A=[[0,3],[2,0]]" and "A^(-1)=lambda(ad)(A)" ),then "lambda=

If A=[[0, 3], [2, 0]] and A^(-1)=lambda(adjA) , then lambda=

A= [{:(0,3),(2,0):}] and A^(-1)=lambda (adj, A) then lambda is equal to

A= [{:(0,3),(2,0):}] and A^(-1)=lambda (adj, A) then lambda is equal to

Let A=|{:(,-2,7,sqrt3),(,0,0,-2),(,0,2,0):}| and A^(4)=lambda,I,"then"lambda "is"

Let A=|{:(,-2,7,sqrt3),(,0,0,-2),(,0,2,0):}| and A^(4)=lambda,I,"then"lambda "is"

If A= [[2,3,5],[1,2,1],[0,1,7]] and A^(3)-lambda A^(2)+28A-10=0 , then the value of lambda is

[" 2.Let matrix "A=[[1,y,4],[2,2,z]]" ; if "xyz=2 lambda" and "8x+4y+3z=lambda+28," then "(adj A)" A equals: "],[[" (A) "[[lambda+1,0,0],[0,lambda+1,0],[0,0,lambda+1]]," (B) "[[lambda,0,0],[0,lambda,0],[0,0,lambda]]],[" (C) "[[lambda^(2),0,0],[0,lambda^(2),0],[0,0,lambda^(2)]]," (D) "[[lambda+2,0,0],[0,lambda+2,0],[0,0,lambda+2]]]]

lambda ^ (3) -6 lambda ^ (2) -15 lambda-8 = 0

If the matrix [[0,1,-2],[-1,0,3],[lambda,-3,0]] is singular,then lambda=