Solution

Solution

We compute the parametric vector form of the solutions of Ax=0. The reduced row echelon form of A is

The free variables are x₃ and x₄;the parametric form of the solution set is

Therefore,

Solution

k = 2, -1

Solution

Solution

suppose a, b, c ∈ R are constants so that

Then, after distributing and combining terms, we see that

Since the coefficients of x2 and x3 must be zero, we see that a = 0 and c = 0. But then the above reduces to

so b = 0 as well. Therefore, the only way to write 0 as a linear combination of 1+ x + x^2 , 1−x, and

1 − x^3 is if all the coefficients are zero, which means the set is linearly independent

Solution

We will first compute

Therefore,

is a basis for N(A)

Since the row reduced matrices for [A∣0] and [A^TA∣0] are identical, we can immediately conclude that

and that

is a basis for N(A^TA)

It follows that the nullities of A and A^TA are both 1

The rank-nullity theorem tells

rank of A+nullity of A=3.

Hence the rank of A is 2.Similarly, the rank of A^TA is 2

Solution

Solution

Solution

The Wronskian of S is

Solution

The reduced echelon form of the coefficient matrix is in the form

So the reduced system is x = −2y−4s+3t, z = −s+t.

is a basis of W = 3

Solution

According to this question, we have to calculate the row space of the given matrix:

Solution

Consider the system

once the above matrix has leading 1’s in the first and third columns, we can conclude that the first and third vectors of S form a basis of Span(S). Thus

is a basis for Span(S)

Solution

Solution

1. The existence of 0 is a requirement in the definition.
2. That’s not an axiom, but you can prove it from the axioms. Suppose that z acts like a zero vector, that is to say, v + z = v for every vector v. Then in particular, 0 + z = 0. But z = 0 + z. Therefore, z = 0. Thus there can be only one vector with the properties of a zero vector.
3. It’s only true if you add the hypothesis that x != 0.
4. It’s only true if you add the hypothesis that a != 0.

Solution

The question is, how many different ways and you fill in an m × n matrix with 0’s and 1’s? For example, if you have a 2 × 3 matrix, you have 6 entries, three in the first row and three in the second. Each one of those 6 entries can be 0 or 1. So the total number of ways to fill in the matrix is 2 · 2 · 2 · 2 · 2 · 2 · 2, which is 2^6 . There are mn entries in an m × n matrix, and each can be 0 or 1, so that makes 2^(mn) matrices.

So for m=3,n=4, number of matrices: 2^(3*4) = 4096