phi^7 = \phi\phi^6 = \phi(8\phi+5) &=&8\phi^2 + 5\phi &=&8(\phi+1)+5\phi &=& 13\phi+8 &=& F_7\phi + F_6 \\ \phi^8 = \phi\phi^7 = \phi(13\phi+8) &=&13\phi^2 + 8\phi &=&13(\phi+1)+8\phi &=& 21\phi+13 &=& F_8\phi + F_7 \\ \phi^9 = \phi\phi^8 = \phi(21\phi+13) &=&21\phi^2 + 13\phi &=&21(\phi+1)+13\phi &=& 34\phi+21 &=& F_9\phi + F_8 \\ \phi^{10} = \phi\phi^9 = \phi(34\phi+21) &=&34\phi^2 + 21\phi &=&34(\phi+1)+21\phi &=& 55\phi+34 &=& F_{10}\phi + F_9 \\ \cdots \\ \mathbf{ \phi^n } &&&& && &\mathbf{ =}& \mathbf{ F_n\phi + F_{n-1}} \\ \hline \end{array} }\)

 

 

laugh

heureka 4 hours ago
 #1
avatar+18393 
0

Relationship between n*ϕ and ϕ^n

What is the relationship of nϕ and ϕ^n, when n=1, 2, 3, 4, ..., 15

 

Formula:
\(\begin{array}{rcll} \phi^n &=& F_{n}\phi +F_{n-1}\\ F_{n-2} &=& F_{n}-F_{n-1}\\ \frac{1}{\phi} &=& \phi -1 \\ \end{array}\)

 

List of Fibonacci numbers:

\(\small{ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline F_{-2} & F_{-1} & F_0 & F_1 & F_2 & F_3 & F_4 & F_5 & F_6 & F_7 & F_8 &F_9 &F_{10} &F_{11} &F_{12} &F_{13} &F_{14} &F_{15} \\ \hline -1 & 1 & 0 & 1 & 1 & 2 & 3 & 5 & 8 & 13 & 21 & 34 & 55 & 89 & 144 & 233 & 377 & 610 \\ \hline \end{array} }\)

 

\(\begin{array}{|rcll|} \hline \text{relationship } &=& \frac{n\phi}{\phi^n} \\ &=& \frac{n\phi}{F_{n}\phi +F_{n-1}} \\ &=& \frac{n}{ \frac{F_{n}\phi +F_{n-1} } {\phi} } \\ &=& \frac{n}{ F_{n} + \frac{1}{\phi} F_{n-1} } \\ &=& \frac{n}{ F_{n} + (\phi -1) F_{n-1} } \\ &=& \frac{n}{ F_{n} + F_{n-1}\phi -F_{n-1} } \\ &=& \frac{n}{ F_{n}-F_{n-1} + F_{n-1}\phi } \\ &\mathbf{=}& \mathbf{\frac{n}{ F_{n-2} + F_{n-1}\phi }} \\ \hline \end{array}\)

 

\(\begin{array}{|r|l|} \hline \mathbf{n} & \mathbf{\frac{n\phi}{\phi^n} = \frac{n}{ F_{n-2} + F_{n-1}\phi }} \\ \hline 1 & \frac{ 1 }{ 1+0\phi } = 1 \\ 2 & \frac{ 2 }{ 0+1\phi } = \frac{2}{\phi}= 2(\phi -1) \\ 3 & \frac{ 3 }{ 1+1\phi } \\ 4 & \frac{ 4 }{ 1+2\phi } \\ 5 & \frac{ 5 }{ 2+3\phi } \\ 6 & \frac{ 6 }{ 3+5\phi } \\ 7 & \frac{ 7 }{ 5+8\phi } \\ 8 & \frac{ 8 }{ 8+13\phi } \\ 9 & \frac{ 9 }{ 13+21\phi } \\ 10 & \frac{10 }{ 21+34\phi } \\ 11 & \frac{11 }{ 34+55\phi } \\ 12 & \frac{12 }{ 55+89\phi } \\ 13 & \frac{13 }{ 89+144\phi } \\ 14 & \frac{14 }{ 144+233\phi } \\ 15 & \frac{15 }{ 233+377\phi } \\ \hline \end{array}\)

 

 

laugh

heureka 22-ago-2017
 #2
avatar+18393 
+2

2. -2 A-2(A-2+A-3)-1+2A3(A2+A3)-1+ 3/4 A(A-1+I)-1+ 3/4 A3(A+I)-1

 

