If a-b-c are linearly independent- then -2a-b- 2b-c- 2c-a-a- b- c -

[2a+b, 2b+c, 2c+a]/[a, b, c ]= 2 amp; 1 amp; 0 0 amp; 2 amp; 1 1 amp; 0 amp; 2 =2(4 0) 1(0 1)=8+1=9 ...

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Applications Scalar Triple Product

If a,b,c are ...

Question

If a,b,c are linearly independent, then $\frac{[2 a + b, 2 b + c, 2 c + a]}{[a, b, c]} =$

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\frac{[2 a + b, 2 b + c, 2 c + a]}{[a, b, c]} =

a - 2 b + c, 2 a - b + c

3 a + b + 2 c

\to a, \to b, \to c

\frac{[\to a + \to 2 b - \to c \to 2 a + \to b + \to c \to 4 a - \to b + \to 5 c]}{[\to a \to b \to c]}

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

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

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