If `A^2=8A` + kl where `A=[{:(1,0),(-1,7):}]`, then k is

Find the value of k so that A^2=8A+kI where A=[(1,0),(-1,7)].

The possible values of scalar k such that the matrix A^(-1)-kI is singular where A = [(1,0,2),(0,2,1),(1,0,0)] , are

If A=[(1, 0),(-1 ,7)] , find k such that A^2-8A+k I=O .

A=[{:(1,0,k),(2, 1,3),(k,0,1):}] is invertible for

If [(cos\ (2pi)/7,-sin\ (2pi)/7),(sin\ (2pi)/7,cos\ (2pi)/7)]^k=[(1,0),(0,1)], then the least positive integral value of k , is

For a matrix A of order 3xx3 where A=[(1,4,5),(k,8,8k-6),(1+k^2, 8k+4, 2k+21)] (A) rank of A=2 for k=-1 (B) rank of A=1 for k=-1 (C) rank of A=2 for k=2 (D) rank of A=1 for k=2

If A= [{:(,1,0,2),(,0,2,1),(,2,0,3):}] is a root of polynomial x^(3)-6x^(2)+7x+k=0 then the value of k is

If A= [[1 , 0],[ − 1, 7]] and I=[[1, 0],[ 0 ,1]] , then find k so that A^2=8A+k I

If A=[(3,-4),(1,-1)] prove that A^k=[(1+2k,-4k),(k,1-2k)] where k is any positive integer.