\(\begin{array}{|rcll|} \hline && -2 A^{-2}(A^{-2}+A^{-3})^{-1}+2A^3(A^2+A^3)^{-1}+ \frac34 A(A^{-1}+I)^{-1}+ \frac34 A^3(A+I)^{-1} = \ ?\\ \hline \end{array}\)

 

1.
\(\small{ \begin{array}{|rcll|} \hline &&\mathbf{ -2 A^{-2}(A^{-2}+A^{-3})^{-1}+2A^3(A^2+A^3)^{-1} } \\\\ && & (A^{-2}+A^{-3})^{-1} = [(A^{-1}+I)A^{-2}]^{-1} \\ && & (A^{2}+A^{3})^{-1} = [(A+I)A^{2}]^{-1} \\\\ &=& -2 A^{-2}[(A^{-1}+I)A^{-2}]^{-1}+2A^3[(A+I)A^{2}]^{-1} \quad & | \quad A^{-2} = A^{-1}A^{-1} = (AA)^{-1} = (A^2)^{-1} \\ &=& -2 A^{-2}[(A^{-1}+I)(A^2)^{-1}]^{-1}+2A^3[(A+I)A^{2}]^{-1} \quad & | \quad [(A^{-1}+I)(A^2)^{-1}]^{-1} = [A^2(A^{-1}+I)^{-1}] \\ &=& -2 A^{-2}[A^2(A^{-1}+I)^{-1}]+2A^3[(A+I)A^{2}]^{-1} \\ &=& -2 A^{-2}A^2(A^{-1}+I)^{-1}+2A^3[(A+I)A^{2}]^{-1} \quad & | \quad A^{-2}A^2 = A^{-1}A^{-1}AA = A^{-1}(A^{-1}A)A = A^{-1}(I)A = A^{-1}A = I \\ &=& -2 I(A^{-1}+I)^{-1}+2A^3[(A+I)A^{2}]^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2A^3[(A+I)A^{2}]^{-1} \quad & | \quad [(A+I)A^{2}]^{-1} = [(A^2)^{-1}(A+I)^{-1} ] \\ &=& -2 (A^{-1}+I)^{-1}+2A^3[(A^2)^{-1}(A+I)^{-1} ] \quad & | \quad (A^2)^{-1} = A^{-2} \\ &=& -2 (A^{-1}+I)^{-1}+2A^3A^{-2}(A+I)^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2AA^2A^{-2}(A+I)^{-1} \quad & | \quad A^2A^{-2} = I \\ &=& -2 (A^{-1}+I)^{-1}+2AI(A+I)^{-1} \\ &=&\color{red}{ -2 (A^{-1}+I)^{-1}+2A(A+I)^{-1} } \quad & | \quad A(A+I)^{-1} = [(A+I)A^{-1} ]^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2[(A+I)A^{-1} ]^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2(AA^{-1}+IA^{-1} )^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2(I+A^{-1} )^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2(A^{-1}+I )^{-1} \\ &=& \mathbf{0} \\\\ && \mathbf{ -2 A^{-2}(A^{-2}+A^{-3})^{-1}+2A^3(A^2+A^3)^{-1} } = 0 \\ \hline \end{array} }\)

 

2.
\(\begin{array}{|rcrcl|} \hline && \mathbf{ -2 A^{-2}(A^{-2}+A^{-3})^{-1}+2A^3(A^2+A^3)^{-1} } \\ &=& \color{red}{-2 (A^{-1}+I)^{-1}+2A(A+I)^{-1}} & =& 0 \\ && -2 (A^{-1}+I)^{-1}+2A(A+I)^{-1} & =& 0 \\ && 2 (A^{-1}+I)^{-1} &=& 2A(A+I)^{-1} \\ && \mathbf{(A^{-1}+I)^{-1}} &\mathbf{=}& \mathbf{ A(A+I)^{-1}} \\ \hline \end{array}\)

 

3.

