Question

Prove that$\begin{vmatrix} a^2&bc &ac+c^2 \\\ a^2+ab&b^2 &ac\\\ ab&b^2+bc &c^2 \end{vmatrix}=4a^2b^2c^2$

Answer 1

$\Delta = \begin{vmatrix} a^2&bc &ac+c^2 \\\ a^2+ab&b^2 &ac\\\ ab&b^2+bc &c^2 \end{vmatrix}$

Taking out common factors a,b and c from C1,C2 and C3 we have,
$\Delta=abc\begin{vmatrix} a&c &a+c \\\ a+b&b &a \\\ b&b+c &c \end{vmatrix}$
Applying R2$\rightarrow$R2-R1 and R3$\rightarrow$R3-R1,we have:
$\Delta=abc\begin{vmatrix} a&c &a+c \\\ b&b-c &-c \\\ b-a&b &-a \end{vmatrix}$
Applying R2$\rightarrow$R2+R1,we heve:
$\Delta=abc\begin{vmatrix} a&c &a+c \\\ a+b&b &a \\\ 2b&2b &0 \end{vmatrix}$
Applying R3$\rightarrow$R3+R2,we heve:
$\Delta=abc\begin{vmatrix} a&c &a+c \\\ a+b&b &a \\\ 2b&2b &0 \end{vmatrix}$
$\Delta=2a^2bc\begin{vmatrix} a&c &a+c \\\ a+b&b &a \\\ 2b&2b &0 \end{vmatrix}$
ApplyingC2$\rightarrow$C2-C1, we have:
$\Delta=2a^2bc\begin{vmatrix} a&c-a &a+c \\\ a-b&-a &a \\\ 0&0 &0 \end{vmatrix}$
Expanding along R3,we have:
Δ=2ab2c[a(c-a)+a(a+c)]
=2ab2c[ac-a2+a2+ac]
=2ab2c(2ac)
=4a2b2c2

Hence, the given result is proved.