A/

1) a) not a vector space, axiom 8 fails.

b) yes it's a vector space, it is the xy-plane.

c) not a vector space, axioms 2, 3, 4 and 8 fail.

d) yes it's a vector space.

2/ ( x₁ , x₂ ) x is in R, x₃ = 0 -- it is a 3-dimensional vector with the z-component = 0.

3/ Row reduction on the matrix should lead to a row like ,
so the system is inconsistent. There are no solutions.

B/ 1)

a) dependent set -- 3 vectors in R²	b) independent set -- det (A) = 49 ! 0
c) dependent set -- 4 vectors in R³	d) dependent set -- includes the zero vector

2) The normal vector of the plane we want is u % v which is .

So the equation of the plane is: 8(x – 1) – 7(y – 1) + (z + 1) = 0 or 8x – 7y + z = 0.

C/ 1) a) The row-echelon form of the system matrix is:

these 2 vectors form a basis for the solution space of this homogeneous system.

b) dimension = 2

c) Since there are 5 components in each vector, they come from R ⁵,
and we can show that the set is closed under addition and under scalar multiplication.
That is: add any two solution vectors and we'll still have a solution vector and
multiply any solution vector by a nonzero scalar k, and we'll still have a solution vector.

2) a) a basis for the row space is

b) a basis for the column space is i, j, k the standard basis vectors for R³.

c) Rank = 3 since there are 3 vectors in a basis.

d) Since M is a 3 x 5 matrix, it can only have a maximum rank of 3 -- the # of rows.
Therefore the 5 column vectors form a linearly dependent set since
there are more vectors than the rank or the number of basis vectors.

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