\(\begin{array}{|rcll|} \hline && -2 A^{-2}(A^{-2}+A^{-3})^{-1}+2A^3(A^2+A^3)^{-1}+ \frac34 A(A^{-1}+I)^{-1}+ \frac34 A^3(A+I)^{-1} \\ &=& 0 + \frac34 A(A^{-1}+I)^{-1}+ \frac34 A^3(A+I)^{-1} \\ &=& \frac34 A\underbrace{(A^{-1}+I)^{-1}}_{=A(A+I)^{-1}}+ \frac34 A^3(A+I)^{-1} \\ &=& \frac34 AA(A+I)^{-1}+ \frac34 A^3(A+I)^{-1} \\ &=& \frac34 A^2(A+I)^{-1}+ \frac34 A^3(A+I)^{-1} \\ &=& \frac34 ( A^2 + A^3)(A+I)^{-1} \\ &=& \frac34 A^2(I+A)(A+I)^{-1} \\ &=& \frac34 A^2\underbrace{(I+A)(I+A)^{-1}}_{=I} \\ &=& \frac34 A^2 I \\ &\mathbf{=}& \mathbf{\frac34 A^2 } \\ \hline \end{array}\)

 

laugh

heureka 21-ago-2017
 #2
avatar+18393 
+1

2. -2 A-2(A-2+A-3)-1+2A3(A2+A3)-1+ 3/4 A(A-1+I)-1+ 3/4 A3(A+I)-1

 

\(\begin{array}{|rcll|} \hline && -2 A^{-2}(A^{-2}+A^{-3})^{-1}+2A^3(A^2+A^3)^{-1}+ \frac34 A(A^{-1}+I)^{-1}+ \frac34 A^3(A+I)^{-1} = \ ?\\ \hline \end{array} \)

 

1.
\(\small{ \begin{array}{|rcll|} \hline &&\mathbf{ -2 A^{-2}(A^{-2}+A^{-3})^{-1}+2A^3(A^2+A^3)^{-1} } \\\\ && & (A^{-2}+A^{-3})^{-1} = [(A^{-1}+I)A^{-2}]^{-1} \\ && & (A^{2}+A^{3})^{-1} = [(A+I)A^{2}]^{-1} \\\\ &=& -2 A^{-2}[(A^{-1}+I)A^{-2}]^{-1}+2A^3[(A+I)A^{2}]^{-1} \quad & | \quad A^{-2} = A^{-1}A^{-1} = (AA)^{-1} = (A^2)^{-1} \\ &=& -2 A^{-2}[(A^{-1}+I)(A^2)^{-1}]^{-1}+2A^3[(A+I)A^{2}]^{-1} \quad & | \quad [(A^{-1}+I)(A^2)^{-1}]^{-1} = [A^2(A^{-1}+I)^{-1}] \\ &=& -2 A^{-2}[A^2(A^{-1}+I)^{-1}]+2A^3[(A+I)A^{2}]^{-1} \\ &=& -2 A^{-2}A^2(A^{-1}+I)^{-1}+2A^3[(A+I)A^{2}]^{-1} \quad & | \quad A^{-2}A^2 = A^{-1}A^{-1}AA = A^{-1}(A^{-1}A)A = A^{-1}(I)A = A^{-1}A = I \\ &=& -2 I(A^{-1}+I)^{-1}+2A^3[(A+I)A^{2}]^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2A^3[(A+I)A^{2}]^{-1} \quad & | \quad [(A+I)A^{2}]^{-1} = [(A^2)^{-1}(A+I)^{-1} ] \\ &=& -2 (A^{-1}+I)^{-1}+2A^3[(A^2)^{-1}(A+I)^{-1} ] \quad & | \quad (A^2)^{-1} = A^{-2} \\ &=& -2 (A^{-1}+I)^{-1}+2A^3A^{-2}(A+I)^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2AA^2A^{-2}(A+I)^{-1} \quad & | \quad A^2A^{-2} = I \\ &=& -2 (A^{-1}+I)^{-1}+2AI(A+I)^{-1} \\ &=&\color{red}{ -2 (A^{-1}+I)^{-1}+2A(A+I)^{-1} } \quad & | \quad A(A+I)^{-1} = [(A+I)A^{-1} ]^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2[(A+I)A^{-1} ]^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2(AA^{-1}+IA^{-1} )^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2(I+A^{-1} )^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2(A^{-1}+I )^{-1} \\ &=& \mathbf{0} \\\\ && \mathbf{ -2 A^{-2}(A^{-2}+A^{-3})^{-1}+2A^3(A^2+A^3)^{-1} } = 0 \\ \hline \end{array} }\)

 

2.

