First slide

Introduction to Determinants

Question

$If a, b, c, d are in A.P and \int_{0}^{2} f (x) d x = - 4 where f (x) = |\begin{matrix} x + a & x + b & x + a - c \\ x + b & x + c & x - 1 \\ x + c & x + d & x - b + d \end{matrix}|,$ then the common difference of the A.P may be

Difficult

A

1
B

-1
C

2
D

-2

Solution

$\begin{matrix} Since a, b, c, d are in arithmatic progression \\ b - a = c - b = d - c = k \\ f (x) = |\begin{matrix} x + a & x + b & x + a - c \\ x + b & x + c & x - 1 \\ x + c & x + d & x - b + d \end{matrix}| \\ \cdot R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{2} \\ = |\begin{matrix} x + a & x + b & x + a - c \\ k & k & - 1 + 2 k \\ k & k & 1 + 2 k \end{matrix}| \cdot C_{2} \to C_{2} - C_{1} \\ = |\begin{matrix} x + a & k & x + a - c \\ k & 0 & - 1 + 2 k \\ k & 0 & 1 + 2 k \end{matrix}| \\ = - k (k + 2 k^{2} + k - 2 k^{2}) \\ = - 2 k^{2} \\ \int_{0}^{2} - 2 k^{2} d x = - 4 \\ 4 k^{2} = 4 \\ k = \pm 1 \end{matrix}$

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