Proof using properties of determinants

Question

I have to use properties of determinants to show that

$$\left| \begin{array}{ccc} b^2+c^2 & ab & ac \\ ab & a^2+c^2 & bc \\ ac & bc & a^2+b^2 \end{array} \right| = 4a^2b^2c^2$$

I am tempted to start by reducing the above expression to

By dividing respective rows by a,b,c and multiplying respective columns by a,b,c, $$\left| \begin{array}{ccc} b^2+c^2 & b^2 & c^2 \\ a^2 & c^2+a^2 & c^2 \\ a^2 & b^2 & a^2+b^2 \end{array} \right|$$

and it's a road-bock ahead.

Please provide a hint. I would appreciate if you don't provide the whole solution, but just a hint.

Answer 1

You have to compute

$$ D=\left| \begin{array}{ccc} y+z & y & z \\ x & x+z & z \\ x & y & x+y \end{array} \right| $$

where $x=a^2,y=b^2,z=c^2$. You can proceed as follows :

$$ \begin{array}{lcl} D &=& \frac{1}{x}\left| \begin{array}{ccc} xy+xz & xy & xz \\ x & x+z & z \\ x & y & x+y \end{array} \right| \\ &=& \frac{1}{x}\left| \begin{array}{ccc} 0 & xy-y(x+z)-z(y) & xz-y(z)-z(x+y) \\ x & x+z & z \\ x & y & x+y \end{array} \right| \\ &=& -2yz\left| \begin{array}{ccc} 0 & 1 & 1 \\ 1 & x+z & z \\ 1 & y & x+y \end{array} \right| \\ \end{array} $$

Since you insist on not being provided the full solution, I stop here.

Answer 2

Hint: from the first situation of the determinant you need to calculate, notice that (for example) $b^2+c^2=(b+c)^2-2bc$. Than remember the property of addings columns\rows to each other.

Answer 3

If we don't want to use the definition of the determinant, we can obtain the final result from the second matrix you posted by adding/subtracting rows and columns; the goal is to eliminate the cells with sums of squares. I started by subtracting first and se cond column from the third one.