full answer: in exercises 117126, find the determinant of the matrix. expand by | studysoup
in exercises 117126, find the determinant of the matrix. expand by cofactors using the row or column that appears to make the computations easiest. 3
0
6
0
0
8
1
3
4
1
8
4
0
2
2
1
10.1 sequences polynomials: easy to evaluate, integrate, diﬀerentiate general functions: -a sequence is a list of #s -sequences are written as: -many series correspond to functions formally, the sequence converges to l if for every positive number , there corresponds an integer n such that for all
finding the correct answer here is important, but perhaps equally important is finding it using the method of matrix reduction, since those are the instructions.the augmented matrix is4 -2 -5 -514 -1 -2 -364 -2 4 -24you'll have to forgive the poor formatting. this answer box doesn't have a proper math symbol editor.using elementary row operations, the reduced row echelon form of the augmented matrix is1 0 0 -60 1 0 60 0 1 3from which we conclude that the system has a unique solution (-6,6,3). 1) p(x) = 1/x2 + 1 if x ≤ 3 it is clear that x= 3 is included, so you have to use this function.2) loo k at table d, and compare the given information with the data written inside the table. it's not clear what it is you want as an output. if it's a row vector, there's no concatenation involved, you just simply reshape your matrix:
a = [0 1 1 1;
 1 0 1 0;
 1 0 0 1]
 v = reshape(a, 1, []) %reshape a into one row. since matlab is column major, this is done column-wise to find: limitsnote: first check the form of imit. used this method if the limit is satisfying any one from 7 indeterminate form.in this case, indeterminate form is formula used: = 1so, by using the above formula, we have = = therefore, = Notice that the matrix A is upper-triangle, so its determinant is a  
product   
of diagonal elements:  
  
det(A) = 1*2*3 = 6.  
  
Moreover, det(A) is   
non-zero.  
Therefore the columns of A are linearly independent and   
therefore  
they constitutive a basis for R^3, so R^3 = Col(A).  
  
Hence   
any base of R^3 fits to be a base for Col(A),   
for instance the columns of   
A:  
  
(1,0,0), (2,2,0), (3,3,3)  
  
or   
  
(1,0,0), (0,1,0),   
(0,0,1)  
  
are the base.  
  
The dimension of Col(A) is the dimension of   
R^3 and therefore is equal  
to 3. applying and factorization, we get, also changing by reversing n and dnote: - try to factorize that term which help in reducing expression. >> [x,y] = ismember(a,b);
>> [~,z] = sort(y(x));
>> v = find(x);
>> r = reshape(v(z),2,[]).'
r =
 2 9
 4 8
 5 7 perhaps what you are looking for is in objective and nonlinear constraints in the same function.
alan weiss
matlab mathematical toolbox documentation we know that- - - - - (i)taking 3 as common, we get3; solving only bracket term first,here a = 2a, b = 3b; putting values in eq.i}applying }}ans: - 3}note: must memorize cubes upto 12 [@cssc](/u/cssc), another option would be to take the regex way.

first, it finds the index of the 5th pipe.

then, it takes the substring until that index.

give this a try:

inputstr.substring(0, regex.matches(inputstr, "\|")(4).index + 1)