Grade 12Algebra

Show that |[(b+c)2,c2,b2]
[c2,(c+a)2,a2] = 2(ab+bc+ca)3
[b2,a2,(a+b)2]|

Anapoorini

10 Years agoGrade 12

10 Years ago

With substitutions x = 1/a, y=1/b, z=1/c we wish to prove that the determinant

$\begin{vmatrix} (y+z)^2 & y^2 & z^2 \\ x^2& (x+z)^2) & z^2 \\ x^2 & y^2 & (x+y)^2) \end{vmatrix}$ = 2xyz(x+y+z)³

Using R1 $\rightarrow$ R1-R2 and R3 $\rightarrow$ R3-R2 and taking (x+y+z) common from R1 and R3 we get the determinant as

$(x+y+z)^2 )\begin{vmatrix} y+z-x& y-x-z & 0\\ x^2& (x+z)^2 & z^2 \\ 0 & y-x-z & x+y-z \end{vmatrix}$

Now with C2 $\rightarrow$ C2 – C1 – C3 and taking 2 common from C2 we get

$2(x+y+z)^2) \begin{vmatrix} y+z-x & -z & 0\\ x^2 & xz & z^2 \\ 0&-x & x+y-z \end{vmatrix}$

Now adding x*R1 to R2 we simplify to

$2(x+y+z)^2) \begin{vmatrix} y+z-x & -z & 0\\ x(y+z) & 0 & z^2 \\ 0&-x & x+y-z \end{vmatrix}$

The determinant on expanding by the usual rule gives $xyz(x+y+z)$

and hence the original determinant = $2xyz(x+y+z)^3$ as required

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