Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove that if the system of equations $x+2ay + az = 0$,$x+3by+bz = 0$ ,$x+4cy+cz = 0$ has a non-zero solution then a,b,c are in harmonic progression.

I am looking for a suitable approach for this problem.

share|cite|improve this question
First of all, ask yourself what it means for the matrix of coefficients that this system of equations has a non-zero solution. – Raskolnikov Dec 5 '10 at 14:32
up vote 6 down vote accepted

Write these equations in determinants form and use the following operations:

  • $C_{2} \to C_{2} -2C_{3}$

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

When these two operations are done, then the resulting determinant should 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 give $$ \frac{2}{b} = \frac{1}{a} + \frac{1}{c}$$

share|cite|improve this answer

OK, Chandru1 solved it, so I'm gonna explain the way I would have approached it. (Which is actually the same approach up to ordering my equations differently.)

1/ A homogeneous system of equations has a non-zero solution if the determinant is zero. Therefore look at this equation:

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

2/ Since we want to check for a harmonic progression, we need to consider the inverses of $a,b$ and $c$. If one of these coefficients is zero but not the others, there can't be a harmonic progression, but you can also show that in these cases, the system of equations can't have a non-zero solution, so they are ruled out anyway. If two or more of these coefficients are zero, the system is underdetermined and there is an infinity of solutions. So, it seems that there is a small loophole here in the question.

Now, assuming that all coefficients are different from zero, we can rewrite the determinant as

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

Subtracting the second row from the first and third 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 developed with respect to the third column to get as an end result:


share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.