Matrix Algebra Help: Solve for p1, p2, and p3

[maarrcc1994]

Hopefully someone can help me solve this, I'm usually quite good at matrix algebra but for some reason I cannot solve this equation.

p1+4p2+8p3=26
5p1+7p2=38
8P1+12p2+2p3=66

If somebody could help me with the values of p1 p2 p3 that would be a great help :)

 

[MarkFL]

Are you wanting to use Gaussian elimination?

If so, your first step is to set up the augmented matrix:

$$\left[\begin{array}{ccc|c} 1&4&8&26\\ 5&7&0&38\\ 4&6&1&33 \end{array} \right]$$

Note: I divided the last equation by 2 for the sake of simplcity.

Do you know what the next step is?

 

[maarrcc1994]

I've only been taught the inversion method so that's what I usually use, I tried with cramers rule but still couldn't seem to get it right, could you show me using inversion? :)

 

[MarkFL]

Show what you did with the inversion method so we can see where you are stuck and then we can go from there. :D

edit: I've got to run now, but there are many here much better at matrix algebra than me, and so perhaps someone will help in the meantime. To be honest, it is Gaussian elimination that I find to be the most straightforward method, and which is the only one I recall how to do without looking things up. :D

 

[maarrcc1994]

|1 4 8 | |p1| 26
|5 7 0 | |p2| = 38
|8 12 2 | |p3| 66

The determinant=
1*[7*2] - [12*0]=14
-4*[5*2] - [8*0]=-40
8*[5*12] - [7*8]= 32

14--40-32=22

I found the determinant to be 22 so I'm hoping that part was right.Then i moved on to finding each cofactor-

| 7 0| |5 0| |5 7|
+ |12 2| - |8 12| + |8 12|

- |4 8| + |1 8| - | 1 4 |
|12 2| |8 2| | 8 12|

+ | 4 8| - |1 8 | + | 1 4|
| 7 0| | 5 0 | | 5 7|| 14 -60 4 |
| 88 -62 20 |
| -56 40 -13 |Then I inverted the results

|14 88 -56 |
1/22 |-60 -62 40 |
| 4 20 -13 |

From here the answers were way off so I've obviously miscalculated disastrously

apologies for the formatting, it seems I'm as bad as this as I am at maths

 

[I like Serena]

Hi maarrcc1994! Welcome to MHB! :)

|1 4 8 | |p1| 26
|5 7 0 | |p2| = 38
|8 12 2 | |p3| 66

The determinant=
1*([7*2] - [12*0])=14
-4*([5*2] - [8*0])=-40
8*([5*12] - [7*8])= 32

Up till here it is all correct! ;)
(Except for a couple of parentheses that I've added in the quote.)
Note that with the -4 in the second line, you have properly picked up on the sign.

14--40-32=22

I found the determinant to be 22 so I'm hoping that part was right.

You already had all the signs in place, so I'm afraid you shouldn't introduce more minus signs.
The determinant should be just the sum: $D=14-40+32=6$.

Then i moved on to finding each cofactor-

      | 7 0|    |5  0|    |5  7|   
 +    |12 2|  - |8 12|  + |8 12|

The approach is good, but your second submatrix should have a 2 instead of a 12, yielding -10 instead of -60.

  -     |4  8|  + |1  8|    - | 1  4 |
        |12 2|    |8  2|      | 8  12|

  +     | 4  8|  - |1  8 |  + | 1  4|
        | 7  0|    |5  0 |    | 5  7|| 14  -60  4  |      
| 88  -62  20 |
| -56  40 -13 |

The rest is correct, but with the correction I mentioned, it should be:
\begin{bmatrix}
14 & -10 & 4 \\
88 & -62 & 20 \\
-56 & 40 & -13 \\
\end{bmatrix}

Then I inverted the results

             |14   88  -56 |
1/22         |-60 -62  40  |
             | 4   20  -13 |

From here the answers were way off so I've obviously miscalculated disastrously

So this should be (transposing and multiplying with the inverse of the determinant):
$$\frac 1 6
\begin{bmatrix}
14 & 88 & -56 \\
-10 & -62 & 40 \\
4 & 20 & -13 \\
\end{bmatrix}$$

apologies for the formatting, it seems I'm as bad as this as I am at maths

For reference, when you click Reply with Quote, you can see how I formatted what I did.

 

[maarrcc1994]

Its really clear where I went wrong now, thank you so much! :)