math hellp. linear equations (solving with matrices)

**jellosalade** · March 7th 2009, 12:55 PM

Consider the system of linear equations:

2x + 4z = -2
-x + 2y - 3z = 4
x + 2y + z = k

Row- reduce the augmented matrix for the given system and then answer each of the following questions:

a) for what value(s) of k does the system of linear equations have a unique solution?
b) for what value(s) of k does the system of linear equations have no solution?
c) for what value(s) of k does the system of linear equations have infinitely many solutions?
d) for k where the system has infinitely many solutions, solve the system.

Where im having the most trouble is the row-reducing part. i was able to the x variables put in a way so that there was only 1 leading 1 in the column. however, im struggling ith the y and z. as well having the variable k there it is confusing me. please help if you can.

**zhupolongjoe** · March 7th 2009, 06:31 PM

So the augmented matrix is:

2 0 4 -2
-1 2 -3 4
1 2 1 k

I assume you know how to row reduce? The k can be treated just as any constant....i.e. "do things to it"

First things first, divide the first row by 2. Then start adding multiples of the first row to the second (since the first entry of row 2 is -1, the choice should be obvious). Continue through to row 3 after you have entry (2,1) as 0. Then after that, work backwards. Don't let k trick you, treat it like a number but remember you can put things like "k+2" if you need.

**HallsofIvy** · March 8th 2009, 06:26 AM

Originally Posted by jellosalade

Consider the system of linear equations:

2x + 4z = -2
-x + 2y - 3z = 4
x + 2y + z = k

Row- reduce the augmented matrix for the given system and then answer each of the following questions:

a) for what value(s) of k does the system of linear equations have a unique solution?
b) for what value(s) of k does the system of linear equations have no solution?
c) for what value(s) of k does the system of linear equations have infinitely many solutions?
d) for k where the system has infinitely many solutions, solve the system.

Where im having the most trouble is the row-reducing part. i was able to the x variables put in a way so that there was only 1 leading 1 in the column. however, im struggling ith the y and z. as well having the variable k there it is confusing me. please help if you can.

$\begin{bmatrix}2 & 0 & 4 & -2\\-1 & 2 & -3 & 4 \\1 & 2 & 1 & k\end{bmatrix}$
You say you got the first column correct so I presume you got to
$\begin{bmatrix}1 & 0 & 2 & -1\\0 & 2 & -1 & 3\\0 & 2 & -1 & k+1\end{bmatrix}$
I recommend you not worry about getting a "1" in the second column, second row, or about getting "0"s above the main diagonal. It is sufficient to row reduce to "triangular form" ("0"s below the main diagonal). Subtract the second row from the third row.

**Soroban** · March 8th 2009, 08:15 AM

Hello, jellosalade!

I changed the order of the equations.
. . Yes, we can do that!

Consider the system of linear equations:
. . $\begin{array}{ccc}x + 2y + z &=& k \\<br /> \text{-}x + 2y - 3z &=& 4 \\ 2x \qquad + 4z &=& \text{-}2 \end{array}$

Row-reduce the augmented matrix for the given system and then answer:

a) for what value(s) of $k$ does the system have a unique solution?
b) for what value(s) of $k$ does the system have no solution?
c) for what value(s) of $k$ does the system have infinitely many solutions?
d) for $k$ where the system has infinitely many solutions, solve the system.

We have: . $\left[\begin{array}{ccc|c} 1&2&1&k \\ \text{-}1&2&\text{-}3&4 \\ 2&0&4&\text{-}2 \end{array}\right]$

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

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

(a) The matrix has a row-of-zeros: it has no unique solution.

(b) If $2-k \neq 0\quad\Rightarrow\quad k \neq 2$ , the system has no solution.

(c) If $k = 2$ , it has infinitely many solutions.

(d) Since $k = 2$ :

we have: . $\left[\begin{array}{ccc|c}1&2&1&2 \\ 0&4&\text{-}2&6 \\ 0&0&0&0 \end{array}\right]$

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

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

We have: . $\begin{array}{ccc}x + 2z &=& \text{-}1 \\ 2y-z &=& 3\end{array}$

. . Hence: . $\begin{array}{ccc}x &=& \text{-}1 - 2z \\ y &=& \frac{3}{2} + \frac{1}{2}x \\ z &=& z\end{array}$

On the right, replace $z$ with a parameter $t.$

. . . $\begin{Bmatrix}x &=& \text{-}1-2t \\ y &=& \frac{3}{2} + \frac{1}{3}t \\ z &=& t \end{Bmatrix}$

These parametric equations represent the infinitely many solutions,
. . one for each value of $t.$

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