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Using properties of determinants prove that: $\left|\begin{array}{lll} 1 & \mathrm{~b}+\mathrm{c} & \mathrm{b}^{2}+\mathrm{c}^{2} \\ 1 & \mathrm{c}+\mathrm{a} & \mathrm{c}^{2}+\mathrm{a}^{2} \\ 1 & \mathrm{a}+\mathrm{b} & \mathrm{a}^{2}+\mathrm{b}^{2} \end{array}\right|=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})$
Using properties of determinants prove that: $\left|\begin{array}{lll} 1 & \mathrm{~b}+\mathrm{c} & \mathrm{b}^{2}+\mathrm{c}^{2} \\ 1 & \mathrm{c}+\mathrm{a} & \mathrm{c}^{2}+\mathrm{a}^{2} \\ 1 & \mathrm{a}+\mathrm{b} & \mathrm{a}^{2}+\mathrm{b}^{2} \end{array}\right|=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})$
CBSE | Class 12 | Determinants | Exercise 6B | Maths | RS Aggarwal
Using properties of determinants prove that: $\left|\begin{array}{lll} 1 & \mathrm{~b}+\mathrm{c} & \mathrm{b}^{2}+\mathrm{c}^{2} \\ 1 & \mathrm{c}+\mathrm{a} & \mathrm{c}^{2}+\mathrm{a}^{2} \\ 1 & \mathrm{a}+\mathrm{b} & \mathrm{a}^{2}+\mathrm{b}^{2} \end{array}\right|=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})$

Solution:

$\begin{array}{l}
\left|\begin{array}{llll}
1 & \mathrm{~b}+\mathrm{c} & \mathrm{b}^{2}+\mathrm{c}^{2} \\
1 & \mathrm{c}+\mathrm{a} & \mathrm{c}^{2}+\mathrm{a}^{2} \\
1 & \mathrm{a}+\mathrm{b} & \mathrm{a}^{2}+\mathrm{b}^{2}
\end{array}\right| \\
=\left|\begin{array}{lll}
0 & \mathrm{~b}-\mathrm{a} & \mathrm{b}^{2}-\mathrm{a}^{2} \\
0 & \mathrm{c}-\mathrm{b} & \mathrm{c}^{2}-\mathrm{b}^{2} \\
1 & \mathrm{a}+\mathrm{b} & \mathrm{a}^{2}+\mathrm{b}^{2}
\end{array}\right|\left[\mathrm{R}_{1}^{\prime}=\mathrm{R}_{1}-\mathrm{R}_{2} \& \mathrm{R}_{2}^{\prime}=\mathrm{R}_{2}-\mathrm{R}_{3}\right] \\
=\left|\begin{array}{lll}
0 & \mathrm{~b}-\mathrm{a} & (\mathrm{b}-\mathrm{a})(\mathrm{b}+\mathrm{a}) \\
0 & \mathrm{c}-\mathrm{b} & (\mathrm{c}-\mathrm{b})(\mathrm{c}+\mathrm{b}) \\
1 & \mathrm{a}+\mathrm{b} & \mathrm{a}^{2}+\mathrm{b}^{2}
\end{array}\right| \\
=(\mathrm{b}-\mathrm{a})(\mathrm{c}-\mathrm{b})\left|\begin{array}{lll}
0 & 1 & \mathrm{~b}+\mathrm{a} \\
0 & 1 & \mathrm{c}+\mathrm{b} \\
1 & \mathrm{a}+\mathrm{b} & \mathrm{a}^{2}+\mathrm{b}^{2}
\end{array}\right|\left[\mathrm{R}_{1}^{\prime}=\mathrm{R}_{1} /(\mathrm{b}-\mathrm{a}) \& \mathrm{R}_{2}^{\prime}=\mathrm{R}_{2} /(\mathrm{c}-\mathrm{b})\right] \\
=(\mathrm{b}-\mathrm{a})(\mathrm{c}-\mathrm{b})[0+0+1\{(\mathrm{c}+\mathrm{b})-(\mathrm{b}+\mathrm{a})\}][\text { expansion by first column }] \\
=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})
\end{array}$

i

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