If $M_k = \begin{bmatrix} k & k-1 \ k-1 & k \end{bmatrix}$ where k is a natural number, then what is $|M_1| + |M_2| + |M_3| + \dots + |M_{50}|$ equal to ?

Question

Prepp Team · Accepted Answer

Matrix Determinant Calculation

The matrix is given by $M_k = \begin{bmatrix} k & k-1 \\ k-1 & k \end{bmatrix}$. The determinant of a 2x2 matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is $ad - bc$. Therefore, the determinant of $M_k$, denoted as $|M_k|$, is calculated as: $ |M_k| = (k \times k) - ((k-1) \times (k-1)) $ $ |M_k| = k^2 - (k^2 - 2k + 1) $ $ |M_k| = k^2 - k^2 + 2k - 1 $ $ |M_k| = 2k - 1 $

Summation of Determinants

The question asks for the sum $S = |M_1| + |M_2| + |M_3| + \dots + |M_{50}|$. Substituting the expression for $|M_k|$, we get: $ S = \sum_{k=1}^{50} |M_k| = \sum_{k=1}^{50} (2k - 1) $

Applying Summation Formula

The expression $\sum_{k=1}^{n} (2k - 1)$ represents the sum of the first $n$ odd natural numbers, which is known to be equal to $n^2$. In this case, $n = 50$.