\(\begin{array}{|rcrcl|} \hline && \mathbf{ -2 A^{-2}(A^{-2}+A^{-3})^{-1}+2A^3(A^2+A^3)^{-1} } \\ &=& \color{red}{-2 (A^{-1}+I)^{-1}+2A(A+I)^{-1}} & =& 0 \\ && -2 (A^{-1}+I)^{-1}+2A(A+I)^{-1} & =& 0 \\ && 2 (A^{-1}+I)^{-1} &=& 2A(A+I)^{-1} \\ && \mathbf{(A^{-1}+I)^{-1}} &\mathbf{=}& \mathbf{ A(A+I)^{-1}} \\ \hline \end{array}\)

 

3.

\(\begin{array}{|rcll|} \hline && -2 A^{-2}(A^{-2}+A^{-3})^{-1}+2A^3(A^2+A^3)^{-1}+ \frac34 A(A^{-1}+I)^{-1}+ \frac34 A^3(A+I)^{-1} \\ &=& 0 + \frac34 A(A^{-1}+I)^{-1}+ \frac34 A^3(A+I)^{-1} \\ &=& \frac34 A\underbrace{(A^{-1}+I)^{-1}}_{=A(A+I)^{-1}}+ \frac34 A^3(A+I)^{-1} \\ &=& \frac34 AA(A+I)^{-1}+ \frac34 A^3(A+I)^{-1} \\ &=& \frac34 A^2(A+I)^{-1}+ \frac34 A^3(A+I)^{-1} \\ &=& \frac34 ( A^2 + A^3)(A+I)^{-1} \\ &=& \frac34 A^2(I+A)(A+I)^{-1} \\ &=& \frac34 A^2\underbrace{(I+A)(I+A)^{-1}}_{=I} \\ &=& \frac34 A^2 I \\ &\mathbf{=}& \mathbf{\frac34 A^2 } \\ \hline \end{array} \)

 

laugh

heureka 21-ago-2017
 #1
avatar+18393 
+2

1. A3(A3+A2)-1+A2(A2+A)-1+2(A3+A2)-1A4

Question see also:

https://web2.0calc.com/questions/matrix-problem-simplify-as-little-as-possible-each


1.
\(\begin{array}{|rcll|} \hline &&\mathbf{ A^3(A^3+A^2)^{-1} } \\ &=& A^3[ AA^2+IA^2 ]^{-1} \quad & | \quad \text{Identity matrix I } \\ &=& A^3[ (A+I)A^2 ]^{-1} \quad & | \quad \text{Formula:} \quad (AB)^{-1} = B^{-1}A^{-1} \\ &=& A^3[ (A^2)^{-1}(A+I)^{-1} ] \quad & | \quad (A^2)^{-1} = (AA)^{-1} = A^{-1}A^{-1} = (A^{-1})^2 \\ &=& A^3(A^{-1})^2(A+I)^{-1} \\ &=& AA^2(A^{-1})^2(A+I)^{-1} \quad & | \quad A^2(A^{-1})^2 = I \\ &=& AI(A+I)^{-1} \quad & | \quad AI = A \\ &\mathbf{=}& \mathbf{ A(A+I)^{-1} } \\ \hline \end{array}\)

 

2.
\(\begin{array}{|rcll|} \hline &&\mathbf{ A^2(A^2+A)^{-1} } \\ &=& A^2(AA+IA)^{-1} \quad & | \quad \text{Identity matrix I } \\ &=& A^2[(A+I)A]^{-1} \quad & | \quad \text{Formula:} \quad (AB)^{-1} = B^{-1}A^{-1} \\ &=& A^2[A^{-1}(A+I)^{-1} ] \\ &=& AAA^{-1}(A+I)^{-1} \quad & | \quad AA^{-1} = I \\ &=& AI(A+I)^{-1} \quad & | \quad AI = A \\ &\mathbf{=}& \mathbf{ A(A+I)^{-1} } \\ \hline \end{array}\)

 

3.
\(\begin{array}{|rcll|} \hline &&\mathbf{ 2(A^3+A^2)^{-1}A^4 }\\ &=& 2[ A^2A+A^2I ]^{-1}A^4 \quad & | \quad \text{Identity matrix I } \\ &=& 2[ A^2(A+I) ]^{-1}A^4 \quad & | \quad \text{Formula:} \quad (AB)^{-1} = B^{-1}A^{-1} \\ &=& 2[ (A+I)^{-1}(A^2)^{-1} ] A^4 \quad & | \quad (A^2)^{-1} = (AA)^{-1} = A^{-1}A^{-1} = (A^{-1})^2 \\ &=& 2(A+I)^{-1}(A^{-1})^2 A^4 \\ &=& 2(A+I)^{-1}(A^{-1})^2 A^2A^2 \quad & | \quad (A^{-1})^2 A^2 = I \\ &=& 2(A+I)^{-1}IA^2 \quad & | \quad IA^2 = A^2 \\ &\mathbf{=}& \mathbf{ 2(A+I)^{-1}A^2 } \\ \hline \end{array}\)

