If $A$ is nil potent of index 2 , then for any positive integer $\mathrm{n}, \mathrm{A}(\mathrm{I} \

Question

If $A$ is nil potent of index 2 , then for any positive integer $\mathrm{n}, \mathrm{A}(\mathrm{I} \pm \mathrm{A})^{\mathrm{n}}=$\

Professor PC Thomas Classes · Answer

(b)\ $A(I+A)^{n}=A\left[I+n C_{1} A+n C_{2} n^{2}+. .+n C_{n} A^{n}ight]$\ $=\mathrm{A}+\mathrm{nC}_{1} \mathrm{~A}^{2}+\mathrm{nC}_{2} \mathrm{~A}^{3}+\ldots+\mathrm{A}^{\mathrm{n}+1}$\ $A(I-A)^{n}=\left(1^{n}-n C_{1} A+n C_{2}+A^{2}+\ldots-A^{n}ight)$ = A\