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Question Number 210080 by Spillover last updated on 30/Jul/24

Given that  det  [(a,b,c),(d,e,f),(g,h,i) ]=n    find det [((d+2a),(e+2b),(f+2c)),((2a),(2b),(2c)),((4g),(4h),(4i)) ]

$${Given}\:{that}\:\:{det}\:\begin{bmatrix}{{a}}&{{b}}&{{c}}\\{{d}}&{{e}}&{{f}}\\{{g}}&{{h}}&{{i}}\end{bmatrix}={n} \\ $$$$ \\ $$$${find}\:{det}\begin{bmatrix}{{d}+\mathrm{2}{a}}&{{e}+\mathrm{2}{b}}&{{f}+\mathrm{2}{c}}\\{\mathrm{2}{a}}&{\mathrm{2}{b}}&{\mathrm{2}{c}}\\{\mathrm{4}{g}}&{\mathrm{4}{h}}&{\mathrm{4}{i}}\end{bmatrix} \\ $$$$ \\ $$

Commented by Frix last updated on 30/Jul/24

Doesn′t give any “nice” result.

$$\mathrm{Doesn}'\mathrm{t}\:\mathrm{give}\:\mathrm{any}\:``\mathrm{nice}''\:\mathrm{result}. \\ $$

Commented by Spillover last updated on 30/Jul/24

what result did you get?

$${what}\:{result}\:{did}\:{you}\:{get}? \\ $$

Commented by Frix last updated on 30/Jul/24

Sorry I had a typo in my calculation.  The 2^(nd)  det =−8n

$$\mathrm{Sorry}\:\mathrm{I}\:\mathrm{had}\:\mathrm{a}\:\mathrm{typo}\:\mathrm{in}\:\mathrm{my}\:\mathrm{calculation}. \\ $$$$\mathrm{The}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{det}\:=−\mathrm{8}{n} \\ $$

Commented by Spillover last updated on 30/Jul/24

correct

$${correct} \\ $$

Answered by Spillover last updated on 30/Jul/24

det  [(a,b,c),(d,e,f),(g,h,i) ]=n    det  [(a,b,c),(d,e,f),((4g),(4h),(4i)) ]=4n    det  [((2a),(2b),(2c)),(d,e,f),((4g),(4h),(4i)) ]=8n    det  [((2a),(2b),(2c)),((d+2a),(e+2b),(f+2c)),((4g),(4h),(4i)) ]=8n    det  [((2a+d),(2b+e),(2c+f)),((2a),(2b),(2c)),((4g),(4h),(4i)) ]=−8n

$${det}\:\begin{bmatrix}{{a}}&{{b}}&{{c}}\\{{d}}&{{e}}&{{f}}\\{{g}}&{{h}}&{{i}}\end{bmatrix}={n} \\ $$$$ \\ $$$${det}\:\begin{bmatrix}{{a}}&{{b}}&{{c}}\\{{d}}&{{e}}&{{f}}\\{\mathrm{4}{g}}&{\mathrm{4}{h}}&{\mathrm{4}{i}}\end{bmatrix}=\mathrm{4}{n} \\ $$$$ \\ $$$${det}\:\begin{bmatrix}{\mathrm{2}{a}}&{\mathrm{2}{b}}&{\mathrm{2}{c}}\\{{d}}&{{e}}&{{f}}\\{\mathrm{4}{g}}&{\mathrm{4}{h}}&{\mathrm{4}{i}}\end{bmatrix}=\mathrm{8}{n} \\ $$$$ \\ $$$${det}\:\begin{bmatrix}{\mathrm{2}{a}}&{\mathrm{2}{b}}&{\mathrm{2}{c}}\\{{d}+\mathrm{2}{a}}&{{e}+\mathrm{2}{b}}&{{f}+\mathrm{2}{c}}\\{\mathrm{4}{g}}&{\mathrm{4}{h}}&{\mathrm{4}{i}}\end{bmatrix}=\mathrm{8}{n} \\ $$$$ \\ $$$${det}\:\begin{bmatrix}{\mathrm{2}{a}+{d}}&{\mathrm{2}{b}+{e}}&{\mathrm{2}{c}+{f}}\\{\mathrm{2}{a}}&{\mathrm{2}{b}}&{\mathrm{2}{c}}\\{\mathrm{4}{g}}&{\mathrm{4}{h}}&{\mathrm{4}{i}}\end{bmatrix}=−\mathrm{8}{n} \\ $$$$ \\ $$$$ \\ $$

Commented by Frix last updated on 30/Jul/24

Yes.

$$\mathrm{Yes}. \\ $$

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