# Matrix again

• Jan 21st 2013, 05:55 PM
Tutu
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!

$\displaystyle \begin{bmatrix} 1 & 4 & 11 \\1 & 6 & 17 \\1 & 4 & 8 \end{bmatrix}\begin{bmatrix} x \\y \\z \end{bmatrix}$ =

=$\displaystyle \begin{bmatrix} 7 \\9 \\4 \end{bmatrix}$

After Row reduction I got

$\displaystyle \begin{bmatrix} 1 & 4 & 11 \\0 & 2 & 6 \\0 & 0 & -3 \end{bmatrix}$\begin{bmatrix} x \\y \\z \end{bmatrix}[/tex] =

$\displaystyle \begin{bmatrix} 7 \\2 \\-3 \end{bmatrix}$

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!
• Jan 21st 2013, 06:31 PM
Deveno
Re: Matrix again
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.

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.

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.

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
• Jan 21st 2013, 07:58 PM
Soroban
Re: Matrix again
Hello, Tutu!

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

Quote:

$\displaystyle \begin{bmatrix} 1 & 4 & 11 \\1 & 6 & 17 \\1 & 4 & 8 \end{bmatrix}\begin{bmatrix} x \\y \\z \end{bmatrix} \;=\; \begin{bmatrix} 7 \\9 \\4 \end{bmatrix}$

$\displaystyle \text{We have: }\;\left[\begin{array}{ccc|c} 1&4&11&7 \\ 1&6&17 & 9 \\ 1&4&8&4 \end{array}\right]$

$\displaystyle \begin{array}{c}\\R_2-R_1 \\ R_3-R_1 \end{array}\;\left[\begin{array}{ccc|c}1&4&11 & 7 \\ 0&2&6&2 \\ 0&0&\text{-}3 & \text{-}3 \end{array}\right]$

. . $\displaystyle \begin{array}{c}\\ \;\frac{1}{2}R_2 \\ \text{-}\frac{1}{3}R_3 \end{array}\;\;\left[\begin{array}{ccc|c}1&4&11&7 \\ 0&1&3&1 \\ 0&0&1&1 \end{array}\right]$

$\displaystyle \begin{array}{c}R_1-4R_2 \\ R_2-3R_3 \\ \\ \end{array} \left[\begin{array}{ccc|c}1&0&\text{-}1 & 3 \\ 0&1&0&\text{-}2 \\ 0&0&1&1 \end{array}\right]$

$\displaystyle \begin{array}{c}R_1+R_3 \\ \\ \\ \end{array}\;\; \left[\begin{array}{ccc|c}1&0&0&4 \\ 0&1&0&\text{-}2 \\ 0&0&1&1 \end{array}\right]$

$\displaystyle \text{Therefore: }\:\begin{Bmatrix}x &=& 4 \\ y &=& \text{-}2 \\ z &=& 1 \end{Bmatrix}$
• Jan 21st 2013, 09:37 PM
Deveno
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 non-zero, that is not the case here.
• Jan 21st 2013, 09:50 PM
Tutu
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?

$\displaystyle \begin{bmatrix} 1 & 1 & 1 \\2 & 4 & 1 \\2 & 3 & 1 \end{bmatrix}\begin{bmatrix} x \\y \\z \end{bmatrix}$ = $\displaystyle \begin{bmatrix} 6 \\5 \\6 \end{bmatrix}$

I reduced it to
$\displaystyle \begin{bmatrix} 1 & 1 & 1 \\0 & 1 & 0 \\0 & -1 & 0 \end{bmatrix}\begin{bmatrix} x \\y \\z \end{bmatrix}$ = $\displaystyle \begin{bmatrix} 6 \\-1 \\1 \end{bmatrix}$

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.

Thank you so so so much both!
• Jan 21st 2013, 10:36 PM
Deveno
Re: Matrix again
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:

$\displaystyle \begin{bmatrix}1&1&1\\0&2&-1\\0&-1&0 \end{bmatrix} \begin{bmatrix}x\\y\\z \end{bmatrix} = \begin{bmatrix}6\\-7\\1 \end{bmatrix}$

now we can add (1/2)*R2 to R3, and we get:

$\displaystyle \begin{bmatrix}1&1&1\\0&2&-1\\0&0&-\frac{1}{2} \end{bmatrix} \begin{bmatrix}x\\y\\z \end{bmatrix} = \begin{bmatrix}6\\-7\\ -\frac{5}{2} \end{bmatrix}$

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!!!)
• Jan 22nd 2013, 02:10 AM
Tutu
Re: Matrix again
Thank you again!

May I clarify, prior to the first question I asked, how is $\displaystyle \begin{bmatrix} 1 & 4 & 11 \\0 & 2 & 6 \\0 & 0 & -3 \end{bmatrix}$ 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!
• Jan 22nd 2013, 02:28 AM
Deveno
Re: Matrix again
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.