 

summary
\(\begin{array}{|rcll|} \hline && A^3(A^3+A^2)^{-1}+A^2(A^2+A)^{-1}+2(A^3+A^2)^{-1}A^4 \\ &=& \mathbf{ A(A+I)^{-1} } + \mathbf{ A(A+I)^{-1} } + \mathbf{ 2(A+I)^{-1}A^2 } \\ &=& 2A(A+I)^{-1} + 2(A+I)^{-1}A^2 \\ &=& 2A(A+I)^{-1}I + 2(A+I)^{-1}A^2 \quad & | \quad I = A^{-1}A \\ &=& 2A(A+I)^{-1}A^{-1}A + 2(A+I)^{-1}A^2 \\ &=& [A(A+I)^{-1}A^{-1} + (A+I)^{-1}A]\ 2A \quad & | \quad A(A+I)^{-1} = [(A+I)A^{-1} ]^{-1} \\ &=& \{ [(A+I)A^{-1} ]^{-1} A^{-1} + (A+I)^{-1}A \} \ 2A \quad & | \quad (A+I)^{-1}A = [A^{-1}(A+I)]^{-1} \\ &=& \{ [(A+I)A^{-1} ]^{-1} A^{-1} + [A^{-1}(A+I)]^{-1} \} \ 2A \\ &=& [ (AA^{-1}+IA^{-1})^{-1} A^{-1} + (A^{-1}A+A^{-1}I)^{-1} ] \ 2A \quad & | \quad AA^{-1} = A^{-1}A = I \qquad IA^{-1} = A^{-1}I = A^{-1} \\ &=& [ (I+A^{-1})^{-1} A^{-1} + (I+A^{-1})^{-1} ] \ 2A \\ &=& [ (I+A^{-1})^{-1} (A^{-1} + I ) ] \ 2A \\ &=& [ (I+A^{-1})^{-1} (I+A^{-1} ) ] \ 2A \quad & | \quad (I+A^{-1})^{-1} (I+A^{-1} ) = I \\ &=& I 2A \\ &\mathbf{=}& \mathbf{ 2A } \\ \hline \end{array}\)

 

 

laugh

heureka 21-ago-2017
 #1
avatar+18393 
+1

1. A3(A3+A2)-1+A2(A2+A)-1+2(A3+A2)-1A4

Question see also:

http://web2.0calc.com/questions/matrix-problem-simplify-as-little-as-possible-each_1

 

\(\begin{array}{|rcll|} \hline && A^3(A^3+A^2)^{-1}+A^2(A^2+A)^{-1}+2(A^3+A^2)^{-1}A^4 = \ ?\\ \hline \end{array}\)

 

1.

\(\begin{array}{|rcll|} \hline &&\mathbf{ A^3(A^3+A^2)^{-1} } \\ &=& A^3[ AA^2+IA^2 ]^{-1} \quad & | \quad \text{Identity matrix I }\\ &=& A^3[ (A+I)A^2 ]^{-1} \quad & | \quad \text{Formula:} \quad (AB)^{-1} = B^{-1}A^{-1}\\ &=& A^3[ (A^2)^{-1}(A+I)^{-1} ] \quad & | \quad (A^2)^{-1} = (AA)^{-1} = A^{-1}A^{-1} = (A^{-1})^2 \\ &=& A^3(A^{-1})^2(A+I)^{-1} \quad & | \quad \\ &=& AA^2(A^{-1})^2(A+I)^{-1} \quad & | \quad A^2(A^{-1})^2 = I \\ &=& AI(A+I)^{-1} \quad & | \quad AI = A \\ &\mathbf{=}& \mathbf{ A(A+I)^{-1} } \\ \hline \end{array}\)

 

2.

\(\begin{array}{|rcll|} \hline &&\mathbf{ A^2(A^2+A)^{-1} } \\ &=& A^2(AA+IA)^{-1} \quad & | \quad \text{Identity matrix I } \\ &=& A^2[(A+I)A]^{-1} \quad & | \quad \text{Formula:} \quad (AB)^{-1} = B^{-1}A^{-1} \\ &=& A^2[A^{-1}(A+I)^{-1} ] \\ &=& AAA^{-1}(A+I)^{-1} \quad & | \quad AA^{-1} = I \\ &=& AI(A+I)^{-1} \quad & | \quad AI = A \\ &\mathbf{=}& \mathbf{ A(A+I)^{-1} } \\ \hline \end{array} \)

 

3.

