Thread: 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=1-3t, 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!

first of all, you didn't "finish" row-reduction, because you only reduced your matrix to upper-triangular 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

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the "other" question you ask, how do you know "when to use parameters?" depends on the RANK of your original matrix.

row-reduction is one way we find out the rank, after row-reduction, it is the number of non-zero 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 non-singular (the null space is ONLY the 0-vector (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 R³).
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 non-zero
7) we can use Cramer's Rule to solve the system of equations

Hello, Tutu!

There's an error in your first step . . .

. .

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 non-zero, that is not the case here.

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?

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I reduced it to
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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!

again, you don't seem to understand what it is you're doing when you are row-reducing. your matrix is NOT:

1)upper triangular

2)upper row-echelon form

3)reduced row-echelon form


it appears in this case you have also row-reduced incorrectly. i can obtain the same last row you did by taking R3-R2, 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!!!)

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!

it's not in row echelon form because you don't have leading 1's...for example the first non-zero entry in row 2 is 2, not 1, and the first non-zero 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.