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Determinant

Prove that ...

Question

Prove that $∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} & b c & a c + c^{2} a^{2} + a b & b^{2} & a c a b & b^{2} + b c & c^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$ = 4 $a^{2} b^{2} c^{2}$

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Solution

L.H.S.=
$∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} & b c & a c + c^{2} a^{2} + a b & b^{2} & a c a b & b^{2} + b c & c^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$

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

$= a b c ∣ ∣ ∣ ∣ \begin{matrix} 2 a + 2 b & 2 b + 2 c & 2 c + 2 a a + b & b & a b & b + c & c \end{matrix} ∣ ∣ ∣ ∣$ $(R_{1} \to R_{1} + R_{2} + R_{3})$

$= 2 a b c ∣ ∣ ∣ ∣ \begin{matrix} a + b & b + c & c + a a + b & b & a b & b + c & c \end{matrix} ∣ ∣ ∣ ∣$

$= 2 a b c ∣ ∣ ∣ ∣ \begin{matrix} 0 & c & c a + b & b & a b & b + c & c \end{matrix} ∣ ∣ ∣ ∣$ $(R_{1} \to R_{1} - R_{2})$

$= 2 a b c ∣ ∣ ∣ ∣ \begin{matrix} 0 & c & 0 a + b & b & a - b b & b + c & - b \end{matrix} ∣ ∣ ∣ ∣$ $(C_{3} \to C_{3} - C_{2})$

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

$= 2 a b c (2 a b c) = 4 a^{2} b^{2} c^{2}$ = R.H.S.

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Q. The value of

∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} & b c & a c + c^{2} a^{2} + a b & b^{2} & a c a b & b^{2} + b c & c^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

Q. Prove the following :

∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} & b c & a c + c^{2} a^{2} + a b & b^{2} & a c a b & b^{2} + b c & c^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = 4 a^{2} b^{2} c^{2} .

Q. Using properties of determinants, prove the following:

∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} & b c & a c + c^{2} a^{2} + a b & b^{2} & a c a b & b^{2} + b c & c^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = 4 a^{2} b^{2} c^{2}

Q. Prove that :

|\begin{matrix} a^{2} & b c & a c + c^{2} \\ a^{2} + a b & b^{2} & a c \\ a b & b^{2} + b c & c^{2} \end{matrix}| = 4 a^{2} b^{2} c^{2}

a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} \geq a b c (a + b + c)

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