Question and Answers Forum

All Questions      Topic List

Matrices and Determinants Questions

Previous in All Question      Next in All Question      

Previous in Matrices and Determinants      Next in Matrices and Determinants      

Question Number 74323 by arthur.kangdani@gmail.com last updated on 22/Nov/19

Commented by arthur.kangdani@gmail.com last updated on 22/Nov/19

Determine the inverse of the  matrix of AB+C!

$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{inverse}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{matrix}\:\mathrm{of}\:{AB}+{C}! \\ $$

Commented by mathmax by abdo last updated on 22/Nov/19

A.B=  (((4          2)),((6           3)) )  (((2        3)),((5        4)) ) =  (((18         20)),((27           30)) )  A.B +C = (((18       20)),((27       30)) )  + (((6        4)),((2         3)) ) = (((24        24)),((29        33)) ) =M  P_c (x) =det(M−xI)=  determinant (((24−x        24)),((29          33−x)))  =(24−x)(33−x)−29×24 =24×33−24x−33x+x^2 −29×24  =x^2 −57x +24(33−29) =x^2 −57x +96  cayley hamilton ⇒M^2 −57 M +96I =0 ⇒  57M−M^2 =96I ⇒ M×((1/(96))( I−M))=I ⇒  M^(−1) =(1/(96))(I−M) =(1/(96))(   (((1       0)),((0        1)) ) − (((24       24)),((29       33)) ))  =(1/(96))  (((−23           −24)),((−29             −32)) )  = (((−((23)/(96))           −((24)/(96)))),((−((29)/(96))               −((32)/(96)))) )

$${A}.{B}=\:\begin{pmatrix}{\mathrm{4}\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{6}\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{2}\:\:\:\:\:\:\:\:\mathrm{3}}\\{\mathrm{5}\:\:\:\:\:\:\:\:\mathrm{4}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{18}\:\:\:\:\:\:\:\:\:\mathrm{20}}\\{\mathrm{27}\:\:\:\:\:\:\:\:\:\:\:\mathrm{30}}\end{pmatrix} \\ $$$${A}.{B}\:+{C}\:=\begin{pmatrix}{\mathrm{18}\:\:\:\:\:\:\:\mathrm{20}}\\{\mathrm{27}\:\:\:\:\:\:\:\mathrm{30}}\end{pmatrix}\:\:+\begin{pmatrix}{\mathrm{6}\:\:\:\:\:\:\:\:\mathrm{4}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:\mathrm{3}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{24}\:\:\:\:\:\:\:\:\mathrm{24}}\\{\mathrm{29}\:\:\:\:\:\:\:\:\mathrm{33}}\end{pmatrix}\:={M} \\ $$$${P}_{{c}} \left({x}\right)\:={det}\left({M}−{xI}\right)=\:\begin{vmatrix}{\mathrm{24}−{x}\:\:\:\:\:\:\:\:\mathrm{24}}\\{\mathrm{29}\:\:\:\:\:\:\:\:\:\:\mathrm{33}−{x}}\end{vmatrix} \\ $$$$=\left(\mathrm{24}−{x}\right)\left(\mathrm{33}−{x}\right)−\mathrm{29}×\mathrm{24}\:=\mathrm{24}×\mathrm{33}−\mathrm{24}{x}−\mathrm{33}{x}+{x}^{\mathrm{2}} −\mathrm{29}×\mathrm{24} \\ $$$$={x}^{\mathrm{2}} −\mathrm{57}{x}\:+\mathrm{24}\left(\mathrm{33}−\mathrm{29}\right)\:={x}^{\mathrm{2}} −\mathrm{57}{x}\:+\mathrm{96} \\ $$$${cayley}\:{hamilton}\:\Rightarrow{M}^{\mathrm{2}} −\mathrm{57}\:{M}\:+\mathrm{96}{I}\:=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{57}{M}−{M}^{\mathrm{2}} =\mathrm{96}{I}\:\Rightarrow\:{M}×\left(\frac{\mathrm{1}}{\mathrm{96}}\left(\:{I}−{M}\right)\right)={I}\:\Rightarrow \\ $$$${M}^{−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{96}}\left({I}−{M}\right)\:=\frac{\mathrm{1}}{\mathrm{96}}\left(\:\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:−\begin{pmatrix}{\mathrm{24}\:\:\:\:\:\:\:\mathrm{24}}\\{\mathrm{29}\:\:\:\:\:\:\:\mathrm{33}}\end{pmatrix}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{96}}\:\begin{pmatrix}{−\mathrm{23}\:\:\:\:\:\:\:\:\:\:\:−\mathrm{24}}\\{−\mathrm{29}\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{32}}\end{pmatrix}\:\:=\begin{pmatrix}{−\frac{\mathrm{23}}{\mathrm{96}}\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{24}}{\mathrm{96}}}\\{−\frac{\mathrm{29}}{\mathrm{96}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{32}}{\mathrm{96}}}\end{pmatrix} \\ $$

Commented by abdomathmax last updated on 22/Nov/19

error at lime 7  M×(57I−M)=96 I ⇒  M×[(1/(96))( 57 I−M)]=I ⇒  M^(−1) =(1/(96)){  (((57       0)),((0        57)) ) − (((24        24)),((29         33)) )}  =(1/(96))  (((33         −24)),((−29         24)) )=....

$${error}\:{at}\:{lime}\:\mathrm{7}\:\:{M}×\left(\mathrm{57}{I}−{M}\right)=\mathrm{96}\:{I}\:\Rightarrow \\ $$$${M}×\left[\frac{\mathrm{1}}{\mathrm{96}}\left(\:\mathrm{57}\:{I}−{M}\right)\right]={I}\:\Rightarrow \\ $$$${M}^{−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{96}}\left\{\:\begin{pmatrix}{\mathrm{57}\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{57}}\end{pmatrix}\:−\begin{pmatrix}{\mathrm{24}\:\:\:\:\:\:\:\:\mathrm{24}}\\{\mathrm{29}\:\:\:\:\:\:\:\:\:\mathrm{33}}\end{pmatrix}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{96}}\:\begin{pmatrix}{\mathrm{33}\:\:\:\:\:\:\:\:\:−\mathrm{24}}\\{−\mathrm{29}\:\:\:\:\:\:\:\:\:\mathrm{24}}\end{pmatrix}=.... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com