Help ME Please!!!

**Lane** · July 23rd 2006, 06:28 PM

questions on attachment!

**earboth** · July 23rd 2006, 09:30 PM

Originally Posted by Lane

questions on attachment!

Hello, Lane,

use Cramer's rule:

$\left|\begin{array}{cc}5&9\\7&-10\end{array}\right|=5\cdot (-10)-9\cdot7=-113$ . So it's answer C.

Greetings

EB

**earboth** · July 23rd 2006, 09:35 PM

Originally Posted by Lane

questions on attachment!

Hello, Lane,

it's me again.
Use the same method as in prob. nr. 4.:

$\left|\begin{array}{cc}x-9&2\\3&x-4\end{array}\right|=0$

$(x-9)(x-4)-6=0\ \Longleftrightarrow\ x^2-13x+30=0$ . Solving this quadratic equation you should get x= 3 or x = 10.

Greetings

EB

**earboth** · July 23rd 2006, 09:52 PM

Originally Posted by Lane

questions on attachment!

Hello, Lane

(I assume that there is a typo in the first row)

$\begin{array}{r}x+3y+z=-12\\2x+7y-2z=-25\\x-y-6z=15\end{array}$

Calculate: R2-2*R1 and R3-R1. You'll get:

$\begin{array}{r}x+3y+z=-12\\y-4z=-1\\-4y-7z=27\end{array}$

Now calculate R3+4*R2:

$\begin{array}{r}x+3y+z=-12\\y-4z=-1\\-23z=23\end{array}$

So you get z = -1. Now re-substitute this value into equation 2 and afterward plug in the values for y and z into the first equation. You should get: [4, -5, -1]

Greetings

EB

**CaptainBlack** · July 23rd 2006, 10:56 PM

The augmented matrix is equivalent to the system of equations:

$ \begin{array} {ccccc} 1x&+0y&-1z&=&-5\\ 0x&+1y&-3z&=&4\\ 0x&+0y&+0z&=&0 \end{array} $

Now check the candidate solutions to see which satisfy these equations.

The first equation is satisfied by those candidate solutions with $x=t-5,\ z=t$ , and the second by those candidates with $y=3t+4,\ z=t$ , so it looks like (C) is the required answer.

RonL

**CaptainBlack** · July 23rd 2006, 11:07 PM

Let:

$ M=\left[ \begin{array}{ccc} 3&10&-3\\ 1&5&-9\\ -4&-7&1 \end{array} \right] $

A: Swap rows 1 and 2 gives:

$ M'=\left[ \begin{array}{ccc} 1&5&-9\\ 3&10&-3\\ -4&-7&1 \end{array} \right] $

B: Replace row 2 by row 2 plus -3 times row 1

$ M''=\left[ \begin{array}{ccc} 1&5&-9\\ 3-3&10-15&-3+27\\ -4&-7&1 \end{array} \right]=$ $ \left[ \begin{array}{ccc} 1&5&-9\\ 0&-5&25\\ -4&-7&1 \end{array} \right] $

I am sure you can now do part C yourself.

RonL

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