If a + b + c ≠ 0 and `|("a", "b","c"),("b", "c", "a"),("c", "a", "b")|` 0, then prove that a = b = c.

If a + b + c ≠ 0 and abcbcacab|abcbcacab| 0, then prove that a = b = c. - Mathematics

Question

If a + b + c ≠ 0 and $| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix} |$ 0, then prove that a = b = c.

Sum

Solution

Let Δ = $| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix} |$

[Applying R₁ → R₁ + R₂ + R₃]

Δ = $| \begin{matrix} a + b + c & a + b + c & a + b + c \\ b & c & a \\ c & a & b \end{matrix} |$

= $(a + b + c) | \begin{matrix} 1 & 1 & 1 \\ b & c & a \\ c & a & b \end{matrix} |$

[Applying C₁ → C₁ + C₃ and C₂ → C₂ – C₃]

Δ = $(a + b + c) | \begin{matrix} 0 & 0 & 1 \\ b - a & c - a & a \\ c - b & a - b & b \end{matrix} |$

[Expanding along R₁]

= $(a + b + c) [1 (b - a) (a - b) - (c - a) (c - b)$

= $(a + b + c) (ba - b^{2} - a^{2} + ab - c^{2} + cb + ac - ab)$

= $- (a + b + c) (a^{2} + b^{2} + c^{2} - ab - bc - ca)$

= $\frac{- 1}{2} (a + b + c) [2 a^{2} + 2 b^{2} + 2 c^{2} - 2 ab - 2 bc - 2 ca]$

= $- \frac{1}{2} (a + b + c) [(a^{2} + b^{2} - 2 ab) + (b^{2} + c^{2} - 2 bc) + (c^{2} + a^{2} - 2 ac)]$

= $\frac{- 1}{2} (a + b + c) [{(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}]$

Given, Δ = 0

⇒ $\frac{- 1}{2} (a + b + c) [{(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}]$ = 0

⇒ ${(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}$ = 0 ...[∵ a + b + c ≠ 0, given]

⇒ a – b = b – c = c – a = 0

⇒ a = b = c

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Determinant of a Matrix of Order 3 × 3

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Q 21 Q 20 Q 22

Chapter 4: Determinants - Exercise [Page 79]

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NCERT Exemplar Mathematics [English] Class 12

Chapter 4 Determinants
Exercise | Q 21 | Page 79

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Determinant of a Matrix of Order 3 × 3

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If a + b + c ≠ 0 and abcbcacab|abcbcacab| 0, then prove that a = b = c. - Mathematics

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