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Prove that $\left|\begin{array}{lll}1 & a^{2}+b c & a^{3} \\ 1 & b^{2}+c a & b^{3} \\ 1 & c^{2}+a b & c^{3}\end{array}\right|=-(a-b)(b-c)(c-a)\left(a^{2}+b^{2}+c^{2}\right)$
Prove that $\left|\begin{array}{lll}1 & a^{2}+b c & a^{3} \\ 1 & b^{2}+c a & b^{3} \\ 1 & c^{2}+a b & c^{3}\end{array}\right|=-(a-b)(b-c)(c-a)\left(a^{2}+b^{2}+c^{2}\right)$
CBSE | Class 12 | Determinants | Exercise 6B | Maths | RS Aggarwal
Prove that $\left|\begin{array}{lll}1 & a^{2}+b c & a^{3} \\ 1 & b^{2}+c a & b^{3} \\ 1 & c^{2}+a b & c^{3}\end{array}\right|=-(a-b)(b-c)(c-a)\left(a^{2}+b^{2}+c^{2}\right)$

Solution:

Operating $R_{1} \rightarrow R_{1}-R_{2}, R_{2} \rightarrow R_{2}-R_{3}$
$\begin{array}{l}
=\left|\begin{array}{ccc}
0 & \mathrm{a}^{2}+\mathrm{bc}-\mathrm{b}^{2}-\mathrm{ac} & \mathrm{a}^{3}-\mathrm{b}^{3} \\
0 & \mathrm{~b}^{2}+\mathrm{ca}-\mathrm{c}^{2}-\mathrm{ab} & \mathrm{b}^{3}-\mathrm{c}^{3} \\
1 & \mathrm{c}^{2}+\mathrm{ab} & \mathrm{c}^{3}
\end{array}\right| \\
=\left|\begin{array}{ccc}
0 & (\mathrm{a}-\mathrm{b})(\mathrm{a}+\mathrm{b})-\mathrm{c}(\mathrm{a}-\mathrm{b}) & (\mathrm{a}-\mathrm{b})\left(\mathrm{a}^{2}+\mathrm{ab}+\mathrm{b}^{2}\right) \\
0 & (\mathrm{~b}-\mathrm{c})(\mathrm{b}+\mathrm{c})-\mathrm{a}(\mathrm{b}-\mathrm{c}) & (\mathrm{b}-\mathrm{c})\left(\mathrm{b}^{2}+\mathrm{bc}+\mathrm{c}^{2}\right) \\
1 & \mathrm{c}^{2}+\mathrm{ab} & \mathrm{c}^{3}
\end{array}\right|
\end{array}$
Taking (a-b), (b-c) common from $R_{1}, R_{2}$ respectively
$=(a-b)(b-c)\left|\begin{array}{ccc}
0 & a+b-c & a^{2}+a b+b^{2} \\
0 & b+c-a & b^{2}+b c+c^{2} \\
1 & c^{2}+a b & c^{3}
\end{array}\right|$
Operating $R_{1} \rightarrow R_{1^{-}} R_{2}$
$\begin{array}{l}
=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})\left|\begin{array}{ccc}
0 & 2 \mathrm{a}-2 \mathrm{c} & \mathrm{a}^{2}+\mathrm{ab}-\mathrm{b} c-\mathrm{c}^{2} \\
0 & \mathrm{~b}+\mathrm{c}-\mathrm{a} & \mathrm{b}^{2}+\mathrm{b} c+\mathrm{c}^{2} \\
1 & \mathrm{c}^{2}+\mathrm{ab} & \mathrm{c}^{3}
\end{array}\right| \\
=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})\left|\begin{array}{ccc}
0 & 2(\mathrm{a}-\mathrm{c}) & (\mathrm{a}+\mathrm{c})(\mathrm{a}-\mathrm{c})+\mathrm{b}(\mathrm{a}-\mathrm{c}) \\
0 & \mathrm{~b}+\mathrm{c}-\mathrm{a} & \mathrm{b}^{2}+\mathrm{bc}+\mathrm{c}^{2} \\
1 & \mathrm{c}^{2}+\mathrm{ab} & \mathrm{c}^{3}
\end{array}\right|
\end{array}$
Taking $(a-c)$ common from $R_{1}$
$=(a-c)(a-b)(b-c)\left|\begin{array}{ccc}
0 & 2 & a+b+c \\
0 & b+c-a & b^{2}+b c+c^{2} \\
1 & c^{2}+a b & c^{3}
\end{array}\right|$
Expanding with $C_{1}$
$\begin{array}{l}
=(a-c)(a-b)(b-c) \times\left(2 b^{2}+2 b c+2 c^{2}-a b-b^{2}-b c-a c-b c-c^{2}+a^{2}+a b+a c\right) \\
=-(c-a)(b-c)(a-b)\left(a^{2}+b^{2}+c^{2}\right)
\end{array}$
Hence Proved

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