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I am self studying mathematics for Physics by reading book Mathematical methods in Physical Sciences. I am stuck at this problem for days:

Prove the following by appropriate manipulations using Fact 1 to 4; 
do not just evaluate the determinants.

| 1  a  bc |   | 1  a  a^2 |                        | 1  a  a^2 |
| 1  b  ac | = | 1  b  b^2 | = (c - a)(b - a)(c - b)| 0  1  b+a | 
| 1  c  ab |   | 1  c  c^2 |                        | 0  0   1  |

                             = (c - a)(b - a)(c - b)

I can evaluate the first determinant and obtain the result is (c-a)(b-a)(c-b) as above. Also, I can manipulate the second determinant to the third determinant and obtain the result as above. In fact, this can prove above equations.

However, I wonder there is any way to transform first determinant to the second and the third. And is there any way to transform the third determinant back to the first and the second? I have been spend a day to find a way but without success.

Can you suggest or hint me something to overcome this struggle?

Here is the four facts about determinant manipulation that above problem mentioned:

Fact 1: If each element of one row (or one column) of a determinant is multiplied by a number k, the value of the determinant is multiplied by k.

Fact 2: The value of a determinant is zero if

(a) all elements of one row (or column) are zero; or if

(b) two rows (or two columns) are identical; or if

(c) two rows (or two columns) are proportional.

Fact 3: If two rows (or two columns) of a determinant are interchanged, the value of the determinant changes sign.

Fact 4: The value of a determinant is unchanged if

(a) rows are written as columns and columns as rows; or if

(b) we add to each element of one row, k times the corresponding element of another row, where k is any number (and a similar statement for columns).

Thank you very much.

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Transform the first to the second:

Add $(a+b+c)$ times the second column to the third column to get the determinant below:

$ \left| \begin{array}& 1 & a & a^2 + ab + bc + ca \\ 1 & b & b^2 + ab + bc + ca \\ 1 & c & c^2 + ab + bc + ca \end{array} \right| $

Now subtract $(ab + bc + ca)$ times the first column from the third column. Voila.

Doing the exact reverse takes you from the second to the first determinant.

The idea above comes from reasoning that to get $a^2$ only in the $(3,1)$ slot, I must multiply the second column with $a$ and add it to the third.

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  • $\begingroup$ Wow, it's amazing :) How can you see it that way? $\endgroup$ – Linh Sep 28 '15 at 2:43
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To go from second to third ...

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

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

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

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

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