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Prove that the following determinant is equal to $$2abc(a+b+c)^3$$

Using row column operations.

$$ \det \begin{pmatrix} (b+c)^2 & a^2 & a^2 \\ b^2 & (c+a)^2 & b^2 \\ c^2 & c^2 & (a+b)^2 \\ \end{pmatrix} $$ I tried using various row and column operations but without any success.

egreg
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user250085
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    At least present what the determinant calculation you have made. – NoChance Jul 08 '15 at 16:01
  • I tried using various row and column operations but without any success – user250085 Jul 08 '15 at 16:03
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    Why do you need to use row / column operations? Can't you just calculate the determinant directly from the definition? – Arthur Jul 08 '15 at 16:04
  • If we calculate it directly the problem becomes quite lengthy and complex @Arthur – user250085 Jul 08 '15 at 16:05
  • http://math.stackexchange.com/questions/954623/demonstrate-using-determinant-properties-that-the-determinant-of-matrix-a-is-e http://math.stackexchange.com/questions/1889822/how-to-solve-this-determinanat – Martin Sleziak Jan 10 '17 at 05:51

3 Answers3

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$$\det \begin{pmatrix} (b+c)^2 & a^2 &a^2\\ b^2 & (c+a)^2 & b^2\\ c^2&c^2&(a+b)^2 \end{pmatrix}$$

$$\det \begin{pmatrix} (b+c)^2-a^2 & 0 & a^2\\ 0 & (c+a)^2-b^2 & b^2\\ c^2-(a+b)^2 & c^2-(a+b)^2 & (a+b)^2 \end{pmatrix}$$ By $C_1-C_3$, $C_2-C_3$

$$\det \begin{pmatrix} (b+c+a)(b+c-a) & 0 & a^2\\ 0 & (c+a+b)(c+a-b) & b^2\\ (c+a+b)(c-a-b) & (c+a+b)(c-a-b) & (a+b)^2 \end{pmatrix}$$

$$(a+b+c)^2\det \begin{pmatrix} (b+c-a) & 0 & a^2\\ 0 & (c+a-b) & b^2\\ (c-a-b) & (c-a-b) & (a+b)^2 \end{pmatrix}$$ By Taking Common $(a+b+c)$ from each $C_1$ and $C_2$

$$(a+b+c)^2\det \begin{pmatrix} (b+c-a) & 0 & a^2\\ 0 & (c+a-b) & b^2\\ -2b & -2a & 2ab \end{pmatrix}$$ By $R_3-(R_1+R_2)$

$$(a+b+c)^2\det \begin{pmatrix} (b+c-a) & 0 & a^2\\ 0 & (c+a-b) & b(c+a)\\ (c-a-b) & (c-a-b) & 0 \end{pmatrix}$$ By $C_3+bC_2$

Now By expanding determinant we get

$$=2abc(a+b+c)^3$$

Syed Muhammad Asad
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$$\det \begin{pmatrix} (b+c)^2 & a^2 & a^2 \\ b^2 & (c+a)^2 & b^2 \\ c^2 & c^2 & (a+b)^2 \\ \end{pmatrix}$$ Replace $C_1$ by$ C_1−C_2$ and $ C_2$ by $C_2−C_3$, we have,

$$ \det\begin{pmatrix} (b+c)^2-a^2 & 0 & a^2 \\ b^2-(c+a)^2 & (c+a)^2-b^2 & b^2 \\ 0 & c^2-(a+b)^2 & (a+b)^2 \\ \end{pmatrix}$$

$$(a+b+c)^2\det \begin{pmatrix} b+c-a & 0 & a^2 \\ b-c-a & c+a-b & b^2 \\ 0 & c-a-b & (a+b)^2 \\ \end{pmatrix}$$

absolute friend
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    We don't usually post multiple answers to the same question unless they are different in some essential way. I'm not sure that we have a definite rule but I have misgivings about posting first a hint and then following up on it in a different post. Why did you not edit the hint? – Jyrki Lahtonen Jul 08 '15 at 16:50
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Hint: Replace $C_1$ by $C_1-C_2$ and $C_2$ by $C_2-C_3$

absolute friend
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