Prove in harmonic development

Write these formulas in components create and also make use of the adhering to procedures :

• $C_{2} \to C_{2} -2C_{3}$

• $R_{3} \to R_{3} - R_{2}$ and also $R_{2} \to R_{2}-R_{1}$

When these 2 procedures are done, after that the resulting component needs to be :

$$\left|\begin{array}{lll} 1 & 0 & a \\ 0 & b & b-a \\ 0 & 2c-b & c-b \end{array}\right| = b(c-b) - (b-a)(2c-b) =0 $$

which will certainly offer $$ \frac{2}{b} = \frac{1}{a} + \frac{1}{c}$$

OK, Chandru1 addressed it, so I'm gon na clarify the means I would certainly have approached it. (Which is in fact the very same strategy approximately getting my formulas in different ways.)

1/ An uniform system of formulas has a non - absolutely no remedy if the component is absolutely no. Consequently consider this formula :

$$\left|\begin{array}{ccc} 1 & 2a & a \\ 1 & 3b & b \\ 1 & 4c & c \end{array}\right| = 0$$

2/ Since we intend to look for a harmonic development, we require to take into consideration the inverses of $a,b$ and also $c$. If among these coefficients is absolutely no yet not the others, there can not be a harmonic development, yet you can additionally show that in these instances, the system of formulas can not have a non - absolutely no remedy, so they are dismissed anyhow. If 2 or even more of these coefficients are absolutely no, the system is underdetermined and also there is an infinity of remedies. So, it appears that there is a tiny technicality below in the inquiry.

Currently, thinking that all coefficients are various from absolutely no, we can revise the component as

$$\left|\begin{array}{ccc} 1/a & 2 & 1 \\ 1/b & 3 & 1 \\ 1/c & 4 & 1 \end{array}\right| = 0$$

Subtracting the 2nd row from the first and also 3rd row, you get

$$\left|\begin{array}{ccc} 1/a-1/b & -1 & 0 \\ 1/b & 3 & 1 \\ 1/c-1/b & 1 & 0 \end{array}\right| = 0$$

And this can be created relative to the 3rd column to get as an outcome :

$$\frac{1}{a}+\frac{1}{c}=\frac{2}{b}$$