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If A=[1101]A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}, then the determinant of A+A2+A3++A13A + A^2 + A^3 + \cdots + A^{13} is

Solution

Correct Option: 3

For this upper triangular matrix with 11's on the diagonal:

An=[1n01]A^n = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}

This can be checked by induction: multiplying AnA^n by AA increments the top-right entry by 11.


Add A1A^1 through A13A^{13}. The top-right entries are 1,2,3,,131, 2, 3, \ldots, 13:

Sum of top-right entries =1+2++13=13142=91= 1 + 2 + \cdots + 13 = \dfrac{13 \cdot 14}{2} = 91

Diagonal entries: each sum is 1+1++11 + 1 + \cdots + 1 (1313 times) =13= 13


The summed matrix is:

[1391013]\begin{bmatrix} 13 & 91 \\ 0 & 13 \end{bmatrix}

Determinant =13×130×91=169= 13 \times 13 - 0 \times 91 = 169

Answer =169= 169

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