$∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} + 1 & a b & a c a b & b^{2} + 1 & b c a c & b c & c^{2} + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ =$

abc

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a+b+c

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1 + a^{2} + b^{2} + c^{2}

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abc(1+a+b+c)

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Solution

The correct option is C $1 + a^{2} + b^{2} + c^{2}$
On solving the given matrix, wevget
$(a^{2} + 1) [(b^{2} + 1) (c^{2} + 1) - b^{2} c^{2}] - a b [a b c^{2} + a b - a b c^{2}] + a c [a b^{2} c - a c b^{2} - a c]$
$=$ $a^{2} b^{2} + a^{2} c^{2} + a^{2} + b^{2} + c^{2} + 1 - a^{2} b^{2} - a^{2} c^{2}$
$=$ $a^{2} + b^{2} + c^{2} + 1$

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Similar questions

Q. Let

∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} + 1 & a b & a c a b & b^{2} + 1 & b c a c & b c & c^{2} + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = k + a^{2} + b^{2} + c^{2}

then

4 k

Prove that $∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} + 1 & a b & a c a b & b^{2} + 1 & b c a c & b c & c^{2} + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = a^{2} + b^{2} + c^{2} + 1$

∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} + 1 & a b & a c a b & b^{2} + 1 & b c a c & b c & c^{2} + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = f (a, b, c)

f

Q. Show that

b^{2} c^{2} + c^{2} a^{2} + a^{2} b^{2} > a b c (a + b + c)

Q. Solve:

∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} + 1 & a b & a c a b & b^{2} + 1 & b c c a & c b & c^{2} + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 1 + a^{2} + b^{2} + c^{2}

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∣∣ ∣ ∣∣a2+1abacabb2+1bcacbcc2+1∣∣ ∣ ∣∣=

The correct option is C 1+a2+b2+c2On solving the given matrix, wevget(a2+1)[(b2+1)(c2+1)−b2c2]−ab[abc2+ab−abc2]+ac[ab2c−acb2−ac]= a2b2+a2c2+a2+b2+c2+1−a2b2−a2c2= a2+b2+c2+1

$∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} + 1 & a b & a c a b & b^{2} + 1 & b c a c & b c & c^{2} + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ =$