JEE Advanced 2016 Paper 2 · Mathematics · Matrices and Determinants · medium

Let A=beginpmatrix3&-1\\4&-1endpmatrix and B=A+A^2+…+A^10. The sum of all four entries of B is

Single correct option

Let

A=\begin{pmatrix}3&-1\\4&-1\end{pmatrix}

and

B=A+A^{2}+\dots+A^{10}

. The sum of all four entries of

B

Show answer and solution

Answer: (D) $185$

Let

N=A-I=\begin{pmatrix}2&-1\\4&-2\end{pmatrix}

. Then

N^{2}=\begin{pmatrix}4-4&-2+2\\8-8&-4+4\end{pmatrix}=O

, so

N

is nilpotent and, since

I

commutes with

N

, the binomial theorem gives

A^{k}=(I+N)^{k}=I+kN=\begin{pmatrix}1+2k&-k\\4k&1-2k\end{pmatrix}

. The sum of entries of

A^{k}

(1+2k)-k+4k+(1-2k)=2+3k

. Summing over

k=1,\dots,10

\sum(2+3k)=20+3\cdot55=185

. A direct computation of

B

gives

\begin{pmatrix}120&-55\\220&-100\end{pmatrix}

whose entries add up to

185

. Option (A) uses

\sum_{k=1}^{10}k=45

-type slips, and (C) includes the

k=0

term incorrectly.

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