JEE Advanced 2020 Paper 1 · Mathematics · Matrices and Determinants · medium

Let P=beginpmatrix1&1&0\\0&1&1\\0&0&1endpmatrix and Q=[qij]=P^40+P^-40. The value of q13+q22 is

Single correct option

Let

P=\begin{pmatrix}1&1&0\\0&1&1\\0&0&1\end{pmatrix}

and

Q=[q_{ij}]=P^{40}+P^{-40}

. The value of

q_{13}+q_{22}

Show answer and solution

Answer: (B) 1602

Write

P=I+N

with

N=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}

, so

N^3=0

and

N^2

has a single non-zero entry

1

at position

(1,3)

. Since

I

and

N

commute,

P^{n}=I+nN+\binom n2N^2

and

P^{-n}=I-nN+\binom{n+1}{2}N^2

(from

(I+N)^{-1}=I-N+N^2

and the binomial series). Adding for

n=40

Q=2I+\left(\binom{40}{2}+\binom{41}{2}\right)N^2=2I+(780+820)N^2=2I+1600N^2

. Therefore

q_{13}=1600

q_{22}=2

and the sum is

1602

. Using

\binom{40}{2}

twice would give

1560

-type errors, and forgetting the

2I

gives

1600

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