Thread: Matrices

I would really appreciate the help for the question below

Solve the linear equations
2x+3y+z=0
x-y+2z = 0 ( THe first equation is at the top and the second is at the bottom row in matrix form)

and hence solve
2x+3y+z=0
x-y+2z = 0
ax+y-z=0 ( respectively from top to bottom row in the matrix) for all aER.

Pleas ehelp me with this! I solved the first part with parametric equations, I got x=-7/3t and y =t and z=5/3t. However, how do I apply it to the second one? I just went about the sam way for the second one as I did the first one but I got stuck and they have to be related somehow. I only can figure out that both have the same two equations, but I'm a blockhead for anything more.

2.)Consider the system of equations
x+3y+kz=2
kx-2y+3z=k
4x-3y+10z= 5

I reduced it to my very best, I had 4 times Row =2 minus k times Row 3 to change row 2 and then for channging row 3, I had -1 times Row 3 + Row 2.
I got
[tex]\begin{bmatrix} 1 & 3 & k \\0 & 3k-8 & 12-10k \\0 & 0 & 0 \end{bmatrix}\begin{bmatrix} x \\y \\z \end{bmatrix} =

However, the answer had a very different form of doing the row reduction, it had subtracted the multiple of another row form some row. SOrry that I could not be more specific as the answer sheet is not with me now..please could you help me identify where I had gone wrong here?

Appreciate it loads, thank you!

1) we know that any solution to the first two equations is of the form t(-7/3,1,5/3). if this is to be a solution to the third equation as well, we must have:

a(-7/3)t + t - (5/3)t = 0 that is:

t(-7a - 2)/2 = 0.

since it makes no difference if we multiply this by -2 (the RHS is still 0), we can write this as:

t(7a + 2) = 0.

now taking t = 0 works, leading to the solution (0,0,0) which is always a solution of any system of homogeneous equations (ones where the RHS is equal to 0 in every equation).

but if a = -2/7, then the third equation is 0 = 0 no matter what t is, so in this case we have the same solution as with just the first 2 equations: any vector of the form t(-7/3,1,5/3).

(it is worth noting that this is the same set of vectors as the set of vectors of the form s(-7,3,5), since the two vectors in the parentheses only differ by a scalar multiple, so {(-7,3,5)} is a basis for the solution space).

2) here is how i would do the second problem:

i would form the augmented matrix



R₂-->-kR₁+R₂ (subtract k times the first row from the second row):



R₃-->-4R₁+R₃:



R₂-->15R₂



R₃-->(-2-3k)R₃:



R₃-->R₂+R₃:



just to clean things up, let's divide row 2 by -15:



now let's see what we can do with this. with a little algebraic rearrangement, the last row becomes:

. now IF k ≠ 1, this means that:

<--only defined when k ≠ -25/3.

the second row is now:



IF k ≠ -2/3, we can divide both sides by 2+3k, to get:

<--again, only defined when k ≠ -25/3

the top row now becomes:

, so:



now, i've done the hard stuff, you can decide what happens when k = 1, -2/3 or -25/3.

Thank you! For the second problem, could you tell me why the way I did it is not preferred?

i don't know exactly what you did (your notation is hard to read), but in general, to change row k, you do one of 3 things:

1) add m times another row to it
2) multiply it by a number
3) exchange it with another row

it's not a good idea to add m times row k plus n times row j back to row k. that's too many steps at once, and it's easy to get confused. it's also not a good idea to work on two rows simultaneously. do one operation at a time. in what i did above, i was careful to list every step BEFORE i took it, and only ONE row changed at a time. this makes your work easier to check (but uses a LOT more paper). row-reduction is very labor-intensive, and an early mistake can ruin everything. and you must have made a mistake SOMEWHERE because you ended up with a 0 row at the bottom where i did not.

Thank you so so so much!

I have another really short question, could you help me again?

Make Y the subject in A(Y^(-1))=B
I had
A(Y^(-1))Y = BY
AI = BY
A = BY
A(B^(-1))= = B(B^(-1))Y
Therefore
Y = A(B^(-1))
However the answer is
Y = (A^(-1))B..

Thank you so much for helping me!

.

it's not clear what the "sizes" of the matrices A,Y, and B are.

there also appears to be a type somewhere. i will assume A,B and Y are all invertible nxn matrices.

IF:



then:



so:

so far, you're good.

your next step is wrong, you CANNOT "put a matrix in the middle" (AB)C is usually NOT equal to A(C)B.

in this case, what you have to do is multiply on the left by B^-1:



from this, it follows that:

.

***********************

IF, however, the original equation is:

, where y and b are vectors, then you would have to left-multiply by A^-1, to get:

Originally Posted by tmoria

Should be easy enough from there

btw, if you have given the correct , the answer in your book is wrong

this makes no sense, for vectors OR matrices, division is not defined.

I maybe reading worngly, but is it that you got The answer is different, states that it is

Nontheless, thank you so so much as usual!

Originally Posted by Deveno

this makes no sense, for vectors OR matrices, division is not defined.

Yep, that's why I deleted the post. I had only read Tutu's last post and missed the matrices connection

yes, it appears that either:

a) you stated the problem incorrectly, or

b) there is a typo in the book

it is impossible for me to say which.

The question:

Make Y the subject of


Hopefully I have not typed wrongly!

It doesn't say anything about them being of the same order, that they are square matrices etc. But from what I know of these questions, just solve as if they are square and of the same order..

well, the trouble is, not all square matrices HAVE inverses. for instance, we might have:



then we certainly have:



but there is NO way to express Y as a product of A (to any exponent) and B (to any exponent).

an interesting point from my earlier error though...

treating as ordinary algebra





since

we have

which agrees with your matrix solution.

Is this true in all situations or just peculiar to this one ?