{vmatrix} 1 & 1 & 1 \\ a² & b² & c² \\ a³ & b³ & c³ {vmatrix}= [AP EAPCET 19 May 2024 Shift 2]

Correct Answer is : (a−b)(b−c)(c−a)(ab+bc+ca)(a-b)(b-c)(c-a)(ab+bc+ca)(a−b)(b−c)(c−a)(ab+bc+ca)

Correct Answer is : (a−b)(b−c)(c−a)(ab+bc+ca)(a-b)(b-c)(c-a)(ab+bc+ca)(a−b)(b−c)(c−a)(ab+bc+ca)

Test Index

Matrices and Determinants

Section: Mathematics

Question : 44 of 48

Marks: +1, -0

\begin{vmatrix} 1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix}=

Solution:

We have,

\begin{vmatrix} 1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix}

\;C_2 \longrightarrow C_2-C_1 \;\text{ and }\; C_3 \longrightarrow C_3-C_1

\;= \begin{vmatrix} 1 & 0 & 0 \\ a^2 & b^2-a^2 & c^2-a^2 \\ a^3 & b^3-a^3 & c^3-a^3 \end{vmatrix}

\;=(b-a)(c-a)

\; \begin{vmatrix} 1 & 0 & 0 \\ a^2 & (b+a) & (c+a) \\ a^3 & (b^2+a^2+ab) & (c^2+a^2+ac) \end{vmatrix}

\;=(b-a)(c-a)

\; \begin{vmatrix} b+a & c+a \\ b^2+a^2+ab & c^2+a^2+ac \end{vmatrix}

\;= [(b-a)(c-a)[b c^2 + a^2 b + a b c + a c^2 + a^3 \dots

\;+a^2 c - b^2 c - a^3 - a^2 b - a^2 c - a b c - a b^2]

\;=(b-a)(c-a)[b c^2 + a c^2 - b^2 c - a b^2]

\;=(b-a)(c-a)[b c(c-b)

\;+a(c-b)(c+b)]

\;=(a-b)(b-c)(c-a)(ab+bc+ca)

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