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Question-143718




Question Number 143718 by cherokeesay last updated on 17/Jun/21
Commented by TheHoneyCat last updated on 17/Jun/21
Hi, what do you mean by ′′∣B∣′′?  it looks like you are talking about a norm but which one?
$${Hi},\:{what}\:{do}\:{you}\:{mean}\:{by}\:''\mid{B}\mid''? \\ $$$${it}\:{looks}\:{like}\:{you}\:{are}\:{talking}\:{about}\:{a}\:{norm}\:{but}\:{which}\:{one}? \\ $$
Commented by TheHoneyCat last updated on 17/Jun/21
Q_2   ED= [(1,X),(X,1) ]× [(2,y),(1,(−y)) ]= [((2+X),((1−X)y)),((2X+1),((X−1)y)) ]  DE= [(2,y),(1,(−y)) ]× [(1,X),(X,1) ]= [((2+yX),(2X+y)),((1−yX),(X−y)) ]  the first term (m_(0,0) ) imposes y=1 to get DE=ED  the one under (m_(1,0) ) imposes 2X+1=1−X⇔3X=0⇔X=0  checking (m_(0,1) ) with these conditions, we get 1=−1  which is obviously false    so ∀(X,y)∈C^2  ED≠DE
$${Q}_{\mathrm{2}} \\ $$$${ED}=\begin{bmatrix}{\mathrm{1}}&{{X}}\\{{X}}&{\mathrm{1}}\end{bmatrix}×\begin{bmatrix}{\mathrm{2}}&{{y}}\\{\mathrm{1}}&{−{y}}\end{bmatrix}=\begin{bmatrix}{\mathrm{2}+{X}}&{\left(\mathrm{1}−{X}\right){y}}\\{\mathrm{2}{X}+\mathrm{1}}&{\left({X}−\mathrm{1}\right){y}}\end{bmatrix} \\ $$$${DE}=\begin{bmatrix}{\mathrm{2}}&{{y}}\\{\mathrm{1}}&{−{y}}\end{bmatrix}×\begin{bmatrix}{\mathrm{1}}&{{X}}\\{{X}}&{\mathrm{1}}\end{bmatrix}=\begin{bmatrix}{\mathrm{2}+{yX}}&{\mathrm{2}{X}+{y}}\\{\mathrm{1}−{yX}}&{{X}−{y}}\end{bmatrix} \\ $$$$\mathrm{the}\:\mathrm{first}\:\mathrm{term}\:\left({m}_{\mathrm{0},\mathrm{0}} \right)\:\mathrm{imposes}\:{y}=\mathrm{1}\:\mathrm{to}\:\mathrm{get}\:\mathrm{D}{E}={ED} \\ $$$$\mathrm{the}\:\mathrm{one}\:\mathrm{under}\:\left({m}_{\mathrm{1},\mathrm{0}} \right)\:\mathrm{imposes}\:\mathrm{2}{X}+\mathrm{1}=\mathrm{1}−{X}\Leftrightarrow\mathrm{3}{X}=\mathrm{0}\Leftrightarrow{X}=\mathrm{0} \\ $$$$\mathrm{checking}\:\left({m}_{\mathrm{0},\mathrm{1}} \right)\:\mathrm{with}\:\mathrm{these}\:\mathrm{conditions},\:\mathrm{we}\:\mathrm{get}\:\mathrm{1}=−\mathrm{1} \\ $$$$\mathrm{which}\:\mathrm{is}\:\mathrm{obviously}\:\mathrm{false} \\ $$$$ \\ $$$$\mathrm{so}\:\forall\left({X},{y}\right)\in\mathbb{C}^{\mathrm{2}} \:{ED}\neq{DE} \\ $$
Commented by Olaf_Thorendsen last updated on 17/Jun/21
In some countries, France for example,  det(B) is sometimes noted ∣B∣.
$$\mathrm{In}\:\mathrm{some}\:\mathrm{countries},\:\mathrm{France}\:\mathrm{for}\:\mathrm{example}, \\ $$$$\mathrm{det}\left(\mathrm{B}\right)\:\mathrm{is}\:\mathrm{sometimes}\:\mathrm{noted}\:\mid\mathrm{B}\mid. \\ $$$$ \\ $$
Commented by TheHoneyCat last updated on 17/Jun/21
well, I think this notation only concers the expressions  it is used to do explicit calculations or block calculations    At least that′s how we do it in my School (which is in France of course, otherwise I would not have any idea)  we use detM when M is not explicit    But maybe is is used is some other places
$$\mathrm{well},\:\mathrm{I}\:\mathrm{think}\:\mathrm{this}\:\mathrm{notation}\:\mathrm{only}\:\mathrm{concers}\:\mathrm{the}\:\mathrm{expressions} \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{used}\:\mathrm{to}\:\mathrm{do}\:\mathrm{explicit}\:\mathrm{calculations}\:\mathrm{or}\:\mathrm{block}\:\mathrm{calculations} \\ $$$$ \\ $$$$\mathrm{At}\:\mathrm{least}\:\mathrm{that}'\mathrm{s}\:\mathrm{how}\:\mathrm{we}\:\mathrm{do}\:\mathrm{it}\:\mathrm{in}\:\mathrm{my}\:\mathrm{School}\:\left(\mathrm{which}\:\mathrm{is}\:\mathrm{in}\:\mathrm{France}\:\mathrm{of}\:\mathrm{course},\:\mathrm{otherwise}\:\mathrm{I}\:\mathrm{would}\:\mathrm{not}\:\mathrm{have}\:\mathrm{any}\:\mathrm{idea}\right) \\ $$$$\mathrm{we}\:\mathrm{use}\:\mathrm{det}{M}\:\mathrm{when}\:{M}\:\mathrm{is}\:\mathrm{not}\:\mathrm{explicit} \\ $$$$ \\ $$$$\mathrm{But}\:\mathrm{maybe}\:\mathrm{is}\:\mathrm{is}\:\mathrm{used}\:\mathrm{is}\:\mathrm{some}\:\mathrm{other}\:\mathrm{places} \\ $$
Commented by TheHoneyCat last updated on 17/Jun/21
det(B)=(−2i)(6+2i)−4×1  =−12i+4−4  =−12i
$$\mathrm{det}\left({B}\right)=\left(−\mathrm{2}{i}\right)\left(\mathrm{6}+\mathrm{2}{i}\right)−\mathrm{4}×\mathrm{1} \\ $$$$=−\mathrm{12}{i}+\mathrm{4}−\mathrm{4} \\ $$$$=−\mathrm{12}{i} \\ $$

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