GATE Overflow Test Series | Linear Algebra | Test 1 | Question: 14

Question

The value of $\begin{vmatrix}
a^{2} & a & bc\\
b^{2} & b & ca\\
c^{2} & c & ab
\end{vmatrix} + \begin{vmatrix}
1 & 1 & 1\\
a^{2} & b^{2} & c^{2}\\
a^{3} & b^{3} & c^{3}
\end{vmatrix} =$ ________

Answer 1

We can take the first one and do some transformation operations.

Lets $\mid A \mid = \begin{vmatrix}
a^{2} & a & bc \\
b^{2} & b & ca \\
c^{2} & c & ab
\end{vmatrix}$

By multiplying $R_{1}, R_{2}, R_{3}$ by $a, b$ and $c$ respectively we get

$\mid A \mid =  \dfrac{1}{abc} \begin{vmatrix}
a^{3} & a^{2} & abc \\
b^{3} & b^{2} & abc \\
c^{3} & c^{2} & abc
\end{vmatrix}$

$ \implies \mid A \mid =  \dfrac{abc}{abc} \begin{vmatrix}
a^{3} & a^{2} & 1 \\
b^{3} & b^{2} & 1 \\
c^{3} & c^{2} & 1
\end{vmatrix}$

$ \implies \mid A \mid = - \begin{vmatrix}
1 & a^{2} & a^{3} \\
1& b^{2} & b^{3} \\
1 & c^{2} & c^{3}
\end{vmatrix}$

$ \implies \mid A \mid = - \begin{vmatrix}
1 & 1 & 1 \\
a^{2} & b^{2} & b^{2} \\
a^{3} & b^{3} & c^{3}
\end{vmatrix}$   [By changing rows into columns]

$\therefore \begin{vmatrix}
a^{2} & a & bc \\
b^{2} & b & ca \\
c^{2} & c & ab
\end{vmatrix} + \begin{vmatrix}
1 & 1 & 1 \\
a^{2} & b^{2}  & c^{2} \\
a^{3} & b^{3}  & c^{3}
\end{vmatrix} = 0$

$\textbf{Short Trick:}$ Put $a = b = c = 1.$

So, the correct answer is $0.$