Prove the following determinant Prove the following:
$$\left|
\begin{matrix}
(b+c)&a&a \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|=4abc$$
My Attempt:
$$\left|
\begin{matrix}
(b+c)&a&a \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
Using $R_1\to R_1+R_2+R_3$
$$\left |
\begin{matrix}
2(b+c)&2(a+c)&2(a+b) \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
Taking common $2$ from $R_1$
 $$2\left|
\begin{matrix}
(b+c)&(a+c)&(a+b) \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
How do I proceed further?
 A: You can use the rule of Sarrus in this case:
$$\left|
\begin{matrix}
(b+c)&a&a \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
\begin{align}
&=(b+c)(c+a)(a+b) +abc +abc 
- c(c+a)a - cb(b+c) -ab(a+b)\\
&=a^2 b + a^2 c + a b^2 + 2 a b c + a c^2 + b^2 c + b c^2 +2abc -c^2a-a^2c-c^2b-b^2c -a^2b-b^2a\\
&=4abc.
\end{align}
A: Let me give a solution without using the rule of Sarrus for fun. $\require{cancel}$
\begin{align}
& \begin{vmatrix}
b+c&a&a \\
b&c+a&b \\
c&c&a+b \\
\end{vmatrix} \\
&= \frac{1}{abc}\, \begin{vmatrix}
a(b+c)&ab&ca \\
ab&b(c+a)&bc \\
ca&bc&c(a+b) \\
\end{vmatrix} \\
&= \frac{1}{abc}\, \begin{vmatrix}
0&-2bc&-2bc \\
ab&b(c+a)&bc \\
ca&bc&c(a+b) \\
\end{vmatrix} \tag{$R_1 \to R_1 - R_2 - R_3$} \\
&= \frac{1}{abc}\, \begin{vmatrix}
0&-2bc&0 \\
ab&b(c+a)&-2ab \\
ca&bc&0 \\
\end{vmatrix} \tag{$C_3 \to C_3 - C_1 - C_2$} \\
&= \frac{1}{a\cancel{bc}}\, [-(-2\cancel{bc})]\, \begin{vmatrix}
ab&-2ab \\
ca&0 \\
\end{vmatrix} \\
&= \frac{2}{\cancel{a}} \, [-(-2ab)(c\cancel{a})] \\
&= 4abc
\end{align}
A: The determinant of the given matrix is a symmetric polynomial with degree $3$ in the variables $a,b,c$. By the Laplace expansion the determinant equals zero if $a,b$ or $c$ equal zero, hence the determinant is a constant multiple of $abc$. In order to find which multiple, it is enough to evaluate the determinant at $(a,b,c)=(1,2,3)$, for instance:
$$\det\begin{pmatrix}5 & 1 & 1\\ 2 & 4 & 2 \\ 3 & 3 & 3 \end{pmatrix}=6\det\begin{pmatrix}5 & 1 & 1\\ 1 & 2 & 1 \\ 1 & 1 & 1 \end{pmatrix}=6\det\begin{pmatrix}4 & 0 & 0\\ 0 & 1 & 0 \\ 1 & 1 & 1 \end{pmatrix}=24.$$
This leads to $\det M(a,b,c) = 4abc$ as wanted.
A: Yet another argument: It is easy to check that
\begin{equation}
\left(\begin{array}{ccc}
b+c & a & a \\ b & c+a & b \\ c & c & a+b
\end{array}\right)
= L N ,
\end{equation}
where the matrices $L$ and $N$ are defined by
\begin{equation}
L = \left(\begin{array}{ccc}
0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0
\end{array}\right)
\qquad \text{and} \qquad
N = \left(\begin{array}{ccc}
0 & c & b \\ c & 0 & a \\ b & a & 0
\end{array}\right)
.
\end{equation}
Since the determinant is multiplicative, it thus suffices to prove that $\det L = 2$ and $\det N = 2abc$. This is easy.
A: In the case $3 \times 3$ you might want to use the rule of Sarrus.
