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.