... e−1, f(0, 1) = −e−1and te−t≤ 2e 2 fort ≥ 2. Therefore |f(x, y)| ≤ (x 2 + y 2 )e−x 2 −y 2 ≤ 2e 2 < e−1for (x, y) /∈M = {(u, v) : u 2 + v 2 ≤ 2} and f cannot attain its minimum ... b). From∂f∂x(x, y) =2x(1 − x 2 + y 2 )e−x 2 −y 2 we get(1) x(1 − x 2 + y 2 ) = 0.Similarly (2) y(1 + x 2 − y 2 ) = 0.All solutions (x, y) of the system (1), (2) are (0, 0), (0, 1), (0, ... n matrixA =1 1 1 1 . . . 11 2 2 2 . . . 2 1 2 1 1 . . . 11 2 1 2 . . . 2 . . . . . . . . . . . . . . . . . . . .1 2 1 2 . . . . . .?Solution. Denote...