esame scritto II intermedio PDF

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Secondo parziale...

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2

3

  x + ky + 2t = 0   y + kz + t = 0    x + (k + 1)y + kz + (k + 3)t = 2 − k.

R4,5

U +W

k



2 k+1 k2 − 1 0



 x + ky + 2t = 0    y + kz + t = 0    x + (k + 1)y + kz + (k + 3)t = 4 − k.

R5,4

U +W

4×4

0

4×4

4

k

0 1

2

R4

U = L((1, 0, k, 0), (1, 0, 1, 0))

  x + ky + 2t = 1   y + kz + t = 2    x + (k + 1)y + kz + (k + 3)t = 3.

   x+y+z =0   y+z =0    x − z = 0

R4,5

U ∩W

k

(1, 1)

A=



1 k −k 1



k

  x + ky + 2t = 0   y + kz + t = 0    x + (k + 1)y + kz + (k + 3)t = 4 − k.

R4,5

U +W

 x + 3y = 0 x − y + 4z = −1

R4

4×4

4

U = {(x, y, z, t) : x + y − z = 0 =

0 1

2

  x + ky + 2t = 0   y + kz + t = 0    x + (k + 1)y + kz + (k + 3)t = 2 − k.

R6,3

U ∩W

4×4

4

3

R4

5

2

U = {(x, y, z, t) : 3x+2y−4z = 0 = x+y}

 x + ky + 2t = 1    y + kz + t = 2    x + (k + 1)y + kz + (k + 3)t = 3.

R5,4

U +W

4×4

3

1 2

2...