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Summation of Determinant

Prove that a ...

Question

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

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Solution

$Let LHS = Δ = |a^{2} + 1 a b a c a b b^{2} + 1 b c c a c b c^{2} + 1| = (a b c) |a + \frac{1}{a} b c a b + \frac{1}{b} c a b c + \frac{1}{c}| [Taking out a, b and c common from R_{1}, R_{2} and R_{3}] = (a b c) |a + \frac{1}{a} b c - \frac{1}{a} \frac{1}{b} 0 - \frac{1}{a} 0 \frac{1}{c}| [Applying R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{1}] = (a b c) (\frac{1}{a b c}) |a^{2} + 1 b^{2} c^{2} - 1 1 0 - 1 0 1| [Applying C_{1} \to a C_{1}, C_{2} \to b C_{2} and C_{3} \to c C_{3}] = |a^{2} + 1 b^{2} c^{2} - 1 1 0 - 1 0 1| = (- 1) |b^{2} c^{2} 1 0| + (1) |a^{2} + 1 b^{2} - 1 1| [Expanding along R_{3}] = (- 1) (- c^{2}) + (a^{2} + 1 + b^{2}) = (a^{2} + 1 + b^{2} + c^{2}) = (a^{2} + b^{2} + c^{2} + 1) = RHS$

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Q. Using properties of determination, prove that

∣ ∣ ∣ ∣ ∣ \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. Using properties of determinants, prove that

∣ ∣ ∣ ∣ ∣ \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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⎛ ⎜ ⎝ \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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