Prove the following determinant

Question

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?

Answer 1

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}

Answer 2

In the case $3 \times 3$ you might want to use the rule of Sarrus.

Answer 3

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}

Answer 4

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.

Answer 5

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.