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 141400 by Raffaqet last updated on 18/May/21

fjnd inverse of matrix [(2,(−5)),(1,3) ]

$${fjnd}\:{inverse}\:{of}\:{matrix}\begin{bmatrix}{\mathrm{2}}&{−\mathrm{5}}\\{\mathrm{1}}&{\mathrm{3}}\end{bmatrix} \\ $$

Answered by iloveisrael last updated on 18/May/21

A^(−1) = (1/(6−(−5))) [(( 3 5)),((−1 2)) ] A^(−1) = [((3/11 5/11)),((−1/11 2/11)) ]

$$\:{A}^{−\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{6}−\left(−\mathrm{5}\right)}\:\begin{bmatrix}{\:\:\:\:\mathrm{3}\:\:\:\:\:\mathrm{5}}\\{−\mathrm{1}\:\:\:\:\:\mathrm{2}}\end{bmatrix} \\ $$$${A}^{−\mathrm{1}} \:=\:\begin{bmatrix}{\mathrm{3}/\mathrm{11}\:\:\:\:\:\:\:\:\:\mathrm{5}/\mathrm{11}}\\{−\mathrm{1}/\mathrm{11}\:\:\:\:\:\mathrm{2}/\mathrm{11}}\end{bmatrix} \\ $$

Answered by EDWIN88 last updated on 18/May/21

Cayley −Hamilton theorem ∣λI−A∣ = 0 ⇒p(λ)=λ^2 −tr .A+ det(A) =0 p(A) = A^2 −5A +11I = 0 ⇒A−5I = −11A^(−1) ⇒A^(−1) = (1/(11)) { (((5 0)),((0 5)) ) − (((2 −5)),((1 3)) ) } ⇒A^(−1) = (1/(11)) ((( 3 5)),((−1 2)) ) ⋇

$$\:\mathrm{Cayley}\:−\mathrm{Hamilton}\:\mathrm{theorem} \\ $$$$\:\mid\lambda\mathrm{I}−\mathrm{A}\mid\:=\:\mathrm{0}\:\Rightarrow\mathrm{p}\left(\lambda\right)=\lambda^{\mathrm{2}} −\mathrm{tr}\:.\mathrm{A}+\:\mathrm{det}\left(\mathrm{A}\right)\:=\mathrm{0} \\ $$$$\mathrm{p}\left(\mathrm{A}\right)\:=\:\mathrm{A}^{\mathrm{2}} −\mathrm{5A}\:+\mathrm{11I}\:=\:\mathrm{0} \\ $$$$\Rightarrow\mathrm{A}−\mathrm{5I}\:=\:−\mathrm{11A}^{−\mathrm{1}} \\ $$$$\Rightarrow\mathrm{A}^{−\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{11}}\:\left\{\:\begin{pmatrix}{\mathrm{5}\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\mathrm{5}}\end{pmatrix}\:−\:\begin{pmatrix}{\mathrm{2}\:\:\:\:\:\:−\mathrm{5}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\mathrm{3}}\end{pmatrix}\:\right\} \\ $$$$\Rightarrow\mathrm{A}^{−\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{11}}\:\begin{pmatrix}{\:\:\:\:\mathrm{3}\:\:\:\:\:\mathrm{5}}\\{−\mathrm{1}\:\:\:\:\:\mathrm{2}}\end{pmatrix}\:\divideontimes \\ $$

Answered by mathmax by abdo last updated on 19/May/21

p_c (A) =det (A−xI)= determinant (((2−x −5)),((1 3−x)))=(2−x)(3−x)+5 =(x−2)(x−3)+5 =x^2 −5x+6 +5 =x^2 −5x +11 cayley hamilton ⇒A^2 −5A +11I =0 ⇒A^2 −5A =−11 I ⇒ −(1/(11))(A^2 −5A) =I ⇒−(1/(11))A×(A−5I)=I ⇒ A^(−1) =−(1/(11))(A−5I)=(1/(11))(5I−A)=(5/(11)) (((1 0)),((0 1)) )−(1/(11)) (((2 −5)),((1 3)) ) = ((((3/(11)) (5/(11)))),((−(1/(11)) (2/(11)))) )

$$\mathrm{p}_{\mathrm{c}} \left(\mathrm{A}\right)\:=\mathrm{det}\:\left(\mathrm{A}−\mathrm{xI}\right)=\begin{vmatrix}{\mathrm{2}−\mathrm{x}\:\:\:\:\:−\mathrm{5}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}−\mathrm{x}}\end{vmatrix}=\left(\mathrm{2}−\mathrm{x}\right)\left(\mathrm{3}−\mathrm{x}\right)+\mathrm{5} \\ $$$$=\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{x}−\mathrm{3}\right)+\mathrm{5}\:=\mathrm{x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{6}\:+\mathrm{5}\:=\mathrm{x}^{\mathrm{2}} −\mathrm{5x}\:+\mathrm{11} \\ $$$$\mathrm{cayley}\:\mathrm{hamilton}\:\Rightarrow\mathrm{A}^{\mathrm{2}} −\mathrm{5A}\:+\mathrm{11I}\:=\mathrm{0}\:\Rightarrow\mathrm{A}^{\mathrm{2}} −\mathrm{5A}\:=−\mathrm{11}\:\mathrm{I}\:\Rightarrow \\ $$$$−\frac{\mathrm{1}}{\mathrm{11}}\left(\mathrm{A}^{\mathrm{2}} −\mathrm{5A}\right)\:=\mathrm{I}\:\Rightarrow−\frac{\mathrm{1}}{\mathrm{11}}\mathrm{A}×\left(\mathrm{A}−\mathrm{5I}\right)=\mathrm{I}\:\Rightarrow \\ $$$$\mathrm{A}^{−\mathrm{1}} \:=−\frac{\mathrm{1}}{\mathrm{11}}\left(\mathrm{A}−\mathrm{5I}\right)=\frac{\mathrm{1}}{\mathrm{11}}\left(\mathrm{5I}−\mathrm{A}\right)=\frac{\mathrm{5}}{\mathrm{11}}\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}−\frac{\mathrm{1}}{\mathrm{11}}\begin{pmatrix}{\mathrm{2}\:\:\:\:\:\:\:\:−\mathrm{5}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}}\end{pmatrix} \\ $$$$=\begin{pmatrix}{\frac{\mathrm{3}}{\mathrm{11}}\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{5}}{\mathrm{11}}}\\{−\frac{\mathrm{1}}{\mathrm{11}}\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}}{\mathrm{11}}}\end{pmatrix} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com