Matrices and Gaussian Method

[Jaskaran]

Hi all, I have a problem that I can't seem to figure out using the Gaussia method!

Write the agumented matrix for the system

\(\displaystyle \begin{bmatrix}3x + 2y - 5z = 17 \\ 2x - 3y +2z = -8 \\ 4x +5y - 3z = 20 \end{bmatrix}\)

I know the variables, but it's almost impractical to convert \(\displaystyle R_1_1\) into a 1 without converting the rest into fractions. :!:

Any help?

Also, how do I find the equation of a parabola whose vertex is the point (2,5) and whose y-intercept is the point (0,1)? My book simply does not have such example!

Thank you,

Jaskaran.

 

[Deleted member 4993]

Hi all, I have a problem that I can't seem to figure out using the Gaussia method!

Write the agumented matrix for the system

\(\displaystyle \begin{bmatrix}3x + 2y - 5z = 17 \\ 2x - 3y +2z = -8 \\ 4x +5y - 3z = 20 \end{bmatrix}\)

I know the variables, but it's almost impractical to convert \(\displaystyle R_1_1\) into a 1 without converting the rest into fractions. <<< So what is the problem with that??!! :!:

Any help?

Also, how do I find the equation of a parabola whose vertex is the point (2,5) and whose y-intercept is the point (0,1)? My book simply does not have such example!

What is the general equation of a parabola whose vertex is at (h,k)?

Thank you,

Jaskaran.

 

[mmm4444bot]

... it's almost impractical ... without ... fractions

Do you need help with fractions?

On the second exercise, you could substitute the xy-coordinates of three known points on the parabola into the standard quadratic equation (replacing zero with y, as shown).

y = ax^2 + bx + c

Solve the resulting system of three equations for a, b, and c.

(Hint: use symmetry about the line x = 2 to find the third point.)

 

[Jaskaran]

The generation equation for a parabola whose vertex is (h,k) is \(\displaystyle f(x)=af(x-h)+k\)

I know what h and k are, so I get \(\displaystyle f(x)=af(x-2)+5\)

My question is, is the y intercept the y-value in this case? So would I get \(\displaystyle 1 = (x-2) + 5\)?

y = ax^2 + bx + c

Also, since vertex which also has the equation of \(\displaystyle x = -b/2a\), would I need to know what b and a are? How do I figure this out?

 

[Deleted member 4993]

The generation equation for a parabola whose vertex is (h,k) is \(\displaystyle f(x)=a\cdot (x-h)^2+k\)

I know what h and k are, so I get \(\displaystyle f(x)=a(x-2)^2+5\)

My question is, is the y intercept the y-value in this case? So would I get \(\displaystyle 1 = a(0-2)^2 + 5\)

Now find 'a'
?
posting.php?mode=quote&f=9&p=123674#

y = ax^2 + bx + c

Also, since vertex which also has the equation of \(\displaystyle x = -b/2a\), would I need to know what b and a are? How do I figure this out?

 

[soroban]

Hello, Jaskaran!

Did your teacher tell you that the only way to get a "1" is to divide?

Sometimes we cannot avoid fractions, but we can minimize their occurance.

Write the agumented matrix for the system

. . \(\displaystyle \begin{array}{ccc}3x + 2y - 5z &=& 17 \\ 2x - 3y +2z &=& \text{-}8 \\ 4x +5y - 3z &=& 20 \end{array}\)

\(\displaystyle \text{We have: }\;\left[\begin{array}{ccc|c} 3 & 2 & \text{-}5 & 17 \\ 2 & \text{-}3 & 2 & -8 \\ 4 & 5 & \text{-}3 & 20 \end{array}\right]\)

\(\displaystyle \begin{array}{c}R_1-R_2 \\ \\ \\ \end{array} \left[\begin{array}{ccc|c}1 & 5 & \text{-}7 & 25 \\ 2 & \text{-}3 & 2 & \text{-}8 \\ 4 & 5 & \text{-}3 & 20 \end{array}\right]\)

\(\displaystyle \begin{array}{c} \\ R_2-2R_1 \\ R_3-4R_1\end{array} \left[\begin{array}{ccc|c}1 & 5 & \text{-}7 & 25 \\ 0 & \text{-}13 & 16 & \text{-}58 \\ 0 & \text{-}15 & 25 & \text{-}80 \end{array}\right]\)

\(\displaystyle \begin{array}{c} \\ R_2-R_3 \\ \\ \end{array} \left[\begin{array}{ccc|c}1& 5 & \text{-}7 & 25 \\ 0 & 2 & \text{-}9 & 22 \\ 0 & \text{-}15 & 25 & \text{-}80 \end{array}\right]\)

. . . \(\displaystyle \begin{array}{c}\\ \frac{1}{2}R_2 \\ \\ \end{array} \left[\begin{array}{ccc|c} 1 & 5 & \text{-}7 & 25 \\ 0 & 1 & \text{-}\frac{9}{2} & 11 \\ 0 & \text{-}15 & 25 & \text{-}80 \end{array}\right]\)

\(\displaystyle \begin{array}{c}R_1-5R_2 \\ \\ \\ R_3 + 15R_2\end{array} \left[\begin{array}{ccc|c}1 & 0 & \frac{31}{2} & \text{-}30 \\ \\[-3mm] 0 & 1 & \text{-}\frac{9}{2} & 11 \\ \\[-3mm] 0 & 0 & \text{-}\frac{85}{2} & 85 \end{array}\right]\)

. . .\(\displaystyle \begin{array}{c}\\ \\ \text{-}\frac{2}{85}R_3 \end{array} \left[\begin{array}{ccc|c} 1 & 0 & \frac{31}{2} & \text{-}30 \\ \\[-3mm] 0 & 1 & \text{-}\frac{9}{2} & 11 \\ 0 & 0 & 1 & \text{-}2 \end{array}\right]\)

\(\displaystyle \begin{array}{c}R_1-\frac{31}{2}R_3 \\ R_2 + \frac{9}{2}R_2 \\ \\ \end{array} \left[\begin{array}{ccc|c}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & \text{-}2 \end{array}\right]\)


\(\displaystyle \text{Therefore: }\;\begin{Bmatrix}x &=& 1 \\ y &=& 2 \\ z &=& \text{-}2 \end{Bmatrix}\)