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Mathematics Question on Determinants

By using properties of determinants,show that:
(i)∣a−b−c2a2a 2bb−c−a2b 2c2cc−a−b∣\begin{vmatrix} a-b-c & 2a & 2a\\\ 2b & b-c-a & 2b \\\ 2c&2c&c-a-b\end{vmatrix}=(a+b+c)3
(ii)∣x+y+2zxy zy+z+2xy zxz+x+2y∣\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

(i) △=∣a−b−c2a2a 2bb−c−a2b 2c2cc−a−b∣\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∣a+b+ca+b+ca+b+c 2bb−c−a2b 2c2cc−a−b∣\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)∣111 2bb−c−a2b 2c2cc−a−b∣\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)∣100 2b−a+b+c0 2c0−(a+b+c)∣\begin{vmatrix} 1 & 0 & 0\\\ 2b & -a+b+c & 0 \\\ 2c&0&-(a+b+c)\end{vmatrix}

=(a+b+c)3∣100 2b−10 2c0−1∣\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.

(ii) △=∣x+y+2zxy zy+z+2xy zxz+x+2y∣\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:

△=∣2(x+y+z)xy 2(x+y+z)y+z+2xy 2(x+y+z)xz+x+2y∣\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)∣1xy 1y+z+2xy 1xz+x+2y∣\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)∣1xy 0x+y+zy 00x+y+z∣\begin{vmatrix} 1 & x & y\\\ 0 &x+y+z & y\\\ 0 &0&x+y+z \end{vmatrix}

△=2(x+y+z)3∣1xy 010 001∣\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.