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in Linear Inequations by (185 points)
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if the set of vectors {(a, b, 1),(b, 1,a),(1,a, b)} linearly dependent in vector space R³ show that a=b=1 and a+b+1=0 Where a, b €R

1 Answer

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Best answer

\(\begin{vmatrix} a&b&1\\b&1&a\\1&a&b\end{vmatrix} = 0\)

⇒ \(a(b - a^2) - b(b^2 - a) + 1 (ab - 1) = 0\)

⇒ \(ab - a^3 - b^3 + ab + ab - 1 = 0\)

⇒ \(a^3 + b^3 - 3ab + 1 = 0\)

⇒ \((a + b)^2 - 3ab(a+b) - 3ab + 1 = 0\)

⇒ \((a + b)^3 + 1^3 - 3ab ((a + b) + 1) = 0\)

⇒ \((a + b + 1)^3 = 0\)

⇒ \(a + b +1 = 0\)

or if \(a = b = 1\)

then \(a^3 + b^3 + 1 - 3ab = 3 - 3 = 0\)

So, either \(a = b = 1\) or \(a + b + 1 = 0\).

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