\(\begin{array}{|rcll|} \hline &&\mathbf{ 2(A^3+A^2)^{-1}A^4 }\\ &=& 2[ A^2A+A^2I ]^{-1}A^4 \quad & | \quad \text{Identity matrix I } \\ &=& 2[ A^2(A+I) ]^{-1}A^4 \quad & | \quad \text{Formula:} \quad (AB)^{-1} = B^{-1}A^{-1} \\ &=& 2[ (A+I)^{-1}(A^2)^{-1} ] A^4 \quad & | \quad (A^2)^{-1} = (AA)^{-1} = A^{-1}A^{-1} = (A^{-1})^2 \\ &=& 2(A+I)^{-1}(A^{-1})^2 A^4 \\ &=& 2(A+I)^{-1}(A^{-1})^2 A^2A^2 \quad & | \quad (A^{-1})^2 A^2 = I \\ &=& 2(A+I)^{-1}IA^2 \quad & | \quad IA^2 = A^2 \\ &\mathbf{=}& \mathbf{ 2(A+I)^{-1}A^2 } \\ \hline \end{array}\)

 

summary

\(\begin{array}{|rcll|} \hline && A^3(A^3+A^2)^{-1}+A^2(A^2+A)^{-1}+2(A^3+A^2)^{-1}A^4 \\ &=& \mathbf{ A(A+I)^{-1} } + \mathbf{ A(A+I)^{-1} } + \mathbf{ 2(A+I)^{-1}A^2 } \\ &=& 2A(A+I)^{-1} + 2(A+I)^{-1}A^2 \\ &=& 2A(A+I)^{-1}I + 2(A+I)^{-1}A^2 \quad & | \quad I = A^{-1}A \\ &=& 2A(A+I)^{-1}A^{-1}A + 2(A+I)^{-1}A^2 \\ &=& [A(A+I)^{-1}A^{-1} + (A+I)^{-1}A]\ 2A \quad & | \quad A(A+I)^{-1} = [(A+I)A^{-1} ]^{-1} \\ &=& \{ [(A+I)A^{-1} ]^{-1} A^{-1} + (A+I)^{-1}A \} \ 2A \quad & | \quad (A+I)^{-1}A = [A^{-1}(A+I)]^{-1} \\ &=& \{ [(A+I)A^{-1} ]^{-1} A^{-1} + [A^{-1}(A+I)]^{-1} \} \ 2A \\ &=& [ (AA^{-1}+IA^{-1})^{-1} A^{-1} + (A^{-1}A+A^{-1}I)^{-1} ] \ 2A \quad & | \quad AA^{-1} = A^{-1}A = I \qquad IA^{-1} = A^{-1}I = A^{-1} \\ &=& [ (I+A^{-1})^{-1} A^{-1} + (I+A^{-1})^{-1} ] \ 2A \\ &=& [ (I+A^{-1})^{-1} (A^{-1} + I ) ] \ 2A \\ &=& [ (I+A^{-1})^{-1} (I+A^{-1} ) ] \ 2A \quad & | \quad (I+A^{-1})^{-1} (I+A^{-1} ) = I \\ &=& I 2A \\ &\mathbf{=}& \mathbf{ 2A } \\ \hline \end{array} \)

 

laugh

heureka 21-ago-2017
 #3
avatar+18393 
0

Matrix Problem: A \(\mathbf{\in \mu}\) (2x2).

Find the possible conditions and possible cases in the real numbers for which A2 = 0

In linear algebra, a nilpotent matrix is a square matrix A such that

\({\displaystyle A^{k}=0\,}\)

for some positive integer k.

