1. ## Matrices and Determinants

1) Use Gauss-Jordan elimination to solve the following system of equations.
3x+5y=7
6x-y=-8
a: x=2, y=1 b: x=5, y=6 c: x=3,y=-1 d: x=-1, y=2

2)Use Gauss-Jordan elimination to solve the following system of equations.
x+y+z=2
2x-3y+z=-11
-x+2y-z=8
a: x=5, y=1, z=6 b: x=3, y=-1, z=2 c: x= 1/3, y= 10/3, z= -5/3 d: x= 1,y=3, z=2

3)If you were to use Gauss-Jordan elimination on the following linear system, what would the augmented matrix be?
3w-x=2y+z-4
9x-y+z=10
4w+3y-z=7
12x+17=2y-z+6

4) Solve the following system of equations.
x-2y=5
-3x+2y=7
a: x=11, y=11/2 b: x=-6, y=-11/2 c: x=3, y=1 d: x=3, y=-1

5)Solve the following system of equations.
x+2y-6=z
3y-2z=7
4+3x=2y-5z
a: x=7/4, y= 3/2, z= -5/4 b: x=2, y=1, z=2 c: x=2/3, y=3/2, z= -23/6
d: x=0, y=5, z=4

6) Use Cramer's rule to solve the following system of equations.
5x+3y=7
4x+5y=3
a: x=4, y=1 b: x=2, y=5 c: x=2, y= -1 d: x= -2, y= -1

I would appreciate any help that I can get with these questions.

2. Originally Posted by pyrogurl6989
...
6) Use Cramer's rule to solve the following system of equations.
5x+3y=7
4x+5y=3
a: x=4, y=1 b: x=2, y=5 c: x=2, y= -1 d: x= -2, y= -1
Hello,

1. calculate the main determinant to see if there is an unique solution ornot:

$D=\begin{array}{|cc|} 5 & 3\\4 & 5 \end{array}= 5 \cdot 5 - 3 \cdot 4 = 13$. Because D is not zero there exist an unique solution.

2. now calculate the determinants containing the constants:

$D_x=\begin{array}{|cc|} 7 & 3\\3 & 5 \end{array}= 7 \cdot 5 - 3 \cdot 3 = 26$

Solution: $x=\frac{D_x}{D}=2$

$D_y=\begin{array}{|cc|} 5 & 7\\4 & 3 \end{array}= 5 \cdot 3 - 7 \cdot 4 = -13$

Solution: $y=\frac{D_y}{D}=-1$

EB

3. Hello, pyrogurl6989!

Do you know Gauss-Jordan Elimination?

1) Use Gauss-Jordan elimination to solve: . $\begin{array}{cc}3x + 5y & =\:7 \\6x - y & =\:\text{-}8\end{array}$

We have: . $\begin{vmatrix}\:3 & 5 & | & 7\: \\ \:6 & \text{-}1 & | & \text{-}8\:\end{vmatrix}$

$\begin{array}{cc} \\R_3\!-\!2\!\cdot\!R_2\end{array}
\begin{vmatrix}\:3 & 5 & | & 7\: \\ \:0 & \text{-}11 & | & \text{-}22\:\end{vmatrix}$

. . $\begin{array}{cc} \\ -\frac{1}{11}R_2\end{array}
\begin{vmatrix}\:3 & 5 & | & 7\: \\ \:0 & 1 & | & 2\:\end{vmatrix}$

$\begin{array}{ccc}R_1\!-\!5\!\cdot\!R_2 \\ \\ \end{array}
\begin{vmatrix}\:3 & 0 & | & \text{-}3\: \\ 0 & 1 & | & 2\:\end{vmatrix}$

. . $\begin{array}{ccc}\frac{1}{3}R_1 \\ \\ \end{array}
\begin{vmatrix}\:1 & 0 & | & -1\: \\ \:0 & 1 & | & 2\:\end{vmatrix}$

Answer: . $x = -1,\:y = 2$ . . . answer (d)

3) If you were to use Gauss-Jordan elimination on the following linear system,

what would the augmented matrix be? . $\begin{array}{cccc}3w-x & = & 2y+z-4 \\ 9x-y+z & = & 10 \\ 4w+3y-z & = & 7 \\ 12x+17 & = &2y-z+6\end{array}$

First, get it into "standard form":

. . $\begin{array}{cccc}3w - x - 2y - z \\ \qquad 9x - \;y + z \\ 4w \quad\qquad 3y - z \\ \quad\;\;\; 12x - 2y + z \end{array}
\begin{array}{cccc}= \\ = \\ = \\ = \end{array}
\begin{array}{cccc} \text{-}4 \\ 10 \\ 7 \\ \text{-}11\end{array}
$

The augmented matrix is: . $\begin{vmatrix}\:3 & \text{-}1 & \text{-}2 & \text{-}1 & | & \text{-}4 \\ \:0 & 9 & \text{-}1 & 1 & | & 10 \\ \:4 & 0 & 3 & \text{-}1 & | & 7 \\ \:0 & 12 & \text{-}2 & 1 & | & \text{-}11\end{vmatrix}$

4. I was taught how to do Gauss-Jordan elimination in school but I don't understand how to do it. Ive been showed a few times but I still cant figure it out.

5. Originally Posted by pyrogurl6989
I was taught how to do Gauss-Jordan elimination in school but I don't understand how to do it. Ive been showed a few times but I still cant figure it out.
Hello,

I'm going to show how to do Gauss elimination in (as I hope) small steps. Maybe this will help a little bit further on:

The general aim is to transform the matrix into one where the main diagonal consists of ones. Then you can get directly the solution in the last column.

(By the way: Most calculators use the command rref(matrix) to do this elimination. rref means "reduced row equelon form")

Code:
         x + y + z = 2
2x -3y + z = -11
-x +2y - z = 8

|  1   1   1  |   2 |
|  2  -3   1  | -11 |
| -1   2  -1  |   8 |

|  1   1   1  |   2 |
|  0  -5  -1  | -15 |         R2-2*R1
|  0   3   0  |  10 |         R3 + R1

|  1   1   1  |   2   |
|  0   1  1/5 |  3    |        (R2)/(-5)
|  0   1   0  |  10/3 |        (R3)/3

|  1   1   1  |   2   |
|  0   1  1/5 |   3   |
|  0   0 -1/5 |  1/3  |         R3 - R2|

|  1   1   1  |   2    |
|  0   1   0  |  10/3  |       R2 + R3
|  0   0 -1/5 |  1/3   |

|  1   1   1  |   2   |
|  0   1   0  | 10/3  |
|  0   0   1  | -5/3  |         (R3)*(-5)
Code:
Now you work back from the last rows to eliminate
the ones in the 1rst row:

|  1   0   0  |  1/3  |        R1 - R2 - R3
|  0   1   0  | 10/3  |
|  0   0   1  | -5/3  |
So it is the solution c: x = 1/3, y = 10/3 , z = -5/3

A personal remark: It takes some time to do this method error-free and fairly fast. So I'll leave the other problems for you.

EB