System of Linear Equations

[kais]

Hey guys, ive got a math question that im stuck on the start, its goes

Consider the system of equations

x + 2y - z = -3
3x + 5y +kz = -4
9x + (k+13)y+ 6z = 9

Express these equations as an augmented matrix.
(b) Show that this matrix can be row-reduced to

1 2 -1 / -3
0 1 -k -3 / -5
0 0 k^2 -2k/ 5k+11

what im lost about is what to do with both the k variables in the system of equations. im guessing that i have to put them on the answer side before i put it into an augmented matrix but im not sure. Any help would be appreciated, cheers!

 

[stapel]

Consider the system of equations

x + 2y - z = -3
3x + 5y +kz = -4
9x + (k+13)y+ 6z = 9

(b) Show that this matrix can be row-reduced to

1 2 -1 / -3
0 1 -k -3 / -5
0 0 k^2 -2k/ 5k+11

what im lost about is what to do with both the k variables in the system of equations.

The "k" is not a variable; the x, y, and z are the variables. The "k" is just another constant, like the 2, 9, etc. You'll deal with the "k" just the same as you'll deal with all the other constants. The only difference will be that you won'd be able to simplify as much, because you don't have a numerical value for the constant "k".

 

[kais]

The "k" is not a variable; the x, y, and z are the variables. The "k" is just another constant, like the 2, 9, etc. You'll deal with the "k" just the same as you'll deal with all the other constants. The only difference will be that you won'd be able to simplify as much, because you don't have a numerical value for the constant "k".

ah i see, now if i put that into an augmented matrix shouldn't it be

1	2	-1	/-3
3	5	k	/4
9	(k+13)	6	/9

problem being im sure you cant have brackets in a matrice, but im not sure how to get them out of the augmented matrix. Any tips on how i can or videos would be awesome.

cheers in advance

 

[stapel]

ah i see, now if i put that into an augmented matrix shouldn't it be

1	2	-1	/-3
3	5	k	/4
9	(k+13)	6	/9

Yes, this is correct. Then start doing row operations. I'd probably start like this:

. . .-3R₂ + R₃ gives new R₃:

. . . . .\(\displaystyle \left[ \begin{array}{ccc|c}1&2&-1&-3\\3&5&k&4\\0&k\, -\, 2&6\, -\, 3k&-3 \end{array} \right]\)

. . .-3R₁ + R₂ gives new R₂:

. . . . .\(\displaystyle \left[ \begin{array}{ccc|c}1&2&-1&-3\\0&-1&k\, +\, 3&13\\0&k\, -\, 2&6\, -\, 3k&-3 \end{array} \right]\)

. . .2R₂ + R₁ gives new R₁:

. . . . .\(\displaystyle \left[ \begin{array}{ccc|c}1&0&2k\, +\, 5&23 \\0&-1&k\, +\, 3&13\\0&k\, -\, 2&6\, -\, 3k&-3 \end{array} \right]\)

. . .(k - 2)R₂ + R₃ gives new R₃:

. . . . .\(\displaystyle \left[ \begin{array}{ccc|c}1&0&2k\, +\, 5&23 \\0&-1&k\, +\, 3&13\\0&0&k^2\, -\, 2k&13k\, -\, 29 \end{array} \right]\)

(Check my work!!)

You can use whatever steps you like. I would strongly suggest, however, that you use lots of scratch paper, so you can have the space to write out all of your steps explicitly. It is amazingly easy to mess of the computations on these things! :shock:

problem being im sure you cant have brackets in a matrix...

Why not? What's wrong with putting parentheses or other grouping symbols around grouped characters?

 

[kais]

ahh got it now, thanks a lot for your help all