find the matrix `x` such that `ax=b` where `a=|[1,2],[-1,3]| , b=|[0,1],[2,4]|`
solution: ` x = {:((-4/5,-1),(2/5,1)):}`
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). a - ii, b - iv, c - i, d - iii find zeroes of the following polynomials (with steps ) (i) p(x)=6x^2-3-7x (ii) f(x)=2-x^2 let put a(x2+4)+(bx+c)(2x+1) =17put 2x+1=0a=4by equating the coefficient of x2 and constant term,a+2b=04+2b=0b=-24a+c=174×4+c=17c=1from the equation(1), we get, ?solution 
step 1:
we have to draw the graph of form r to r. 
step 2:
the graph of f is the set of all points (x,f(x)) with x in the domain of f.
we take x in [-4,4]
and we will find the separate values of y for separate intervals.
 
 
 

 
......(1)
 

 

 

using x values along x-axis and along y-axis we get the graph of given function form r to r.
step 3:
the graph is as follow y = 3x

It can be observed that x = −1, y = −3 and x = 1, y = 3 are solutions of the above equation. Therefore, the solution table is as follows.



| x | -1 | 1 |
| y | -3 | 3 |

The graph of the above equation is constructed as follows.

![](https://www.sarthaks.com/?qa=blob&qa_blobid=15681428413536225923) 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 the range of the function f(x) = |x -&nbsp;1| is formula :- therefore ,put -2x dx = dt (i) co-ordinates of point p are `((1xx17+2xx(-1))/(1+2),(1xx10+2xx1)/(1+2))` `=((17-8)/3,(10+2)/3)` `=(9/3,12/3)` `=(3,4)` (ii) `op=sqrt((0-3)^2+(0-4)^2)` `op=sqrt(9+16)` `op=sqrt25` `op=5` units (iii) let ab be divided by the point p(0, y) lying on y-axis in the ratio `therefore (0,y)=((kxx17+1xx(-4))/(k+1),(kxx10+1xx1)/(k+1))` `=>(0,y)=((17k-4)/(k+1),(10k+1)/(k+1))` `=>0=(17k-4)/(k+1)` `=>17k-4=0` `=>k=4/17` thus, the ratio in which the y-axis divide the line ab is 4: 17.