 

If A is a nilpotent matrix then det(A) = 0

\(\begin{array}{|rcll|} \hline det(A) &=& \begin{vmatrix} a & b \\ c&d \end{vmatrix} = ad-bc = 0 \\\\ \Rightarrow \qquad \mathbf{bc} & \mathbf{=} & \mathbf{ad} \\ \hline \end{array}\)

 

\(A^2 = 0 \\ \begin{array}{|rcll|} \hline A^2 &=& \begin{pmatrix} a & b \\ c&d \end{pmatrix} \cdot \begin{pmatrix} a & b \\ c&d \end{pmatrix} \\\\ &=& \begin{pmatrix} a^2+bc & b\cdot (a+d) \\ c\cdot (a+d) & d^2+bc \end{pmatrix} \quad & | \quad \mathbf{bc=ad} \\\\ &=& \begin{pmatrix} a^2+ad & b\cdot (a+d) \\ c\cdot (a+d) & d^2+ad \end{pmatrix} \\\\ &=& \begin{pmatrix} a\cdot (a+d) & b\cdot (a+d) \\ c\cdot (a+d) & d\cdot(a+d) \end{pmatrix} \\\\ A^2 &=& \underbrace{(a+d)}_{=0}\cdot \underbrace{\begin{pmatrix} a & b \\ c & d \end{pmatrix}}_{A\ne 0} = 0\\\\ a+d &=& 0 \\ \mathbf{a} &\mathbf{=}& \mathbf{-d} \\ \hline \end{array}\)

 

if \(A\ne 0 \) and \(A^2 = 0\), then \( a = -d\) and \(det(A) = 0\).

 

\(\text{Example } 1:\\ \begin{array}{|rcll|} \hline A &=& \begin{pmatrix} 0 & 1 \\ 0&0 \end{pmatrix} \\ det(A) &=& \begin{vmatrix} 0 & 1 \\ 0&0 \end{vmatrix} = 0\cdot 0 - 0 \cdot 1 = 0 \\ a &=& -d \\ 0 &=& 0\ \checkmark \\\\ A^2 &=& \begin{pmatrix} 0 & 1 \\ 0&0 \end{pmatrix} \cdot \begin{pmatrix} 0 & 1 \\ 0&0 \end{pmatrix} = 0\ \checkmark \\ \hline \end{array}\)

 

\(\text{Example } 2:\\ \begin{array}{|rcll|} \hline A &=& \begin{pmatrix} 4 & 8 \\ -2 & -4 \end{pmatrix} \\ det(A) &=& \begin{vmatrix} 4 & 8 \\ -2 & -4 \end{vmatrix} = 4\cdot(-4) - (-2) \cdot 8 = -16+16 = 0 \\ a &=& -d \\ 4 &=& -(-4) \\ 4 &=&4\ \checkmark \\\\ A^2 &=& \begin{pmatrix} 4 & 8 \\ -2 & -4 \end{pmatrix} \cdot \begin{pmatrix} 4 & 8 \\ -2 & -4 \end{pmatrix} = 0\ \checkmark \\ \hline \end{array}\)

 

 

laugh

heureka 18-ago-2017
 #2
avatar+18393 
+3

x|x|=2x+1 

The lines are absolute value

 

\(\begin{array}{|rcll|} \hline x|x| &=& 2x+1 \quad & | \quad \text{square both sides} \\ x^2\cdot |x|^2 &=& (2x+1)^2 \quad & | \quad |x|^2 = x^2 \\ x^2\cdot x^2 &=& (2x+1)^2 \\ x^4 &=& 4x^2+4x+1 \\ \mathbf{x^4 - 4x^2+4x+1} & \mathbf{=} & \mathbf{0} \\ \hline \end{array} \)

 

solutions of \(x^4 - 4x^2+4x+1 = 0\)

see: http://www.wolframalpha.com/input/?i=x%5E4-4x%5E2-4x-1%3D0

\(\begin{array}{rcll} x_1 &=& -1 \\ x_2 &=& 1 - \sqrt{2} \\ x_3 &=& 1 + \sqrt{2} \\ \end{array}\)

 

Solutions of \(x|x|=2x+1\):

\(x_1 = -1 \\ \begin{array}{|rcll|} \hline (-1)\cdot |-1| & \overset{?}{=} & 2\cdot (-1)+1 \\ (-1)\cdot 1 & \overset{?}{=} & -2+1 \\ -1 & \overset{!}{=} & -1\ \checkmark \\ \hline \end{array} \)

x = -1 is a solution

 

\(x_2 = 1 - \sqrt{2} \\ \begin{array}{|rcll|} \hline (1 - \sqrt{2})\cdot |1 - \sqrt{2}| & \overset{?}{=} & 2\cdot (1 - \sqrt{2})+1 \\ (-0.41421356237)\cdot |-0.41421356237| & \overset{?}{=} & 2\cdot (-0.41421356237)+1 \\ (-0.41421356237)\cdot 0.41421356237 & \overset{?}{=} & -0.82842712475 + 1 \\ -0.17157287525 & \ne & 0.17157287525 \\ \hline \end{array}\)

