Matematică, întrebare adresată de RedTMS, 8 ani în urmă

Aratati ca M este parte stabila a multimii E in raport cu legea de compozitie specificata:
a) adunarea matricilor
M = \left[\begin{array}{ccc}x&2y\\3y&4x\end{array}\right] , x, y (apartine) N \\\\E = M_{2} (R)
b) inmultirea matricilor
M = \left[\begin{array}{ccc}2x&4y\\5y&2x\end{array}\right] , x, y (apartine) N \\E = M_{2} (R)


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Răspunsuri la întrebare

Răspuns de florin3364
1

Răspuns:

Explicație pas cu pas:

Fie M si N ∈ E

a)

M = \left[\begin{array}{cc}x&2y\\3y&4x\end{array}\right] \\\\\\N = \left[\begin{array}{cc}z&2t\\3t&4z\end{array}\right] \\\\\\

Notam a = x+z  si b = y+t. Cum x,y,z,t ∈ N, atunci si a,b ∈ N

M + N =  \left[\begin{array}{cc}x&2y\\3y&4x\end{array}\right] + \left[\begin{array}{cc}z&2t\\3t&4z\end{array}\right]  = \\\\\\=\left[\begin{array}{cc}x+z&2y+2t\\3y+3t&4x+4z\end{array}\right]  = \\\\\\=\left[\begin{array}{cc}(x+z)&2(y+t)\\3(y+t)&4(x+z)\end{array}\right]  =\\\\\\=\left[\begin{array}{cc}a&2b\\3b&4a\end{array}\right]

deci M + N ∈ E

Rezulta M este parte stabila a lui E in raport cu adunarea matricilor

b)

M = \left[\begin{array}{cc}2x&4y\\5y&2x\end{array}\right] \\\\\\N = \left[\begin{array}{cc}2z&4t\\5t&2z\end{array}\right] \\\\\\

Notam a = 2xy + 10yt  si b = 2xt + 2yz. Cum x,y,z,t ∈ N, atunci si a,b ∈ N

M * N =  \left[\begin{array}{cc}2x&4y\\5y&2x\end{array}\right] * \left[\begin{array}{cc}2z&4t\\5t&2z\end{array}\right]  = \\\\\\=\left[\begin{array}{cc}2x*2z+4y*5t&2x*4t + 4y*2z\\5y*2z + 2x*5t&5y*4t + 2x*2z\end{array}\right]  = \\\\\\=\left[\begin{array}{cc}2(2xz + 10yt)&4(2xt + 2yz)\\5(2yz + 2xt)&2(10yt + 2xz)\end{array}\right]  =\\\\\\=\left[\begin{array}{cc}2a&4b\\5b&2a\end{array}\right]

deci M * N ∈ E

Rezulta M este parte stabila a lui E in raport cu inmultirea matricilor


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