First slide

Introduction to Determinants

Question

If $a_{1}, a_{2}, a_{3}, \dots, a_{12}$ are in A.P. and $Δ_{1} = |\begin{matrix} a_{1} a_{5} & a_{1} & a_{2} \\ a_{2} a_{6} & a_{2} & a_{3} \\ a_{3} a_{7} & a_{3} & a_{4} \end{matrix}| Δ_{2} = |\begin{matrix} a_{2} a_{10} & a_{2} & a_{3} \\ a_{3} a_{11} & a_{3} & a_{4} \\ a_{3} a_{12} & a_{4} & a_{5} \end{matrix}|$ then $Δ_{1} : Δ_{2} =$ ____________ .

Moderate

Solution

$Δ_{1} = |\begin{matrix} a_{1}^{2} + 4 a_{1} d & a_{1} & d \\ a_{2}^{2} + 4 a_{2} d & a_{2} & d \\ a_{3}^{2} + 4 a_{3} d & a_{3} & d \end{matrix}|, [C_{3} \to C_{3} - C_{2}]$
where d is the common difference of A.P.
$\begin{array}{l} = d |\begin{array}{lll} a_{1}^{2} & a_{1} & 1 \\ a_{2}^{2} & a_{2} & 1 \\ a_{3}^{2} & a_{3} & 1 \end{array}| + 4 d |\begin{array}{lll} a_{1} & a_{1} & d \\ a_{2} & a_{2} & d \\ a_{3} & a_{3} & d \end{array}| \\ = d (a_{1} - a_{2}) (a_{2} - a_{3}) (a_{3} - a_{1}) = - 2 d^{4} \end{array}$
Similarly, $Δ_{2} = - 2 d^{4}$

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