0

Guest

Courses New

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

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

Grade:12

1 Answers

mycroft holmes
272 Points
7 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\rightarrowR3-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\rightarrowC2 – 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

Think You Can Provide A Better Answer ?

Other Related Questions on Algebra

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More