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Find-matrix-B-if-given-AB-BA-0-0-0-0-where-A-5-3-5-3-and-B-0-0-0-0-




Question Number 214713 by efronzo1 last updated on 17/Dec/24
  Find matrix B if given AB=BA= (((0  0)),((0  0)) )    where A=  (((5   3)),((5   3)) ) and B ≠  (((0   0)),((0   0)) )
$$\:\:\mathrm{Find}\:\mathrm{matrix}\:\mathrm{B}\:\mathrm{if}\:\mathrm{given}\:\mathrm{AB}=\mathrm{BA}=\begin{pmatrix}{\mathrm{0}\:\:\mathrm{0}}\\{\mathrm{0}\:\:\mathrm{0}}\end{pmatrix} \\ $$$$\:\:\mathrm{where}\:\mathrm{A}=\:\begin{pmatrix}{\mathrm{5}\:\:\:\mathrm{3}}\\{\mathrm{5}\:\:\:\mathrm{3}}\end{pmatrix}\:\mathrm{and}\:\mathrm{B}\:\neq\:\begin{pmatrix}{\mathrm{0}\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\mathrm{0}}\end{pmatrix} \\ $$$$ \\ $$
Answered by golsendro last updated on 18/Dec/24
 det(A)= 15−15=0   so one of possible matrix B is adj(A)    B= adj(A) =  (((    3       −3)),((−5          5)) )   verification      (((5     3)),((5     3)) )  (((    3    −3)),((−5        5)) ) =  (((0     0)),((0     0)) )
$$\:\mathrm{det}\left(\mathrm{A}\right)=\:\mathrm{15}−\mathrm{15}=\mathrm{0} \\ $$$$\:\mathrm{so}\:\mathrm{one}\:\mathrm{of}\:\mathrm{possible}\:\mathrm{matrix}\:\mathrm{B}\:\mathrm{is}\:\mathrm{adj}\left(\mathrm{A}\right) \\ $$$$\:\:\mathrm{B}=\:\mathrm{adj}\left(\mathrm{A}\right)\:=\:\begin{pmatrix}{\:\:\:\:\mathrm{3}\:\:\:\:\:\:\:−\mathrm{3}}\\{−\mathrm{5}\:\:\:\:\:\:\:\:\:\:\mathrm{5}}\end{pmatrix} \\ $$$$\:\mathrm{verification}\: \\ $$$$\:\:\begin{pmatrix}{\mathrm{5}\:\:\:\:\:\mathrm{3}}\\{\mathrm{5}\:\:\:\:\:\mathrm{3}}\end{pmatrix}\:\begin{pmatrix}{\:\:\:\:\mathrm{3}\:\:\:\:−\mathrm{3}}\\{−\mathrm{5}\:\:\:\:\:\:\:\:\mathrm{5}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\mathrm{0}}\end{pmatrix} \\ $$
Answered by didoumaths last updated on 17/Dec/24
B= (((     3),(     3)),((−5),(−5)) )
$${B}=\begin{pmatrix}{\:\:\:\:\:\mathrm{3}}&{\:\:\:\:\:\mathrm{3}}\\{−\mathrm{5}}&{−\mathrm{5}}\end{pmatrix} \\ $$
Commented by som(math1967) last updated on 18/Dec/24
 BA= ((3,3),((−5),(−5)) ) ((5,3),(5,3) )= (((30),(18)),((−50),(−30)) )  ≠ ((0,0),(0,0) )
$$\:{BA}=\begin{pmatrix}{\mathrm{3}}&{\mathrm{3}}\\{−\mathrm{5}}&{−\mathrm{5}}\end{pmatrix}\begin{pmatrix}{\mathrm{5}}&{\mathrm{3}}\\{\mathrm{5}}&{\mathrm{3}}\end{pmatrix}=\begin{pmatrix}{\mathrm{30}}&{\mathrm{18}}\\{−\mathrm{50}}&{−\mathrm{30}}\end{pmatrix} \\ $$$$\neq\begin{pmatrix}{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}\end{pmatrix} \\ $$

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