A= ⎡⎢⎣100010001⎤⎥⎦which of the following is true?
A is a periodic matrix with period 1
A is an idempotent matrix
A=⎡⎢⎣100010001⎤⎥⎦is identity matrix and we know that I=I2=I3 and so on.
I=I2=Ik+1
Now for Ik+1=I we get k =0 which is not a positive integer.
Now to get the least positive integer we will equate Ik+1 to next higher power i.e. I2
So, IK+1=I2
i.e., k+1=2
So k=1.
Since K=1 is the least integer satisfying the condition. So above matrix is a periodic matrix with period1. So option A is correct.
Also we know that for Ak+1=A, if k=1 satisfies the mentioned condition then the matrix is also called as idempotent matrix. So the above matrix is an idempotent matrix. So option (b) is also correct.