Question

1.)Solve the system of equations using matrices. Use Gaussian elimination with back-substitution.

x + y + z = -5
x - y + 3z = -1
4x + y + z = -2

A. {( 1, -4, -2)}

B. {( -2, 1, -4)}

C. {( 1, -2, -4)}

D. {( -2, -4, 1)}


2.)Use Gaussian elimination to find the complete solution to the system of equations, or state that none exists.

4x - y + 3z = 12
x + 4y + 6z = -32
5x + 3y + 9z = 20

A. {(8, -7, -2)}

B. {(-8, -7, 9)}

C. no solution

D. {(2, -7, -1)}


3.)A statement Sn about the positive integers is given. Write statements Sk and Sk+1, simpl

Answer 1

3.) A statement Sn about the positive integers is given. Write statements Sk and Sk+1, simplifying Sk+1 completely.

Sn: 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + . . . + n(n + 1) = [n(n + 1)(n + 2)]/3

Answer 2

I'll help you work through the first problem. The second one is the same sort of thing, so you should be able to do it yourself after seeing the first one.

\[x+y+z=-5\]\[x-y+3z=-1\]\[4x+y+z=-2\]To put these into a matrix, we first make sure that all our variables line up (and they already do, so we're good here). Then we make a 3x4 matrix. The first column will have the coefficients in front of the xs in the three equations, the second column the y coefficients, third the z coefficients, and the last column will be the answers. So\[\left[\begin{matrix}1 & 1 & 1 & -5\\ 1 & -1 & 3 & -1 \\ 4 & 1 & 1 & -2\end{matrix}\right]\]Now, there are three things we can do in Gaussian Elimination:
1. Switch two rows
2. Multiply an entire row by a constant
3. Subtract one row from another

The steps are fairly simple also. First we want to make sure that the first number in the first row is a 1 (which again, it already is). This is our first leading 1. We want to make the other two numbers in this column become 0, so, to make that in the second row, we will just subtract the first row from the second row:\[\left[\begin{matrix}1 & 1 & 1 & -5\\ 1-1 & -1-1 & 3-1 & -1+5 \\ 4 & 1 & 1 & -2\end{matrix}\right]\]\[\left[\begin{matrix}1 & 1 & 1 & -5\\ 0 & -2 & 2 & 4 \\ 4 & 1 & 1 & -2\end{matrix}\right]\]Now we need to make the first number in the third row 0 also. In order to do this, we will subtract 4 times the first row from it:\[\left[\begin{matrix}1 & 1 & 1 & -5\\ 0 & -2 & 2 & 4 \\ 4-4 \times 1 & 1-4 \times 1 & 1-4 \times 1 & -2-4 \times -5\end{matrix}\right]\]\[\left[\begin{matrix}1 & 1 & 1 & -5\\ 0 & -2 & 2 & 4 \\ 0 & -3 & -3 & 18\end{matrix}\right]\]Now we have a column that has a leading 1 and all the rest are 0s.

To continue we now need to make sure that the second number in the second row is a 1. It isn't, so in order to fix that we will divide that entire row by -2:\[\left[\begin{matrix}1 & 1 & 1 & -5\\ 0 & \frac{-2}{-2} & \frac{2}{-2} & \frac{4}{-2} \\ 0 & -3 & -3 & 18\end{matrix}\right]\]\[\left[\begin{matrix}1 & 1 & 1 & -5\\ 0 & 1 & -1 & -2 \\ 0 & -3 & -3 & 18\end{matrix}\right]\]Now we need to make the number below our second leading 1 a 0. We can do this by adding 3 times the second row to the third:\[\left[\begin{matrix}1 & 1 & 1 & -5\\ 0 & 1 & -1 & -2 \\ 0 & -3+3 \times 1 & -3+3 \times -1 & 18+3 \times -2\end{matrix}\right]\]\[\left[\begin{matrix}1 & 1 & 1 & -5\\ 0 & 1 & -1 & -2 \\ 0 & 0 & -6 & 12\end{matrix}\right]\]So now we have two leading 1s with 0s beneath them.

To finish Gaussian Elimination for this matrix, we need to make sure that the third element of the third row is a 1. In this case, it if was also a 0 then we would have the impossible equation 0x+0y+0z=C (which would indicate no solution). However, that is not the case so we can continue. We have to divide the third row by -6:\[\left[\begin{matrix}1 & 1 & 1 & -5\\ 0 & 1 & -1 & -2 \\ 0 & 0 & \frac{-6}{-6} & \frac{12}{ -6}\end{matrix}\right]\]\[\left[\begin{matrix}1 & 1 & 1 & -5\\ 0 & 1 & -1 & -2 \\ 0 & 0 & 1 & -2\end{matrix}\right]\]Now we're done with Gaussian Elimination! You can see that each row has a leading 0.

What this matrix represents now is the following system:\[z=-2\]\[y-z=-2\]\[x+y+z=-5\]Just plug in your value of z to find y, then use your values of y and z to find x. That is back-substitution.

Does all this make sense? :)

Answer 3

Yes it does thank you, now could you help me with the third one? Please?

Answer 4

For the third one:

You have\[S _{n}=\frac{n(n+1)(n+2)}{3}\]To get S sub k you just have to substitute in k wherever you see n.

To get\[S_{k+1}\]substitute in k+1 wherever you see n.\[S_{k+1}=\frac{(k+1)(k+1+1)(k+1+2)}{3}\]\[=\frac{(k+1)(k+2)(k+3)}{3}=\frac{1}{3}\prod_{n=1}^{3}(k+n)\](That's as simple a form I can think of anyway).

Answer 5

okay thanks so much.

Answer 6

You're welcome. :)

Answer 7

@aylin @scarletsummy so would the answer for the first problem be 1, -4, -2??