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Question Number 117973 by bemath last updated on 14/Oct/20

consider a non−singular 2×2 square matrix T. If trace (T) =4 and trace (T^2 )=5 what is determinant of the matrix T ?

$$\mathrm{consider}\:\mathrm{a}\:\mathrm{non}−\mathrm{singular}\:\mathrm{2}×\mathrm{2}\: \\ $$$$\mathrm{square}\:\mathrm{matrix}\:\mathrm{T}.\:\mathrm{If}\:\mathrm{trace}\:\left(\mathrm{T}\right)\:=\mathrm{4} \\ $$$$\mathrm{and}\:\mathrm{trace}\:\left(\mathrm{T}^{\mathrm{2}} \right)=\mathrm{5}\:\mathrm{what}\:\mathrm{is}\:\mathrm{determinant} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{matrix}\:\mathrm{T}\:? \\ $$

Answered by bobhans last updated on 14/Oct/20

letting p and q denote the eigenvalues of T we have tr(T)= p+q = 4...(i) then because p^2 and q^2 are the eigenvalues of T^2 it follows that tr(T^2 )=p^2 +q^2 =5...(ii) solving eq (i) & (ii) 2p^2 −8p+11 = 0 ⇒p= ((4±i(√6))/2) ∧ q=((4∓ i(√6))/2) Determinan of T is the product of its eigenvalues , we get ∣T∣ = (((4+i(√6))/2))(((4−i(√6))/2))=((11)/2)

$$\mathrm{letting}\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}\:\mathrm{denote}\:\mathrm{the}\:\mathrm{eigenvalues}\:\mathrm{of}\:\mathrm{T} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{tr}\left(\mathrm{T}\right)=\:\mathrm{p}+\mathrm{q}\:=\:\mathrm{4}...\left(\mathrm{i}\right) \\ $$$$\mathrm{then}\:\mathrm{because}\:\mathrm{p}^{\mathrm{2}} \:\mathrm{and}\:\mathrm{q}^{\mathrm{2}} \:\mathrm{are}\:\mathrm{the}\:\mathrm{eigenvalues} \\ $$$$\mathrm{of}\:\mathrm{T}^{\mathrm{2}} \:\mathrm{it}\:\mathrm{follows}\:\mathrm{that}\:\mathrm{tr}\left(\mathrm{T}^{\mathrm{2}} \right)=\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} =\mathrm{5}...\left(\mathrm{ii}\right) \\ $$$$\mathrm{solving}\:\mathrm{eq}\:\left(\mathrm{i}\right)\:\&\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{2p}^{\mathrm{2}} −\mathrm{8p}+\mathrm{11}\:=\:\mathrm{0}\:\Rightarrow\mathrm{p}=\:\frac{\mathrm{4}\pm{i}\sqrt{\mathrm{6}}}{\mathrm{2}}\:\wedge\:\mathrm{q}=\frac{\mathrm{4}\mp\:{i}\sqrt{\mathrm{6}}}{\mathrm{2}} \\ $$$${D}\mathrm{eterminan}\:\mathrm{of}\:\mathrm{T}\:\mathrm{is}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{its} \\ $$$$\mathrm{eigenvalues}\:,\:\mathrm{we}\:\mathrm{get}\:\mid\mathrm{T}\mid\:=\:\left(\frac{\mathrm{4}+{i}\sqrt{\mathrm{6}}}{\mathrm{2}}\right)\left(\frac{\mathrm{4}−{i}\sqrt{\mathrm{6}}}{\mathrm{2}}\right)=\frac{\mathrm{11}}{\mathrm{2}} \\ $$

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