Prove that $∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} a & b & c & d a & a + b & a + b + c & a + b + c + d a & 2 a + b & 3 a + 2 b + c & 4 a + 3 b + 2 c + d a & 3 a + b & 6 a + 3 b + c & 10 a + 6 b + 3 c + d \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣$ = $a^{4}$

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Solution

Apply reverse operation for making smaller numbers $R_{4} - R_{3}, R_{3} - R_{2}, R_{2} - R_{1}$ .
Again apply $R_{4} - R_{3}, R_{3} - R_{2}$

Apply $R_{4} - R 3$

We get

$△ =$ $∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} a & b & c & d 0 & a & a + b & a + b + c 0 & 0 & a & 2 a + b 0 & 0 & 0 & a \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ = a^{4}$

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Similar questions

∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} a & b & c & d a & a + b & a + b + c & a + b + c + d a & 2 a + b & 3 a + 2 b + c & 4 a + 3 b + 2 c + d a & 3 a + b & 6 a + 3 b + c & 10 a + 6 b + 3 c + d \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣

a (2 a + b - c) - b (a - 2 b + 2 c)

3 c (- 2 a + 3 b + c)

∣ ∣ ∣ ∣ \begin{matrix} a & a + b & a + b + c 2 a & 3 a + 2 b & 4 a + 3 b + 2 c 3 a & 6 a + 3 b & 10 a + 6 b + 3 c \end{matrix} ∣ ∣ ∣ ∣ = a^{3}

∣ ∣ ∣ ∣ \begin{matrix} a & a + b & a + b + c 2 a & 3 a + 2 b & 4 a + 3 b + 2 c 3 a & 6 a + 3 b & 10 a + 6 b + 3 c \end{matrix} ∣ ∣ ∣ ∣ = a^{3}

∣ ∣ ∣ ∣ \begin{matrix} a & a + b & a + b + c 2 a & 3 a + 2 b & 4 a + 3 b + 2 c 3 a & 6 a + 3 b & 10 a + 6 b + 3 c \end{matrix} ∣ ∣ ∣ ∣ = a^{3}

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