
Matrix again
Hi I really need help!
I did the questions but something went wrong in my working, can someone please show me how to derive the answer with row reduction?! I really am confused when to express my answer in parametric equations!
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After Row reduction I got
\begin{bmatrix} x \\y \\z \end{bmatrix}[/tex] =
For Row two, I took row 2  2 times row 1
FOr Row 3, I took row 3 minus Row 1
So, since I did not get all zeros ir at least three zeros in the bottom entries, I resorted to represent unique solutions in parametric equations, I got y=13t, z=t and x=3+t..however, the answer is in whole numbers for each x, y and z. And when I pressed the matrix into the calculator to solve for x, y and z, I got whole numbers too..
How do I know that I am to use parametric equations to solve? And how do I get the whole numbers in this case?
Thank you so much, your help is much appreciated!

Re: Matrix again
first of all, you didn't "finish" rowreduction, because you only reduced your matrix to uppertriangular form (and not row echelon, or reduced row echelon form, which makes the solutions more transparent).
but look, your last row means that:
0x + 0y  3z = 3 that is:
3z = 3
z = 1 <not a parameter, a VALUE.
your second row means:
0x + 2y + 6z = 2 that is:
2y + 6z = 2
but we already know that z = 1! so this becomes:
2y + 6 = 2
2y = 4
y = 2 <again, not a parameter, an actual number.
finally, your top row means:
x + 4y + 11z = 7
since y = 2, ans z = 1, this is:
x  8 + 11 = 7
x + 3 = 7
x = 4
*************************
the "other" question you ask, how do you know "when to use parameters?" depends on the RANK of your original matrix.
rowreduction is one way we find out the rank, after rowreduction, it is the number of nonzero rows remaining. in this case, the rank of the original matrix is 3.
this means several things, all of which are equivalent:
1) the matrix A in your original problem is invertible
2) the matrix A is nonsingular (the null space is ONLY the 0vector (0,0,0)) that is A(x,y,z) = (0,0,0) implies x = y = z = 0.
3) the column space of A has dimension 3 (it is thus ALL of R^{3}).
4) the columns of A are linearly independent
5) the equation A(x,y,z) = (7,9,4) has a UNIQUE solution
6) the determinant of A is nonzero
7) we can use Cramer's Rule to solve the system of equations

Re: Matrix again
Hello, Tutu!
There's an error in your first step . . .
. .

Re: Matrix again
i don't see any "error", the upper triangular matrix you obtain is the same one he did. this is sufficient to solve the system of equations, since a linear equation in one unknown over a field always has a solution, unless it is of the form:
0x + b = c, with b and c unequal. since the last row is nonzero, that is not the case here.

Re: Matrix again
Thank you so much! I understand it!
I tried it on another question, I got the only 1 out of 3 answers, could you help me to derive the next few?
=
I reduced it to
=
So I got y=1.
However, as I tried to further reduce the matrix like Soroban did, I cannot seem to get the values for x and z individually, I could only get x+z = 7.
Please help me out in solving for x and z!
Thank you so so so much both!

Re: Matrix again
again, you don't seem to understand what it is you're doing when you are rowreducing. your matrix is NOT:
1)upper triangular
2)upper rowechelon form
3)reduced rowechelon form
it appears in this case you have also rowreduced incorrectly. i can obtain the same last row you did by taking R3R2, but its not kosher to then subtract R3 from R2.
instead, you need to subtract 2*R1 from R2, giving:
now we can add (1/2)*R2 to R3, and we get:
again, we have a matrix of rank 3, so we don't need parameters.
the last equation is:
z/2 = 5/2
so z = 5
the second equation is now:
2y  z = 7
2y  5 = 7
2y = 2
y = 1
the first equation is now:
x + y + z = 6
x  1 + 5 = 6
x = 2
we can check this:
2  1 + 5 = 6 (check!)
2(2) + 4(1) + 5 = 4  4 + 5 = 5 (check!!)
2(2) + 3(1) + 5 = 4  3 + 5 = 6 (check!!!)

Re: Matrix again
Thank you again!
May I clarify, prior to the first question I asked, how is not a row echelon?
From what I know, row echelon is when there should be zeros below leading ones, which exists here..
Doe sthat mean that each time I do row reduction, I either change it to role echelon OR reduced row echelon?
Thank you!

Re: Matrix again
it's not in row echelon form because you don't have leading 1's...for example the first nonzero entry in row 2 is 2, not 1, and the first nonzero entry in row 3 is 3, not 1.
to put it in row echelon form, you would have to multiply row 2 by 1/2, and row 3 by 1/3.
to put it in REDUCED row echelon form (rref), you would need to clear out the entries ABOVE the leading 1's as well.