Final answer:
The given matrices are deemed nilpotent by showing that when raised to a certain positive integer power, they result in a zero matrix. This is demonstrated by matrix multiplication, which for the provided matrices, leads to a zero matrix when squared.
option c is the correct
Step-by-step explanation:
The question is about checking whether given matrices are nilpotent. A matrix is called nilpotent if "A" to the power of some positive integer "n" is the zero matrix. Let us take the given matrices and denote them as "A". Assuming a, b, and c are real numbers (elements of R), the given matrices are:
Matrix A = [0 a b]
Matrix B = [0 0]
Matrix C = [0 0 c]
Matrix D = [0 0]
Where each matrix is square, and Matrix D is essentially a one-dimensional zero matrix. To verify that these matrices are nilpotent, we need to show that for some positive integer "n", matrix "A" raised to the "n"th power equals the zero matrix. For the purpose of this question, we can see that raising any of these matrices to the second power (squaring them) will produce a matrix with all entries being zero, provided that the multiplication rule for matrices is followed. Matrix exponentiation by a natural number is basically multiple matrix multiplications.
Starting with Matrix A:
A2 = [0 a b] * [0 a b] = [0 0 0]
For Matrix B and Matrix C, ironically, they are already zero matrices.
So, all given matrices are nilpotent, as they all equal the zero matrix when raised to the second power.