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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Determinants

If $A^k=0$ for some value of k.$(I-A)^P=I+A+A^2+....+A^{k-1}$ thus P is

$(a)\;-1\qquad(b)\;-2\qquad(c)\;3\qquad(d)\;None\;of\;the\;above$

1 Answer

Let $B=I+A+A^2+.....+A^{k-1}$
Post multiply both sides by $(I-A)$ so that
$B(I-A)=(I+A+A^2+.....+A^{k-1})(I-A)$
$\quad\qquad\;\;=I-A+A-A^2+A^2-A^3+....A^{k-1}+A^{k-1}-A^k$
$\quad\qquad\;\;=I-A^k=I$
Since $A^k=0$
$\Rightarrow B=(I-A)^{-1}$
Hence $(I-A)^{-1}=I+A+A^2+......+A^{k-1}$
Thus P=-1
Hence (a) is the correct answer.
answered Nov 25, 2013 by sreemathi.v
 

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