Suppose A and B are non-zero matrices such that A^2=0 and B^3=0(But B^2 does not =0) find in simplst form:
a)(A+B)^2
b)(A+B)^3
c)(A+B)^4
d)(A+B)^5
e)(A+B)^6
To simplify each expression, we will expand them step by step using the binomial theorem.
a) (A+B)^2
Using the binomial theorem, we have:
(A+B)^2 = A^2 + 2AB + B^2
Since A^2 = 0 and B^2 ≠ 0, the expression simplifies to:
(A+B)^2 = 2AB
b) (A+B)^3
Using the binomial theorem, we have:
(A+B)^3 = A^3 + 3A^2B + 3AB^2 + B^3
Since A^2 = 0 and B^3 = 0, the expression further simplifies to:
(A+B)^3 = 3AB^2
c) (A+B)^4
Using the binomial theorem, we have:
(A+B)^4 = A^4 + 4A^3B + 6A^2B^2 + 4AB^3 + B^4
Since A^2 = 0, A^3 = A^2 * A = 0 * A = 0. Moreover, B^3 = 0 and B^4 ≠ 0. Therefore, the expression simplifies to:
(A+B)^4 = 6A^2B^2
d) (A+B)^5
Using the binomial theorem, we have:
(A+B)^5 = A^5 + 5A^4B + 10A^3B^2 + 10A^2B^3 + 5AB^4 + B^5
Following the previous observations, the expression simplifies to:
(A+B)^5 = 10A^2B^3
e) (A+B)^6
Using the binomial theorem, we have:
(A+B)^6 = A^6 + 6A^5B + 15A^4B^2 + 20A^3B^3 + 15A^2B^4 + 6AB^5 + B^6
Similar to the previous steps, the expression simplifies to:
(A+B)^6 = 20A^3B^3