Question

let \(A=\begin{bmatrix} 1 & \frac{1}{51} \\ 0& 1 \end{bmatrix}\) if \(B=\begin{bmatrix} 1 &2\\ -1& -1 \end{bmatrix} A=\begin{bmatrix} -1 &2\\ 1& 1 \end{bmatrix}\) then the sum of all the elements of the matrix \(∑ ^{50}_{n=1}\) Bⁿ is Equal to 

• A. 50
• B. 75
• C. 100
• D. 125

Hint:

When working with matrix multiplication and powers of matrices, look for pat terns and simplifications, especially for special types of matrices like upper or lower triangular matrices. Remember the formula for the sum of the first n integers: n(n+1)

Answer 1

The Correct Option is C

Let: \[ C = \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix}, \quad D = \begin{bmatrix} -1 & -2 \\ 1 & 1 \end{bmatrix}. \]

Compute: \[ DC = \begin{bmatrix} -1 & -2 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I. \] Thus, \(D\) is the inverse of \(C\).

Now, define: \[ B = CAD. \] The general expression for \(B^n\) is: \[ B^n = (CAD)(CAD)(CAD) \cdots (CAD) \, (\text{n-times}). \] Using the properties of matrices: \[ B^n = CA^nD \quad \text{... (1)}. \]

Compute powers of \(A\): \[ A^2 = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}. \] \[ A^3 = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}. \] 
Similarly: \[ A^n = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}. \]

Now substitute \(A^n\) into \(B^n\): \[ B^n = \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & -2 \\ 1 & 1 \end{bmatrix}. \] Simplify: \[ B^n = \begin{bmatrix} 1 & n \\ -1 & -n \end{bmatrix} \begin{bmatrix} -1 & -2 \\ 1 & 1 \end{bmatrix}. \] Expanding: \[ B^n = \begin{bmatrix} n+1 & 2n+2 \\ -n-1 & -2n-1 \end{bmatrix}. \]

Now, compute the sum: \[ \sum_{n=1}^{50} B^n = \begin{bmatrix} \sum_{n=1}^{50} (n+1) & \sum_{n=1}^{50} (2n+2) \\ \sum_{n=1}^{50} (-n-1) & \sum_{n=1}^{50} (-2n-1) \end{bmatrix}. \] Simplifying each term: - \(\sum_{n=1}^{50} (n+1) = 25 + 50 = 75,\) - \(\sum_{n=1}^{50} (2n+2) = 100 + 50 = 150,\) - \(\sum_{n=1}^{50} (-n-1) = -25 - 50 = -75,\) - \(\sum_{n=1}^{50} (-2n-1) = -50 - 50 = -100.\)

The sum of all elements in the resulting matrix is: \[ 75 + 150 - 75 - 100 = 100. \]

Final Answer: The sum of the elements is 100.