Recent content by negation

Alright I was sloppy and careless with words. The matrix has entries aaa, abb ...ann and has zeroes everywhere else.

I admit it was sloppy use of words. The correct word would be 'element'. It's fuzzy to me as to why the vector (1,1,1) was chosen. Could we have chosen (2,2,2)?

Let's stick to " For a matrix in REF, the non-zero row with the entry aaa is to the left of the non-zero row with the entry abb to the left of entry acc to the left of add and so on to ann"

Yea it is. It's probably shaky understanding of the concept and not employing the appropriate terminology.

For a matrix in REF, the non-zero row with the entry aaa is to the left of the non-zero row with the entry abb to the left of entry acc to the left of add and so on to ann Tell me if there's a better way of sub-scripting the entries. Edit: I could have been confused by your question. a3ij is...

For a matrix in REF, the non-zero row with the entry aij is to the left of the non-zero row with the entry a2ij to the left of entry a3ijto the left of a4ij and so on to anij.

Homework Statement Given 2 1 1 0 0 0 1 1 0 0 0 3(i) Show that the rows of A are linearly independent. (ii) Show that the nonzero rows of any matrix in row echelon form are linearly independent.The Attempt at a Solution i) REF gives 1 0 0 | 0 0 1 0 | 0 0 0 1 | 0 0 0 0 | 0 x1 = x2 = x3 =x4 =...

With respect to the question, the solution set has 1 free variable. Thus, the solution set has infinitely many solutions. And so, span {(1; 1; 0);(0; 0; 2);(0; 0; 1);(1; 2; 3)} is a spanning set for R4. Keeping this in mind, I could have equally, instead of expressing [a;b;c;d] as a linear...

I should have known not to complicate the answer by setting 1/k = y. But I have this one. If k = 1, we have 1(1-1) | (-1+1) and this produces 0 = 0. We have an infinite solution set. 0x3 = 0 setting x3 = y we see that via back substituition x2 = 1-2y x1 = [2+2(1-2y)]/-4 :-p (lazy...

I'm unsure if I understood the question rightly. There would hardly be any substance in my workings if my understanding of the question is misguided. I assume the question is asking me to construct a vector whose coordinates are element of integers and subject them to the the condition of...

It is clear now. I was earlier able to work to the part where k(1-k)x3 = (-k+1) simplifying produces x3 = 1/k but was confused as to what I should do with what I had especially since it conflicted with what my answer key showed. In response to your question, x3 = 1/k. If x3 = y, 1/k = y. ∴ x3...

Homework Statement Is {(1; 1; 0);(0; 0; 2);(0; 0; 1);(1; 2; 3)} a spanning set for R3The Attempt at a SolutionThis is supposed to be easy but the answer sheet might be wrong. The answer I have says it is and then proceed to say that (0;0;2) is linearly independent. But it isn't because...

Homework Statement For a set of vectors in R3, is the set of vectors all of whose coordinates are integers a subspace?The Attempt at a Solution I do not exactly understand if I should be looking for a violation or a universal proof. If x,y, z \in Z then x,y,z can be writted as...

I'm still not capturing the full essence of part (1). Maybe some simple examples? part(3): if k ≠ 1 and k ≠ 0 then we can pick any values of k? But what is the corollary from this?

Homework Statement -4x1 +2x2 =2 4x1 -3x2 -2x3 =-3 2x1 -x2 +(k - k2)x3 =-k Find the values of k for which the system has 1) unique solution, 2) infinitely many solution and 3) no solution The Attempt at a Solution In REF: the matrix is -4 2 0 | 2 0 -1 -2 | -1 0 0 k(1-k)...