JEE Advanced 2023 Paper 2 · Mathematics · Matrices and Determinants · easy

Let M=beginpmatrix1&2\\0&1endpmatrix and let N=M^50-I. The sum of all entries of N is

Numerical answer

Let

M=\begin{pmatrix}1&2\\0&1\end{pmatrix}

and let

N=M^{50}-I

. The sum of all entries of

N

Show answer and solution

Answer: 100

M=I+E

with

E=\begin{pmatrix}0&2\\0&0\end{pmatrix}

E^2=0

, so

M^{n}=I+nE=\begin{pmatrix}1&2n\\0&1\end{pmatrix}

. For

n=50

N=M^{50}-I=\begin{pmatrix}0&100\\0&0\end{pmatrix}

and the sum of entries is

100

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