very confused matrices

**i_zz_y_ill** · May 3rd 2008, 07:46 AM

STUCK ON PART (III) hellp

(i) find the inverse of 4 1 k , k doesnt =5
3 2 5
8 5 13
DONE
(ii) solve the smultaneous equation

4x + y + 7z = 12
3x + 2y + 5z = m
8x + 5y +13z = 0
giving x,y,z in terms of m,,,DONE i got x= -6-11m
y= -6+2m
z=6+6m correct i think
(iii) find the value of p for which the simultaneous equations have solutions, and fn general solution in this case

4x+y+5z=12
3x+2y+5z=p
8x+5y+13z=0

how does it have solutions??,ithought this was just det doesnt = 0,,confused as to how im supposed to do this part when k doesnt = 5 either,,,,,,help pls

**Soroban** · May 3rd 2008, 08:50 AM

Hello, i_zz_y_ill!

(iii) Find the value of $p$ for which the system has solutions.

. . $\begin{array}{ccc}4x+y+5z&=& 12 \\<br /> 3x+2y+5z&=&p \\ 8x+5y+13z& =& 0 \end{array}$

We have: . $\left| \begin{array}{ccc|c} 4 & 1 & 5 & 12 \\ 3 & 2 & 5 & p \\ 8 & 5 & 13 & 0 \end{array}\right|$

$\begin{array}{c}R_1-R_2 \\ \\ R_3-2R_1\end{array}\left|\begin{array}{ccc|c}1 & \text{-}1 & 0 & 12-p \\ 3 & 2 & 5 & p \\ 0 & 3 & 3 & \text{-}24 \end{array}\right|$

$\begin{array}{c}\\ R_2 - 3R_1 \\ R_3\div3 \end{array}\left|\begin{array}{ccc|c}1 & \text{-}1 & 0 & 12-p \\ 0 & 5 & 5 & 4p-36 \\ 0 & 1 & 1 & \text{-}8 \end{array}\right|$

$\begin{array}{c}\text{Switch} \\R_2\text{ and }R_3\\ \end{array}<br /> \left|\begin{array}{ccc|c}1 & \text{-}1 & 0 & 12-p \\ 0 & 1 & 1 & \text{-}8 \\ 0 & 5 & 5 & 4p-36 \end{array}\right|$

$\begin{array}{c}R_1+R_2 \\ \\ R_3-5R_2\end{array}<br /> \left|\begin{array}{ccc|c}1 & 0 & 1 & 4-p \\ 0 & 1 & 1 & \text{-}8 \\<br /> 0 & 0 & 0 & 4p+4 \end{array}\right|$

The last row of the matrix is all zeros.
To have a solution, the last term must also be zero. **

. . $4p + 4 \:=\:0\quad\Rightarrow\quad\boxed{ p \:=\:-1}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We have: . $\begin{array}{ccc}x + z &=& 5 \\ y + z &=&\text{-}8 \end{array} \quad\Rightarrow\quad \begin{array}{ccc} x &=& 5 - z \\ y &=& \text{-}8-z \\ z&=&z \end{array}$

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

And we have: . $\begin{Bmatrix}x &=& 5-t \\ y &=&\text{-}8-t \\ z &=& t \end{Bmatrix}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

If the last row were something like: . $0\;\;0\;\;0\;\;|\;\;3$

. . the system would have no solution . . . remember?

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