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Given-the-matrix-A-1-1-1-0-2-1-2-3-0-and-B-3-3-1-2-2-1-4-5-2-find-the-matrix-product-AB-and-BA-state-the-relationship-between-A-and-B-find-also-the-matri




Question Number 75099 by Rio Michael last updated on 07/Dec/19
Given the matrix   A =  ((1,(−1),1),(0,2,( −1)),(2,3,0) )  and B=  ((3,3,(−1)),((−2),(−2),1),((−4),(−5),2) )  find the matrix product AB and BA  state the relationship between A and B  find also the matrix product BM, where M= ((8),((−7)),(1) )  Hence solve the system of equations:    x−y + z = 8,         2y −z =−7,    2x + 3y = 1.
$${Given}\:{the}\:{matrix}\: \\ $$$${A}\:=\:\begin{pmatrix}{\mathrm{1}}&{−\mathrm{1}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{2}}&{\:−\mathrm{1}}\\{\mathrm{2}}&{\mathrm{3}}&{\mathrm{0}}\end{pmatrix}\:\:{and}\:{B}=\:\begin{pmatrix}{\mathrm{3}}&{\mathrm{3}}&{−\mathrm{1}}\\{−\mathrm{2}}&{−\mathrm{2}}&{\mathrm{1}}\\{−\mathrm{4}}&{−\mathrm{5}}&{\mathrm{2}}\end{pmatrix} \\ $$$${find}\:{the}\:{matrix}\:{product}\:{AB}\:{and}\:{BA} \\ $$$${state}\:{the}\:{relationship}\:{between}\:{A}\:{and}\:{B} \\ $$$${find}\:{also}\:{the}\:{matrix}\:{product}\:{BM},\:{where}\:{M}=\begin{pmatrix}{\mathrm{8}}\\{−\mathrm{7}}\\{\mathrm{1}}\end{pmatrix} \\ $$$${Hence}\:{solve}\:{the}\:{system}\:{of}\:{equations}: \\ $$$$\:\:{x}−{y}\:+\:{z}\:=\:\mathrm{8}, \\ $$$$\:\:\:\:\:\:\:\mathrm{2}{y}\:−{z}\:=−\mathrm{7}, \\ $$$$\:\:\mathrm{2}{x}\:+\:\mathrm{3}{y}\:=\:\mathrm{1}. \\ $$
Commented by kaivan.ahmadi last updated on 07/Dec/19
Commented by kaivan.ahmadi last updated on 07/Dec/19
Commented by kaivan.ahmadi last updated on 07/Dec/19
Commented by Rio Michael last updated on 07/Dec/19
nice sir thanks
$${nice}\:{sir}\:{thanks} \\ $$
Answered by Kunal12588 last updated on 07/Dec/19
A= [((   1),(−1),(   1)),((   0),(   2),(−1)),((   2),(   3),(   0)) ], B= [((   3),(   3),(−1)),((−2),(−2),(   1)),((−4),(−5),(   2)) ]  AB= [((3+2−4),(3+2−5),(−1−1+2)),((0−4+4),(0−4+5),(     0+2−2)),((6−6+0),(6−6+0),(−2+3+0)) ]  ⇒AB= [(1,0,0),(0,1,0),(0,0,1) ]= I  BA= [((     3+0−2),(−3+6−3),(    3−3+0)),((−2+0+2),(     2−4+3),(−2+2+0)),((−4+0+4),(  4−10+6),(−4+5+0)) ]  ⇒BA= [(1,0,0),(0,1,0),(0,0,1) ]= I  ∵AB=BA=I⇒A^(−1) =B & B^(−1) =A  M= [((   8)),((−7)),((   1)) ]  BM= [((   3),(   3),(−1)),((−2),(−2),(   1)),((−4),(−5),(   2)) ]_(3×3)  [((   8)),((−7)),((   1)) ]_(3×1)   ⇒BM= [((    24−21−1)),((−16+14+1)),((−32+35+2)) ]= [((   2)),((−1)),((   5)) ]  given eq^n  can be written as   [((   1),(−1),(   1)),((   0),(   2),(−1)),((   2),(   3),(   0)) ] [(x),(y),(z) ]= [((   8)),((−7)),((   1)) ]  ⇒AX=M     ; with X= [(x),(y),(z) ]  ⇒A^(−1) AX=A^(−1) M  ⇒IX=A^(−1) M  ⇒X=BM      [∵ A^(−1) =B proved above]  ⇒ [(x),(y),(z) ]= [((   2)),((−1)),((   5)) ]  ⇒x=2,y=−1 & z=5
