Let A=[\beginarraycc1 & \frac151 0 & 1\endarray] . If B=[\beginarraycc1 & 2 -1 & -1\endarray] A[\beginarraycc-1 & -2 1 & 1\endarray] , then the sum of all the elements of the matrix \sum_n=1^50 B^n is equal to

Matrices and Determinants PYQ — JEE Main 2023 · 12 Apr · Shift 1 · Q24 with Solution

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Correct answer: C

Let $P=\begin{pmatrix}1&2\\-1&-1\end{pmatrix},\qquad Q=\begin{pmatrix}-1&-2\\1&1\end{pmatrix}.$

We check that

\begin{pmatrix}1&0\\0&1\end{pmatrix}=I.$$ So $Q=P^{-1}$. Hence $$B=PAP^{-1}.$$ Therefore, $$B^n=PA^nP^{-1}.$$ Thus, $$\sum_{n=1}^{50} B^n = P\left(\sum_{n=1}^{50} A^n\right)P^{-1}.$$ --- 2. **Compute $A^n$** Given $$A=\begin{pmatrix}1&\frac{1}{51}\\0&1\end{pmatrix}=I+N,$$ where $$N=\begin{pmatrix}0&\frac{1}{51}\\0&0\end{pmatrix},\qquad N^2=0.$$ So, $$A^n=(I+N)^n=I+nN=\begin{pmatrix}1&\frac{n}{51}\\0&1\end{pmatrix}.$$ Hence, $$\sum_{n=1}^{50} A^n= \begin{pmatrix} \sum_{n=1}^{50}1 & \sum_{n=1}^{50}\frac{n}{51}\\ 0 & \sum_{n=1}^{50}1 \end{pmatrix}.

Now, $\sum_{n=1}^{50}1=50,$ and $\sum_{n=1}^{50}\frac{n}{51}=\frac{1}{51}\cdot \frac{50\cdot 51}{2}=25.$

Therefore, $S:=\sum_{n=1}^{50} A^n=\begin{pmatrix}50&25\\0&50\end{pmatrix}.$

First compute $PS$ :

=\begin{pmatrix}50&125\\-50&-75\end{pmatrix}.$$ Now multiply by $P^{-1}=Q$: $$PSP^{-1}=\begin{pmatrix}50&125\\-50&-75\end{pmatrix}\begin{pmatrix}-1&-2\\1&1\end{pmatrix}.

Compute entries:

\begin{aligned} (1,1)&=50(-1)+125(1)=75,\\ (1,2)&=50(-2)+125(1)=25,\\ (2,1)&=(-50)(-1)+(-75)(1)=-25,\\ (2,2)&=(-50)(-2)+(-75)(1)=25. \end{aligned}

So, $\sum_{n=1}^{50} B^n=\begin{pmatrix}75&25\\-25&25\end{pmatrix}.$

Sum of all entries: $75+25-25+25=100.$

The required sum is $\boxed{100}.$ So the correct option is C.

Stored correct answer: C

Derived answer: C

They agree.