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Properties of Determinants

If D=| a 2 ...

Question

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

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Solution

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

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Q. Using properties of determinants, prove that:

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

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∣ ∣ ∣ ∣ ∣ \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}

|\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}

Q. Show that: det

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