Răspuns:
Explicație pas cu pas:
Fie M si N ∈ E
a)
[tex]M = \left[\begin{array}{cc}x&2y\\3y&4x\end{array}\right] \\\\\\N = \left[\begin{array}{cc}z&2t\\3t&4z\end{array}\right] \\\\\\[/tex]
Notam a = x+z si b = y+t. Cum x,y,z,t ∈ N, atunci si a,b ∈ N
[tex]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][/tex]
deci M + N ∈ E
Rezulta M este parte stabila a lui E in raport cu adunarea matricilor
b)
[tex]M = \left[\begin{array}{cc}2x&4y\\5y&2x\end{array}\right] \\\\\\N = \left[\begin{array}{cc}2z&4t\\5t&2z\end{array}\right] \\\\\\[/tex]
Notam a = 2xy + 10yt si b = 2xt + 2yz. Cum x,y,z,t ∈ N, atunci si a,b ∈ N
[tex]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][/tex]
deci M * N ∈ E
Rezulta M este parte stabila a lui E in raport cu inmultirea matricilor