$${A}=\begin{bmatrix}{\:\:\:\mathrm{1}}&{−\mathrm{1}}&{\:\:\:\mathrm{1}}\\{\:\:\:\mathrm{0}}&{\:\:\:\mathrm{2}}&{−\mathrm{1}}\\{\:\:\:\mathrm{2}}&{\:\:\:\mathrm{3}}&{\:\:\:\mathrm{0}}\end{bmatrix},\:{B}=\begin{bmatrix}{\:\:\:\mathrm{3}}&{\:\:\:\mathrm{3}}&{−\mathrm{1}}\\{−\mathrm{2}}&{−\mathrm{2}}&{\:\:\:\mathrm{1}}\\{−\mathrm{4}}&{−\mathrm{5}}&{\:\:\:\mathrm{2}}\end{bmatrix} \\ $$$${AB}=\begin{bmatrix}{\mathrm{3}+\mathrm{2}−\mathrm{4}}&{\mathrm{3}+\mathrm{2}−\mathrm{5}}&{−\mathrm{1}−\mathrm{1}+\mathrm{2}}\\{\mathrm{0}−\mathrm{4}+\mathrm{4}}&{\mathrm{0}−\mathrm{4}+\mathrm{5}}&{\:\:\:\:\:\mathrm{0}+\mathrm{2}−\mathrm{2}}\\{\mathrm{6}−\mathrm{6}+\mathrm{0}}&{\mathrm{6}−\mathrm{6}+\mathrm{0}}&{−\mathrm{2}+\mathrm{3}+\mathrm{0}}\end{bmatrix} \\ $$$$\Rightarrow{AB}=\begin{bmatrix}{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}=\:{I} \\ $$$${BA}=\begin{bmatrix}{\:\:\:\:\:\mathrm{3}+\mathrm{0}−\mathrm{2}}&{−\mathrm{3}+\mathrm{6}−\mathrm{3}}&{\:\:\:\:\mathrm{3}−\mathrm{3}+\mathrm{0}}\\{−\mathrm{2}+\mathrm{0}+\mathrm{2}}&{\:\:\:\:\:\mathrm{2}−\mathrm{4}+\mathrm{3}}&{−\mathrm{2}+\mathrm{2}+\mathrm{0}}\\{−\mathrm{4}+\mathrm{0}+\mathrm{4}}&{\:\:\mathrm{4}−\mathrm{10}+\mathrm{6}}&{−\mathrm{4}+\mathrm{5}+\mathrm{0}}\end{bmatrix} \\ $$$$\Rightarrow{BA}=\begin{bmatrix}{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}=\:{I} \\ $$$$\because{AB}={BA}={I}\Rightarrow{A}^{−\mathrm{1}} ={B}\:\&\:{B}^{−\mathrm{1}} ={A} \\ $$$${M}=\begin{bmatrix}{\:\:\:\mathrm{8}}\\{−\mathrm{7}}\\{\:\:\:\mathrm{1}}\end{bmatrix} \\ $$$${BM}=\begin{bmatrix}{\:\:\:\mathrm{3}}&{\:\:\:\mathrm{3}}&{−\mathrm{1}}\\{−\mathrm{2}}&{−\mathrm{2}}&{\:\:\:\mathrm{1}}\\{−\mathrm{4}}&{−\mathrm{5}}&{\:\:\:\mathrm{2}}\end{bmatrix}_{\mathrm{3}×\mathrm{3}} \begin{bmatrix}{\:\:\:\mathrm{8}}\\{−\mathrm{7}}\\{\:\:\:\mathrm{1}}\end{bmatrix}_{\mathrm{3}×\mathrm{1}} \\ $$$$\Rightarrow{BM}=\begin{bmatrix}{\:\:\:\:\mathrm{24}−\mathrm{21}−\mathrm{1}}\\{−\mathrm{16}+\mathrm{14}+\mathrm{1}}\\{−\mathrm{32}+\mathrm{35}+\mathrm{2}}\end{bmatrix}=\begin{bmatrix}{\:\:\:\mathrm{2}}\\{−\mathrm{1}}\\{\:\:\:\mathrm{5}}\end{bmatrix} \\ $$$${given}\:{eq}^{{n}} \:{can}\:{be}\:{written}\:{as} \\ $$$$\begin{bmatrix}{\:\:\:\mathrm{1}}&{−\mathrm{1}}&{\:\:\:\mathrm{1}}\\{\:\:\:\mathrm{0}}&{\:\:\:\mathrm{2}}&{−\mathrm{1}}\\{\:\:\:\mathrm{2}}&{\:\:\:\mathrm{3}}&{\:\:\:\mathrm{0}}\end{bmatrix}\begin{bmatrix}{{x}}\\{{y}}\\{{z}}\end{bmatrix}=\begin{bmatrix}{\:\:\:\mathrm{8}}\\{−\mathrm{7}}\\{\:\:\:\mathrm{1}}\end{bmatrix} \\ $$$$\Rightarrow{AX}={M}\:\:\:\:\:;\:{with}\:{X}=\begin{bmatrix}{{x}}\\{{y}}\\{{z}}\end{bmatrix} \\ $$$$\Rightarrow{A}^{−\mathrm{1}} {AX}={A}^{−\mathrm{1}} {M} \\ $$$$\Rightarrow{IX}={A}^{−\mathrm{1}} {M} \\ $$$$\Rightarrow{X}={BM}\:\:\:\:\:\:\left[\because\:{A}^{−\mathrm{1}} ={B}\:{proved}\:{above}\right] \\ $$$$\Rightarrow\begin{bmatrix}{{x}}\\{{y}}\\{{z}}\end{bmatrix}=\begin{bmatrix}{\:\:\:\mathrm{2}}\\{−\mathrm{1}}\\{\:\:\:\mathrm{5}}\end{bmatrix} \\ $$$$\Rightarrow{x}=\mathrm{2},{y}=−\mathrm{1}\:\&\:{z}=\mathrm{5} \\ $$
Commented by Rio Michael last updated on 07/Dec/19
thanks so much sir
$${thanks}\:{so}\:{much}\:{sir} \\ $$

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