Solve the determinant.

Question

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.

Answer 1

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

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

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

Answer 2

Hint: Replace $C_1$ by $C_1-C_2$ and $C_2$ by $C_2-C_3$

Answer 3

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

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

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