Matrices: my solutions aren't matching the back of the book

[Feliath Dunami]

I am currently working on matrices and determinants and I am having a lot of trouble figuring out what to do for different ones. I understand the elementary row operations, how to set up the problem, and how they work, but I have trouble figuring out what of the elementary ops to use and in what order. Usually I get partway through the problem and then I get stumped.

So my question is, is there a step-by-step way to go about solving these matrices, or do you just have to randomly try different things to see what works?

For example, the following augmented matrix:

|1 0 -3 -10|
|1 -2 2 0|
|1 -2 0 -7 |

|1 0 -3 -10|
- R1 + R2|0 -2 5 10|
|1 -2 0 -7 |


|1 0 -3 -10|
-1/2R2|0 1 (-5/2) -5|
|1 -2 0 -7 |


|1 0 -3 -10|
|0 1 (-5/2) -5|
-R1 + R3|0 -2 3 3 |

|1 0 -3 -10|
|0 1 (-5/2) -5|
2*R2 + R3|0 0 -2 -7 |

|1 0 -3 -10|
|0 1 (-5/2) -5|
1/2R3|0 0 1 7/2 |

But I checked the answer and that's like... way off.

So what am I doing wrong? Am I just not seeing something or making some stupid mistake?

Thanks

 

[stapel]

I have trouble figuring out what of the elementary ops to use and in what order.

There is no one "right" order. The "right" order is the one you're comfortable with, that gets you to the correct answer in the end.

But I checked the answer and that's like... way off.

Until you get the completed solution (x = whatever, y = whatever else, etc), you can't really compare solutions. Everything you did so far was fine (though I would have avoided fractions until absolutely necessary); keep going:

Take (5/2)R₃, and add to R₂:

. . . . .\(\displaystyle \L \left[\begin{array}{ccc|c}1&0&-3&-10\\0&1&0&\frac{15}{4}\\0&0&1&\frac{7}{2} \end{array}\right]\)

Take (3)R₃, and add to R₁:

. . . . .\(\displaystyle \L \left[\begin{array}{ccc|c}1&0&0&\frac{1}{2}\\0&1&0&\frac{15}{4}\\0&0&1&\frac{7}{2} \end{array}\right]\)

Then check:

. . . . .1x + 0y - 3z = 1(1/2) + 0(15/4) - 3(7/2) = 1/2 + 0 - 21/2 = -20/2 = -10

. . . . .1x - 2y + 2z = 1(1/2) - 2(15/4) + 2(7/2) = 1/2 - 15/2 + 7 = -14/2 + 14/2 = 0

. . . . .1x - 2y + 0z = 1(1/2) - 2(15/4) + 0(7/2) = 1/2 - 15/2 + 0 = -14/2 = -7

So the solution works.

Eliz.

 

[Feliath Dunami]

Ahah- that makes sense now. Thank you so much! I guess I'll just keep plugging away at these then...