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CBSE Class 12-science Answered

eigen vectors
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Asked by dakarapuaditya123 | 26 Dec, 2022, 08:20: PM
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begin mathsize 14px style M a t r i x space A space equals space open square brackets table row 3 1 1 row 1 3 cell negative 1 end cell row 1 cell negative 1 end cell 3 end table close square brackets end style
 
Characterstic equation | A - λ I | = 0
 
where λ is eigen value and I is unit matrix of order ( 3 × 3 )
 
begin mathsize 14px style open vertical bar A space minus space lambda space I close vertical bar space equals space open square brackets table row cell left parenthesis 3 minus lambda right parenthesis end cell 1 1 row 1 cell left parenthesis 3 minus lambda right parenthesis end cell cell negative 1 end cell row 1 cell negative 1 end cell cell left parenthesis 3 minus lambda right parenthesis end cell end table close square brackets space equals space 0 end style
begin mathsize 14px style left parenthesis 3 minus lambda right parenthesis space left square bracket space left parenthesis 3 minus lambda right parenthesis squared minus 1 right square bracket space minus left parenthesis 4 minus lambda right parenthesis plus left parenthesis lambda minus 4 right parenthesis space equals space 0 end style

begin mathsize 14px style left parenthesis 3 minus lambda right parenthesis space left square bracket space lambda squared minus 6 lambda plus 8 space right square bracket space plus space 2 space left parenthesis lambda minus 4 right parenthesis space equals space 0 end style
 
begin mathsize 14px style left parenthesis 3 minus lambda right parenthesis space left parenthesis lambda minus 4 right parenthesis left parenthesis lambda minus 2 right parenthesis space plus space 2 space left parenthesis lambda minus 4 right parenthesis space equals space 0 end style
 
begin mathsize 14px style left parenthesis lambda minus 4 right parenthesis space left square bracket space left parenthesis 3 minus lambda right parenthesis space left parenthesis lambda minus 2 right parenthesis space plus 2 space right square bracket space equals space 0 end style
begin mathsize 14px style left parenthesis lambda minus 4 right parenthesis space left square bracket space lambda squared minus 5 lambda space plus 4 space right square bracket space equals space left parenthesis lambda minus 4 right parenthesis left parenthesis lambda minus 4 right parenthesis left parenthesis lambda minus 1 right parenthesis space equals space 0 end style
Hence eigen values are λ= 1, 4, 4
 
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Let eigen value equation is begin mathsize 14px style A space X space equals space lambda space X end style  and eigen vector begin mathsize 14px style X space equals space open square brackets table row cell x subscript 1 end cell row cell x subscript 2 end cell row cell x subscript 3 end cell end table close square brackets end style
Then we get the matrix equation for λ=1
 
begin mathsize 14px style open square brackets table row 3 1 1 row 1 3 cell negative 1 end cell row 1 cell negative 1 end cell 3 end table close square brackets open square brackets table row cell x subscript 1 end cell row cell x subscript 2 end cell row cell x subscript 3 end cell end table close square brackets space equals space 1 open square brackets table row cell x subscript 1 end cell row cell x subscript 2 end cell row cell x subscript 3 end cell end table close square brackets end style
 
From above matrix equation  , we get
 
begin mathsize 14px style 3 space x subscript 1 space plus space x subscript 2 space plus space x subscript 3 space equals space x subscript 1 end style    or   begin mathsize 14px style 2 space x subscript 1 space plus space x subscript 2 space plus space x subscript 3 space equals space 0 end style    ................................(1)
 
begin mathsize 14px style x subscript 1 space plus space 3 space x subscript 2 space minus space x subscript 3 space equals space x subscript 2 end style    or   begin mathsize 14px style space x subscript 1 space plus space 2 space x subscript 2 space minus space x subscript 3 space equals space 0 end style  .................................(2)
 
begin mathsize 14px style x subscript 1 space minus space x subscript 2 space plus 3 space x subscript 3 space equals space x subscript 3 end style    or   begin mathsize 14px style space x subscript 1 space minus space space x subscript 2 space plus space 2 x subscript 3 space equals space 0 end style  .................................(2)
 

Let x1 = 1
 
By adding eqn.(1) and (2) , we get  3 x1 + 3 x2 = 0   or  x2 = -1
 
If we substitute x1 = 1 , x2 = -1 in any one of above equations , we get x3 = -1
 
Eigen vector for eigen value λ =1 is begin mathsize 14px style open square brackets table row 1 row cell negative 1 end cell row cell negative 1 end cell end table close square brackets end style
 
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Let eigen value equation is begin mathsize 14px style A space X space equals space lambda space X end style  and eigen vector begin mathsize 14px style X space equals space open square brackets table row cell x subscript 1 end cell row cell x subscript 2 end cell row cell x subscript 3 end cell end table close square brackets end style
Then we get the matrix equation for λ=4
 
begin mathsize 14px style open square brackets table row 3 1 1 row 1 3 cell negative 1 end cell row 1 cell negative 1 end cell 3 end table close square brackets open square brackets table row cell x subscript 1 end cell row cell x subscript 2 end cell row cell x subscript 3 end cell end table close square brackets space equals space 4 open square brackets table row cell x subscript 1 end cell row cell x subscript 2 end cell row cell x subscript 3 end cell end table close square brackets end style
 
From above matrix equation  , we get
 
begin mathsize 14px style 3 space x subscript 1 space plus space x subscript 2 space plus space x subscript 3 space equals space 4 x subscript 1 end style    or   begin mathsize 14px style negative space x subscript 1 space plus space x subscript 2 space plus space x subscript 3 space equals space 0 end style    ................................(4)
 
begin mathsize 14px style x subscript 1 space plus space 3 space x subscript 2 space minus space x subscript 3 space equals space 4 space x subscript 2 end style    or   begin mathsize 14px style space x subscript 1 space space minus space x subscript 2 space minus space x subscript 3 space equals space 0 end style  .................................(5)
 
begin mathsize 14px style x subscript 1 space minus space x subscript 2 space plus 3 space x subscript 3 space equals space 4 space x subscript 3 end style    or   begin mathsize 14px style space x subscript 1 space minus space space x subscript 2 space space minus x subscript 3 space equals space 0 end style  .................................(6)
 
 
We see that eqn.(4) , (5) and (6) are not linearly independent , they are one and the same equation

Let x1 = 1 , X2 = 1  then x3 = 0
 
 
Eigen vector for eigen value λ =4 is begin mathsize 14px style open square brackets table row 1 row 1 row 0 end table close square brackets end style





Answered by Thiyagarajan K | 27 Dec, 2022, 12:07: AM

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