questions on attachment!

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- Jul 23rd 2006, 06:28 PM #1

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- Jul 23rd 2006, 09:30 PM #2

- Jul 23rd 2006, 09:35 PM #3
## problem nr. 5 only

Originally Posted by**Lane**

it's me again.

Use the same method as in prob. nr. 4.:

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

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

Greetings

EB

- Jul 23rd 2006, 09:52 PM #4
## problem nr. 2 only

Originally Posted by**Lane**

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

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

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

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

Now calculate R3+4*R2:

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

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

- Jul 23rd 2006, 10:56 PM #5

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## Problem #3 only

The augmented matrix is equivalent to the system of equations:

$\displaystyle

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

RonL

- Jul 23rd 2006, 11:07 PM #6

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## Problem #1 only

Let:

$\displaystyle

M=\left[

\begin{array}{ccc}

3&10&-3\\

1&5&-9\\

-4&-7&1

\end{array} \right]

$

A: Swap rows 1 and 2 gives:

$\displaystyle

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

$\displaystyle

M''=\left[

\begin{array}{ccc}

1&5&-9\\

3-3&10-15&-3+27\\

-4&-7&1

\end{array} \right]=$$\displaystyle

\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