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avatar+846 

only question b please.

after completing a i didnt know what i can do with my answer so i subbed in n=-1 in the original equation and it seemed to work

but im sure thats not how u do it. please explain thanks

 Dec 3, 2018

Best Answer 

 #1
avatar+22500 
+15

inverse matrices

(b)

Hence find the inverse, \(A_n^{-1} \) of \(A_n\) in terms of \(n\)

 

Formula inverse matrice:

\(\begin{array}{|rcll|} \hline A &=& e\cdot \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \\\\ A^{-1} &=& \dfrac{e\cdot \begin{pmatrix} d & -b \\ -c & a \\ \end{pmatrix}}{ det(A)} \qquad det(A) = e^2\cdot(a\cdot d - c\cdot b) \\\\ && \mathbf{ \boxed{ A^{-1} = \dfrac{ \begin{pmatrix} d & -b \\ -c & a \\ \end{pmatrix}}{e\cdot(a\cdot d - c\cdot b)} } } \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline A_n &=& 5^{n-1} \begin{pmatrix} 2n+5 & -2n \\ 2n & 5-2n \\ \end{pmatrix} \\\\ A_n^{-1} &=& \dfrac{ \begin{pmatrix} 5-2n & 2n \\ -2n & 2n+5 \\ \end{pmatrix} } {5^{n-1}\cdot \Big((2n+5)(5-2n) - (2n)(-2n) \Big) } \\\\ &=& \dfrac{ \begin{pmatrix} 5-2n & 2n \\ -2n & 2n+5 \\ \end{pmatrix} } {5^{n-1}\cdot (25-4n^2+4n^2 ) } \\\\ &=& \dfrac{ \begin{pmatrix} 5-2n & 2n \\ -2n & 2n+5 \\ \end{pmatrix} } { 5^{n-1}\cdot 5^2 } \\\\ &=& \dfrac{ \begin{pmatrix} 5-2n & 2n \\ -2n & 2n+5 \\ \end{pmatrix} } { 5^{n-1+2} } \\\\ &=& \dfrac{ \begin{pmatrix} 5-2n & 2n \\ -2n & 2n+5 \\ \end{pmatrix} } { 5^{n+1} } \\\\ \mathbf{A_n^{-1}} & \mathbf{=} & \mathbf{ 5^{-n-1} \begin{pmatrix} 5-2n & 2n \\ -2n & 2n+5 \\ \end{pmatrix} } \\ \hline \end{array}\)

 

laugh

 Dec 3, 2018
 #1
avatar+22500 
+15
Best Answer

inverse matrices

(b)

Hence find the inverse, \(A_n^{-1} \) of \(A_n\) in terms of \(n\)

 

Formula inverse matrice:

\(\begin{array}{|rcll|} \hline A &=& e\cdot \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \\\\ A^{-1} &=& \dfrac{e\cdot \begin{pmatrix} d & -b \\ -c & a \\ \end{pmatrix}}{ det(A)} \qquad det(A) = e^2\cdot(a\cdot d - c\cdot b) \\\\ && \mathbf{ \boxed{ A^{-1} = \dfrac{ \begin{pmatrix} d & -b \\ -c & a \\ \end{pmatrix}}{e\cdot(a\cdot d - c\cdot b)} } } \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline A_n &=& 5^{n-1} \begin{pmatrix} 2n+5 & -2n \\ 2n & 5-2n \\ \end{pmatrix} \\\\ A_n^{-1} &=& \dfrac{ \begin{pmatrix} 5-2n & 2n \\ -2n & 2n+5 \\ \end{pmatrix} } {5^{n-1}\cdot \Big((2n+5)(5-2n) - (2n)(-2n) \Big) } \\\\ &=& \dfrac{ \begin{pmatrix} 5-2n & 2n \\ -2n & 2n+5 \\ \end{pmatrix} } {5^{n-1}\cdot (25-4n^2+4n^2 ) } \\\\ &=& \dfrac{ \begin{pmatrix} 5-2n & 2n \\ -2n & 2n+5 \\ \end{pmatrix} } { 5^{n-1}\cdot 5^2 } \\\\ &=& \dfrac{ \begin{pmatrix} 5-2n & 2n \\ -2n & 2n+5 \\ \end{pmatrix} } { 5^{n-1+2} } \\\\ &=& \dfrac{ \begin{pmatrix} 5-2n & 2n \\ -2n & 2n+5 \\ \end{pmatrix} } { 5^{n+1} } \\\\ \mathbf{A_n^{-1}} & \mathbf{=} & \mathbf{ 5^{-n-1} \begin{pmatrix} 5-2n & 2n \\ -2n & 2n+5 \\ \end{pmatrix} } \\ \hline \end{array}\)

 

laugh

heureka Dec 3, 2018
 #2
avatar+101796 
+1

Nice, heureka !!!

 

cool cool cool

CPhill  Dec 3, 2018
 #3
avatar+22500 
+14

Thank you, CPhill !

 

laugh

heureka  Dec 4, 2018
 #4
avatar+846 
+2

thanks for the detailed explanation. its perfect!!

YEEEEEET  Dec 4, 2018

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