यदि ` A = [ (2,0,-1),(5,1,0),(0,1,3)]` और ` A^(-1) = A^(2) lambda A + mu I`, तब
`lambda` का मान होगा ।

Options

A. `-5`
B. 6
C. `-6`
D. `-7`

Answer

C

Explanation

दिया है , ` A^(-1) = A^(2) + lambda A + mu I`
` rArr I = A^(3) + lambda A^(2) + mu A`
तथा ` A = [{:(2,0,-1),(5,1,0),(0,1,3):}]`
` :. ^(2) = A * A = [{:(2,0,-1),(5,1,0),(0,1,3):}][{:(2,0,-1),(5,1,0),(0,1,3):}]`
` = [{:(4,-1,-5),(15,1,-5),(5,4,9):}]`
और ` A^(3) = A^(2) * A`
` [{:(4,-1,-5),(15,1,-5),(5,4,9):}] [{:(2,0,-1),(5,1,0),(0,1,3):}]`
` [{:(3,-6,-19),(35,-4,-30),(30,13,22):}]`
इसलिए ` A^(3) - 6A^(2) + 11A`
` [{:(3,-6,-19),(35,-4,-30),(30,13,22):}]-6 [{:(4,-1,-5),(15,1,-5),(5,4,9):}]`
` + 11[{:(2,0,-1),(5,1,0),(0,1,3):}]`
` [{:(1,0,0),(0,1,0),(0,0,1):}]=I `
`rArr A^(3) - 6A^(2) + 11A = I`
` rArr A^(-1) I = A^(-1) A^(3) - 6A^(2) + 11A^(-1) A`
` rArr A^(-1) = A^(2) -6A + 11I " "` ...(i)
समी (i ) की `A(-1) = A^(2) + lambda A + mu I` से तुलना
करने पर , ` lambda= -6 , mu = 11`