Let  f (x) = x2  – 5 x + 6  and  A =            1 1 0 2 1 3 2 0 1  , then  f (A) is equal to : (A)                   5 4 4 1 1 10 1 1 3 (B)                   5 4 3 1 1 10 1 1 3 (C)                  5 4 3 0 1 10 0 1 3 (D)      

Asked by vikaskumawat037 | 9th Sep, 2020, 09:58: PM

Expert Answer:

Given: f(x) = x2 - 5x + 6 and A = matrix [2 0 1 // 2 1 3 // 1 -1 0]
straight A equals open square brackets table row 2 0 1 row 2 1 3 row 1 cell negative 1 end cell 0 end table close square brackets
straight f left parenthesis straight A right parenthesis equals straight A squared minus 5 straight A plus 6
straight A squared equals open square brackets table row 2 0 1 row 2 1 3 row 1 cell negative 1 end cell 0 end table close square brackets open square brackets table row 2 0 1 row 2 1 3 row 1 cell negative 1 end cell 0 end table close square brackets equals open square brackets table row 5 cell negative 1 end cell 2 row 9 cell negative 2 end cell 5 row 0 cell negative 1 end cell cell negative 2 end cell end table close square brackets
straight f left parenthesis straight A right parenthesis equals open square brackets table row 5 cell negative 1 end cell 2 row 9 cell negative 2 end cell 5 row 0 cell negative 1 end cell cell negative 2 end cell end table close square brackets minus open square brackets table row 10 0 5 row 10 5 15 row 5 cell negative 5 end cell 0 end table close square brackets plus open square brackets table row 6 0 0 row 0 6 0 row 0 0 6 end table close square brackets
space space space space space space space equals open square brackets table row 1 cell negative 1 end cell cell negative 3 end cell row cell negative 1 end cell cell negative 1 end cell cell negative 10 end cell row cell negative 5 end cell 4 4 end table close square brackets

Answered by Renu Varma | 10th Sep, 2020, 11:51: AM