Question

By using properties of determinants,show that:
(i)\(\begin{vmatrix} a-b-c & 2a & 2a\\ 2b & b-c-a & 2b \\ 2c&2c&c-a-b\end{vmatrix}\)=(a+b+c)³ 
(ii)\(\begin{vmatrix} x+y+2z & x & y\\ z & y+z+2x & y\\ z &x&z+x+2y \end{vmatrix}\)=2(x+y+z)³

Answer 1

(i) △=\(\begin{vmatrix} a-b-c & 2a & 2a\\ 2b & b-c-a & 2b \\ 2c&2c&c-a-b\end{vmatrix}\)

Applying R₁ → R₁ + R₂ + R₃, we have: 

△=I\(\begin{vmatrix} a+b+c & a+b+c & a+b+c\\ 2b & b-c-a & 2b \\ 2c&2c&c-a-b\end{vmatrix}\)

=(a+b+c)\(\begin{vmatrix} 1 & 1 & 1\\ 2b & b-c-a & 2b \\ 2c&2c&c-a-b\end{vmatrix}\)

Applying C₂ → C₂ − C₁, C₃ → C₃ − C₁, we have: 

△=(a+b+c)\(\begin{vmatrix} 1 & 0 & 0\\ 2b & -a+b+c & 0 \\ 2c&0&-(a+b+c)\end{vmatrix}\) 

=(a+b+c)³\(\begin{vmatrix} 1 & 0 & 0\\ 2b & -1 & 0 \\ 2c&0&-1\end{vmatrix}\) 

Expanding along C₃, we have: 

△=(a+b+c)³(-1)(-1)=(a+b+c)³ 

Hence, the given result is proved.

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(ii) △=\(\begin{vmatrix} x+y+2z & x & y\\ z & y+z+2x & y\\ z &x&z+x+2y \end{vmatrix}\)

Applying C₁ → C₁ + C₂ + C₃, we have: 

△=\(\begin{vmatrix} 2(x+y+z) & x & y\\ 2(x+y+z) & y+z+2x & y\\ 2(x+y+z) &x&z+x+2y \end{vmatrix}\) 

=2(x+y+z)\(\begin{vmatrix} 1 & x & y\\ 1 & y+z+2x & y\\ 1 &x&z+x+2y \end{vmatrix}\)

Applying R₂ → R₂ − R₁ and R₃ → R₃ − R₁, we have: 

△=2(x+y+z)\(\begin{vmatrix} 1 & x & y\\ 0 &x+y+z & y\\ 0 &0&x+y+z \end{vmatrix}\) 

△=2(x+y+z)³\(\begin{vmatrix} 1 & x & y\\ 0 &1 & 0\\ 0 &0&1\end{vmatrix}\)

Expanding along R₃, we have:

△=2(x+y+z)³(1)(1-0)=2(x+y+z)³ 

Hence, the given result is proved.