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By using properties of determinants, show that:(i) \left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|=(a-b)(b-c)(c-a) (ii) \left|\begin{array}{lll}1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3}\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)
CBSE | Class 12 | Determinants | Exercise 4.2 | Maths | NCERT
By using properties of determinants, show that:(i) \left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|=(a-b)(b-c)(c-a) (ii) \left|\begin{array}{lll}1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3}\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)

(i) LHS:

\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|

\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1} and \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}

=\left|\begin{array}{ccc}1 & a & a^{2} \\ 0 & b-a & b^{2}-a^{2} \\ 1 & c-a & c^{2}-a^{2}\end{array}\right|

Expanding 1^{\text {st }} column,
=1\left|\begin{array}{cc}b-a & b^{2}-a^{2} \\ c-a & c^{2}-a^{2}\end{array}\right|

Taking (b-a) common from first row,
=(b-a)(c-a)\left|\begin{array}{ll}1 & b+a \\ 1 & c+a\end{array}\right|

Simplifying above expression, we have

=(b-c)(c-a)(c-b)
=(a-b)(b-c)(c-a)
=\mathrm{RHS}
Proved.

(ii) \mathrm{LHS}

\left|\begin{array}{lll}1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3}\end{array}\right|:

C2→C2–C1 and C3→C3–C1
\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1} \text { and } \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1}

=\left|\begin{array}{ccc}1 & 0 & 0 \\ a & b-a & c-a \\ a^{3} & b^{3}-a^{3} & c^{3}-a^{3}\end{array}\right|

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Expanding first row

=1\left|\begin{array}{cc}b-a & c-a \\ (b-a)\left(b^{2}+a^{2}+a b\right) & (c-a)\left(c^{2}+a^{2}+a c\right)\end{array}\right|

=(b-a)(c-a)\left|\begin{array}{cc}1 & 1 \\ \left(b^{2}+a^{2}+a b\right) & \left(c^{2}+a^{2}+a c\right)\end{array}\right|

=(b–a)(c–a)c2+a2+ac–b2–a2–ab
=(b-a)(c-a)\left(c^{2}+a^{2}+a c-b^{2}-a^{2}-a b\right)
=(b–a)(c–a)c2–b2+ac–ab
=(b-a)(c-a)\left(c^{2}-b^{2}+a c-a b\right)
=(b–a)(c–a)[(c–b)(c+b)+a(c–b)]
=(b-a)(c-a)[(c-b)(c+b)+a(c-b)]
=(b–a)(c–a)(c–b)(c+b+a)
=(b-a)(c-a)(c-b)(c+b+a)

=(a-b)(b-c)(c-a)(a+b+c)

=\mathrm{RHS}

Proved

i

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