Prove $\left|\begin{smallmatrix}a&b-c&c+b\\a+c&b&c-a\\a-b&b+a&c\end{smallmatrix}\right|=(a+b+c)(a^2+b^2+c^2)$ without direct evaluation

Question

Prove that $\begin{vmatrix}a&b-c&c+b \\ a+c&b&c-a\\ a-b&b+a&c \end{vmatrix}=(a+b+c)(a^2+b^2+c^2)$ without direct evaluation.

My attempt is as follows:-

$$R_1\rightarrow R_1-R_2$$ $$R_1\rightarrow R_2-R_3$$

$$\begin{vmatrix}-c&-c&b+a \\ c+b&-a&-a\\ a-b&b+a&c \end{vmatrix}$$

$$C_1\rightarrow C_1-C_2$$ $$C_2\rightarrow C_2-C_3$$

$$\begin{vmatrix}0&-c-b-a&b+a \\ a+b+c&0&-a\\ -2b&b+a-c&c \end{vmatrix}$$

$$R_3\rightarrow R_3+R_1$$

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

$$R_1\rightarrow R_1+R_2$$

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

Looks like we can't reduce it anymore, so I expanded:

Expanding with respect to second row as there is one zero in that

$$-(a+b+c)(-(a+b+c)^2+2bc)+a(-2c(a+b+c)-2b(a+b+c)$$

$$(a+b+c)(a^2+b^2+c^2+2ab+2bc+2ca-2bc)+(a+b+c)(-2ca-2ba)$$ $$(a+b+c)(a^2+b^2+c^2)$$

I got the result at the end but is it the good way? Honestly I am not satisified with my way, is there any set of transformations which would be better?

Please help me with your suggestions?

Answer 1

If $a=0$, then the determinant of your matrix is$$b^3+b^2 c+b c^2+c^3=(b+c)(b^2+c^2).$$Otherwise\begin{align}\begin{vmatrix}a & b-c & b+c \\ a+c & b & c-a \\ a-b & a+b & c\end{vmatrix}&=a\begin{vmatrix}1 & \frac{b-c}a & \frac{b+c}a \\ a+c & b & c-a \\ a-b & a+b & c\end{vmatrix}\\&=a\begin{vmatrix}1 & \frac{b-c}{a} & \frac{b+c}a \\ 0 & \frac{a c-b c+c^2}a & \frac{-a^2-a b-b c-c^2}a \\ 0 & \frac{a^2+a c+b^2-b c}a & \frac{-a b+b^2+b c}a\end{vmatrix}\\&=a^3+b^3+c^3+a^2b+a^2c+ab^2+b^2c+ac^2+bc^2\\&=(a+b+c)(a^2+b^2+c^2).\end{align}

Answer 2

I got some very good way to solve this problem:

Multiply first column by $a$

$$\dfrac{1}{a}\begin{vmatrix} a^2&b-c&b+c\\ a^2+ac&b&c-a\\ a^2-ab&a+b&c \end{vmatrix}$$

$$C_1\rightarrow C_1\rightarrow bC_2+cC_3$$

$$\dfrac{1}{a}\begin{vmatrix} a^2+b^2-bc+bc+c^2&b-c&b+c\\ a^2+ac+b^2+c^2-ac&b&c-a\\ a^2-ab+ab+b^2+c^2&a+b&c \end{vmatrix}$$

$$\dfrac{1}{a}\begin{vmatrix} a^2+b^2+c^2&b-c&b+c\\ a^2+b^2+c^2&b&c-a\\ a^2+b^2+c^2&a+b&c \end{vmatrix}$$

$$\dfrac{a^2+b^2+c^2}{a}\begin{vmatrix} 1&b-c&b+c\\ 1&b&c-a\\ 1&a+b&c \end{vmatrix}$$ $$R_1\rightarrow R_1-R_2$$ $$R_2\rightarrow R_2-R_3$$

$$\dfrac{a^2+b^2+c^2}{a}\begin{vmatrix} 0&-c&b+a\\ 0&-a&-a\\ 1&a+b&c \end{vmatrix}$$

$$\dfrac{a^2+b^2+c^2}{a}(ca+ab+a^2)$$

$$(a^2+b^2+c^2)(a+b+c)$$

Hope it will be useful to somebody.

Answer 3

By Gaussian triangularization subtracting a multiple of the first row and from 2,3 and then 2 from 3 yields

for $a \neq 0 ,\neq 0, a-b+c \ne 0 $

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

For $a=0$, rotate b or c to position 1,1.

Even the case $$(a-b+c) =0, $$ to be handled by rotations too, yields determinant

$$a= b-c; \det(m)= 4 b (b^2 - b c + c^2)$$

that has the required form

$$a= b-c; (a+b+c)(a^2+b^2+c^2) = 4 b \left(b^2-b c+c^2\right) $$

Answer 4

Let $$ p=\pmatrix{a\\ b\\ c}=ku, \quad e=\pmatrix{1\\ 1\\ 1} \quad\text{and}\quad [p]_\times=\pmatrix{0&-c&b\\ c&0&-a\\ -b&a&0}, $$ where $k=\sqrt{a^2+b^2+c^2}$, $u$ is a unit vector and $[p]_\times$ denotes the cross product matrix such that $[p]_\times q=p\times q$ for every vector $q$. The matrix in the identity is equal to $$ A=pe^T+[p]_\times=kue^T+k[u]_\times\ . $$ It is the matrix representation of the linear operator $$ Tx=\langle e,x\rangle ku+ku\times x\tag{1} $$ with respect to the standard basis of $\mathbb R^3$. Extend $u$ to an orthonormal basis $\{u,v,w\}$ of $\mathbb R^3$ such that $u\times v=w$. By $(1)$, we have \begin{align*} Tu&=\langle e,u\rangle ku,\\ Tv&=\langle e,v\rangle ku+kw,\\ Tw&=\langle e,w\rangle ku-kv.\\ \end{align*} Therefore, with respect to the ordered basis $\{u,v,w\}$, $T$ has the matrix representation $$ B=\left(\begin{array}{c|cc}k\langle e,u\rangle&k\langle e,v\rangle&k\langle e,w\rangle\\ \hline 0&0&k\\ 0&-k&0 \end{array}\right). $$ It follows that \begin{align*} &\det A=\det T=\det B\\ &=k\langle e,u\rangle \det\pmatrix{0&k\\ -k&0}= \langle e,p\rangle k^2=(a+b+c)(a^2+b^2+c^2). \end{align*}