If $$A^k = 0$$ ($$A$$ is nilpotent with index $$k$$ ), $${\left( {I - A} \right)^p} = I + A + {A^2} + ..... + {A^{k - 1}},$$ thus $$p$$ is,
A.
$$- 1$$
B.
$$ - 2$$
C.
$$\frac{1}{2}$$
D.
None of these
Answer :
$$- 1$$
Solution :
Let $$B = I + A + {A^2} + ..... + {A^{k - 1}}$$
Now multiply both sides by $$\left( {I - A} \right),$$ we get
$$\eqalign{
& B\left( {I - A} \right) = \left( {I + A + {A^2} + ..... + {A^{k - 1}}} \right)\left( {I - A} \right) \cr
& = I - A + A - {A^2} + {A^2} - {A^3} + ..... - {A^{k - 1}} + {A^{k - 1}} - {A^k} \cr
& = I - {A^k} = I,\,\,{\text{since, }}{A^k} = 0 \cr
& \Rightarrow B = {\left( {I - A} \right)^{ - 1}} \cr
& {\text{Hence, }}{\left( {I - A} \right)^{ - 1}} = I + A + {A^2} + ..... + {A^{k - 1}} \cr
& {\text{Thus, }}p = - 1. \cr} $$
Releted MCQ Question on Algebra >> Matrices and Determinants
Releted Question 1
Consider the set $$A$$ of all determinants of order 3 with entries 0 or 1 only. Let $$B$$ be the subset of $$A$$ consisting of all determinants with value 1. Let $$C$$ be the subset of $$A$$ consisting of all determinants with value $$- 1.$$ Then
A.
$$C$$ is empty
B.
$$B$$ has as many elements as $$C$$
C.
$$A = B \cup C$$
D.
$$B$$ has twice as many elements as elements as $$C$$
Let $$a, b, c$$ be the real numbers. Then following system of equations in $$x, y$$ and $$z$$
$$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - \frac{{{z^2}}}{{{c^2}}} = 1,$$ $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1,$$ $$ - \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1$$ has