Recent content by Tim 1234

If you start over, all of your current scores are discarded. The trials consist of 6 independent rolls of three dice. So, you have a set of 6 sums of the roll of three dice. So to get two "18s" you would need to roll three sixes twice in 6 attempts at rolling three dice.

Homework Statement In playing a certain game, your ability scores are determined by six independent rolls of three dice. After each set of six rolls, you are given the choice of keeping your scores or starting over. (a) How many times should you expect to start over in order to get a set of...

Is the hint saying AB can be represented as a linear combination of the vector columns of B where the coefficients for each column are the matrix A. And because A is non-zero and AB=0, the only way for the linear combination AB1+AB2+...+ABn=0 to be valid is if each column of B is composed of...

Okay, all of the information I can deduce from the question: B=[B1, ..., Bn] AB=[AB1, ..., ABn] These were hints given in the question - I can't determine what their relevance is. Also from the question, A is non-singular. From this we know A is not the zero-vector (the zero-vector is...

If Av=0 for any non-zero vector v, then A is singular. Under the assumption that B is not the zero matrix, there exists and x such that v=(Bx) is not 0. If this is the case, and A(Bx)=Av=O, then A is singular. But the question states A is non-singular. So when we assume B not to be the...

Assuming B is not the zero matrix, there is an x such that Bx is not 0. We know by virtue of A being non-singular that A is not the zero matrix. Isn't it possible that there exists an x such that Bx is not 0, but A(Bx)=0?

Denoting the elements of B1 by x1, ..., xn would render all elements of B1 equal to 0? Then we could do this with every column of B, making B an nxn zero matrix? Edit: This problem is from a section before invertibility and determinants.

By theta, I mean the zero vector in R^n. So x1,...,xn are all equal to 0 because A1...An are linearly independent. Doesn't this imply that A is not a zero matrix? If A was a zero matrix, x1, ..., xn would not need to equal 0 such that x1A1+...+xnAn = 0. x1, ..., xn could take on any value...

Homework Statement Suppose A is non-singular (nxn) matrix. Given that AB=(zero matrix), show that B=(zero matrix). Hint: Express A as [A1, A2, ..., An] and Express AB as [AB1, AB2, ..., ABn] Homework EquationsThe Attempt at a Solution [/B] I argued that because A is non-singular, A=[A1...

S={v1, v2, v3} v1=[1, -1], v2=[-2, 2], v3=[3, a] a) For what value(s) a is the set S linearly dependent? b)For what value(s) a can v3 be expressed as a linear combination of v1 and v2? a) p=3 and m=2 3-2=1 free variable Therefore the set has non-trivial solutions and is linearly dependent b)...