Previous in Matrices and Determinants Next in Matrices and Determinants
Question Number 59295 by pooja24 last updated on 07/May/19
$${Let}\:{A}\:{be}\mathrm{3}×\mathrm{3}\:{matrix}\:{with}\:{eigen}\:{values} \\ $$$$\mathrm{1},−\mathrm{1},\mathrm{0}.\:{Then}\:{determinant}\:{of}\:{I}+{A}^{\mathrm{100}\:} =?? \\ $$
Answered by alex041103 last updated on 10/May/19
![basically A=PDP^(−1) where D is a diagonal matrix ⇒A^(100) =PD^(100) P^(−1) also I=PIP^(−1) ⇒M=I+A^(100) =P(I+D^(100) )P^(−1) Then det(M)=det(P)det(I+D^(100) )det(P^(−1) )= =[det(P)det(P^(−1) )]det(I+D^(100) )= =det(PP^(−1) )det(I+D^(100) )= =det(I)det(I+D^(100) )= =det(I+D^(100) )=det(M) We define P as the matrix which gets us from the normal system of coordinates to the system of the coordinates which uses the eigen vectors as basis vectors. Then D is a diagonal matrix with the eigen values in it. ⇒D= [(0,0,0),(0,1,0),(0,0,(−1)) ] ⇒D^(100) = [(0,0,0),(0,1,0),(0,0,1) ] ⇒det(M)= determinant ((1,0,0),(0,2,0),(0,0,2))=4 ⇒det(I+A^(100) )=4](Q59421.png)
$${basically} \\ $$$${A}={PDP}^{−\mathrm{1}} \\ $$$${where}\:{D}\:{is}\:{a}\:{diagonal}\:{matrix} \\ $$$$\Rightarrow{A}^{\mathrm{100}} ={PD}^{\mathrm{100}} {P}^{−\mathrm{1}} \\ $$$${also}\:{I}={PIP}^{−\mathrm{1}} \\ $$$$\Rightarrow{M}={I}+{A}^{\mathrm{100}} ={P}\left({I}+{D}^{\mathrm{100}} \right){P}^{−\mathrm{1}} \\ $$$${Then}\:{det}\left({M}\right)={det}\left({P}\right){det}\left({I}+{D}^{\mathrm{100}} \right){det}\left({P}^{−\mathrm{1}} \right)= \\ $$$$=\left[{det}\left({P}\right){det}\left({P}^{−\mathrm{1}} \right)\right]{det}\left({I}+{D}^{\mathrm{100}} \right)= \\ $$$$={det}\left({PP}^{−\mathrm{1}} \right){det}\left({I}+{D}^{\mathrm{100}} \right)= \\ $$$$={det}\left({I}\right){det}\left({I}+{D}^{\mathrm{100}} \right)= \\ $$$$={det}\left({I}+{D}^{\mathrm{100}} \right)={det}\left({M}\right) \\ $$$${We}\:{define}\:{P}\:{as}\:{the}\:{matrix}\:{which}\:{gets}\: \\ $$$${us}\:{from}\:{the}\:{normal}\:{system}\:{of}\:{coordinates}\:{to}\:{the} \\ $$$${system}\:{of}\:{the}\:{coordinates}\:{which}\:{uses} \\ $$$${the}\:{eigen}\:{vectors}\:{as}\:{basis}\:{vectors}.\:{Then} \\ $$$${D}\:{is}\:{a}\:{diagonal}\:{matrix}\:{with}\:{the}\:{eigen} \\ $$$${values}\:{in}\:{it}. \\ $$$$\Rightarrow{D}=\begin{bmatrix}{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{−\mathrm{1}}\end{bmatrix} \\ $$$$\Rightarrow{D}^{\mathrm{100}} =\begin{bmatrix}{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix} \\ $$$$\Rightarrow{det}\left({M}\right)=\begin{vmatrix}{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{2}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{2}}\end{vmatrix}=\mathrm{4} \\ $$$$\Rightarrow{det}\left({I}+{A}^{\mathrm{100}} \right)=\mathrm{4} \\ $$$$ \\ $$