\(\mathbf{x = 1-\sqrt{2} }\) is not a solution

 

\(x_2 = 1 + \sqrt{2} \\ \begin{array}{|rcll|} \hline (1 + \sqrt{2})\cdot |1 + \sqrt{2}| & \overset{?}{=} & 2\cdot (1 + \sqrt{2})+1 \\ (2.41421356237)\cdot |2.41421356237| & \overset{?}{=} & 2\cdot (2.41421356237)+1 \\ 2.41421356237\cdot 2.41421356237 & \overset{?}{=} & 4.82842712475 + 1 \\ 5.82842712475 & \overset{!}{=} & 5.82842712475\ \checkmark \\ \hline \end{array}\)

\(\mathbf{x = 1+\sqrt{2} }\) is a solution

 

laugh

heureka 18-ago-2017
 #1
avatar+18393 
+1

Two altitudes of a triangle have lengths 12 and 14. What is the longest possible length of the third altitude,

if it is a positive integer?

 

Let h = the length of the third altitude.

Let a, b, and c be the sides corresponding to the altitudes of length 12, 14, and h

 

Let A = Area of the triangle.

Let 2A = a*12
Let 2A = b*14
Let 2A = c*h

 

1.

\(\begin{array}{|rcll|} \hline a &=& \frac{2A}{12} \\ b &=& \frac{2A}{14} \\ c &=& \frac{2A}{h} \\ \hline \end{array}\)

 

2.

A triangle exists with side lengths a, b, and c
if and only if they satisfy the three triangle inequalities:

\(\begin{array}{|lrcll|} \hline (1) & a & < & b + c \\ (2) & b & < & c + a \\ (3) & c & < & a + b \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline (1) & \frac{2A}{12} & < & \frac{2A}{14} + \frac{2A}{h} \quad & | \quad :2A \\ & \frac{1}{12} & < & \frac{1}{14} + \frac{1}{h} \\\\ (2) & \frac{2A}{14} & < & \frac{2A}{h} + \frac{2A}{12} \quad & | \quad :2A \\ & \frac{1}{14} & < & \frac{1}{h} + \frac{1}{12} \\\\ (3) & \frac{2A}{h} & < & \frac{2A}{12} + \frac{2A}{14} \quad & | \quad :2A \\ & \frac{1}{h} & < & \frac{1}{12} + \frac{1}{14} \\ \hline \end{array}\)

 

(1):

\( \begin{array}{|rcll|} \hline \frac{1}{12} & < & \frac{1}{14} + \frac{1}{h} \\ \frac{1}{12}-\frac{1}{14} & < & \frac{1}{h} \\ \frac{12-14}{12\cdot 14} & < & \frac{1}{h} \\ \frac{2}{168} & < & \frac{1}{h} \\ \frac{1}{84} & < & \frac{1}{h} \quad & | \quad \cdot 84h \\ \mathbf{ h } & \mathbf{ < } & \mathbf{ 84 } \\ \hline \end{array} \)

 

(2):

\(\begin{array}{|rcll|} \hline \frac{1}{14} & < & \frac{1}{h} + \frac{1}{12} \\ \frac{1}{14} - \frac{1}{12} & < & \frac{1}{h} \\ \frac{12-14}{14\cdot 12} & < & \frac{1}{h} \\ -\frac{2}{168} & < & \frac{1}{h} \\ -\frac{1}{84} & < & \frac{1}{h} \quad & | \quad \cdot 84h \\ -h & < & 84 \quad & | \quad \cdot (-1) \qquad switch "<" to ">" \\ \mathbf{ h } & \mathbf{ > } & \mathbf{ -84 } \\ \hline \end{array} \)

 

(3):

\(\begin{array}{|rcll|} \hline \frac{1}{h} & < & \frac{1}{12} + \frac{1}{14} \\ \frac{1}{h} & < & \frac{14+12}{12\cdot 14} \\ \frac{1}{h} & < & \frac{26}{168} \\ \frac{1}{h} & < & \frac{13}{84} \quad & | \quad \cdot \frac{84}{13}\ h \\ \frac{84}{13} & < & h \\ \mathbf{ 6.46153846154 } & \mathbf{ < } & \mathbf{ h } \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline 6.46153846154 < h < 84 \\ \hline \end{array}\)

 

The longest possible length of the third altitude, if it is a positive integer is 83

 

laugh

heureka 17-ago-2017