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Monotonically Increasing Functions

Prove ah + ...

Question

Prove $∣ ∣ ∣ ∣ \begin{matrix} a h + b g & g & a b + c h b f + b a & f & h b + b c a f + b c & d & b g + f c \end{matrix} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \begin{matrix} a h + b g & a & h b f + b a & h & b a f + b c & g & f \end{matrix} ∣ ∣ ∣ ∣$

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Solution

We have
$L . H . S = ∣ ∣ ∣ ∣ \begin{matrix} a h + b g & g & a b + c h b f + b a & f & h b + b c a f + b c & c & b g + f c \end{matrix} ∣ ∣ ∣ ∣$

$= b ∣ ∣ ∣ ∣ \begin{matrix} a h + b g & g & a b f + b a & f & h a f + b c & c & b \end{matrix} ∣ ∣ ∣ ∣ + c ∣ ∣ ∣ ∣ \begin{matrix} a h + b g & g & h b f + b a & f & b a f + c & c & f \end{matrix} ∣ ∣ ∣ ∣$

In second determinant applying $C_{1} \to C_{1} - b C_{2} - a C_{3}$
$= ∣ ∣ ∣ ∣ \begin{matrix} a h + b g & b g & a b f + b a & b f & h a f + b c & b c & g \end{matrix} ∣ ∣ ∣ ∣ + c ∣ ∣ ∣ ∣ \begin{matrix} 0 & g & h 0 & f & b 0 & c & f \end{matrix} ∣ ∣ ∣ ∣$
In first determinant applying $C_{2} \to C_{2} - C_{1}$ then
$= ∣ ∣ ∣ ∣ \begin{matrix} a h + b g & - a g & a b f + b a & - b a & h a f + b c & - a f & g \end{matrix} ∣ ∣ ∣ ∣ + 0$ $= ∣ ∣ ∣ ∣ \begin{matrix} a b + b g & a & h b f + b a & h & b a f + b c & g & f \end{matrix} ∣ ∣ ∣ ∣$ $= R . H . S .$

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