1. ## 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} = $\begin{bmatrix} 2 \\-k \\0 \end{bmatrix}$

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?

2. ## Re: Matrices

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

$\left[\begin{array}{ccc|c}1&3&k&2\\k&-2&3&k\\4&-3&10&5 \end{array}\right]$

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

$\left[\begin{array}{ccc|c}1&3&k&2\\0&-2-3k&3-k^2&-k\\4&-3&10&5\end{array}\right]$

R3-->-4R1+R3:

$\left[\begin{array}{ccc|c}1&3&k&2\\0&-2-3k&3-k^2&-k\\0&-15&10-4k&-3\end{array}\right]$

R2-->15R2

$\left[\begin{array}{ccc|c}1&3&k&2\\0&-30-45k&45-15k^2&-15k\\0&-15&10-4k&-3\end{array}\right]$

R3-->(-2-3k)R3:

$\left[\begin{array}{ccc|c}1&3&k&2\\0&-30-45k&45-15k^2&-15k\\0&30+45k&-20-22k+12k^2&6+9k\end{array}\right]$

R3-->R2+R3:

$\left[\begin{array}{ccc|c}1&3&k&2\\0&-30-45k&45-15k^2&-15k\\0&0&25-22k-3k^2&6-6k\end{array}\right]$

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

$\left[\begin{array}{ccc|c}1&3&k&2\\0&2+3k&-3+k^2&k\\0&0&25-22k-3k^2&6-6k\end{array}\right]$

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

$(25+3k)(1-k)z = 6(1-k)$. now IF k ≠ 1, this means that:

$z = \frac{6}{25+3k}$ <--only defined when k ≠ -25/3.

the second row is now:

$(2+3k)y + \frac{-18+6k^2}{25+3k} = k = \frac{25k+3k^2}{25+3k}$

$(2+3k)y = \frac{18+25k-3k^2}{25+3k} = \frac{(9-k)(2+3k)}{25+3k}$ IF k ≠ -2/3, we can divide both sides by 2+3k, to get:

$y = \frac{9-k}{25+3k}$<--again, only defined when k ≠ -25/3

the top row now becomes:

$x + \frac{27-3k}{25+3k} + \frac{6k}{25+3k} = 2$, so:

$x= \frac{50+6k}{25+3k} - \frac{27-3k}{25+3k} - \frac{6k}{25+3k} = \frac{23+3k}{25+3k}$

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

3. ## Re: Matrices

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

4. ## Re: Matrices

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.

5. ## Re: Matrices

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
A(Y^(-1))Y = BY
AI = BY
A = BY
A(B^(-1))= = B(B^(-1))Y
Therefore
Y = A(B^(-1))
Y = (A^(-1))B..

Thank you so much for helping me!

.

7. ## Re: Matrices

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:

$AY^{-1} = B$

then:

$AY^{-1}Y = BY$

so:

$A(Y^{-1}Y) = AI = A = BY$ 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:

$B^{-1}A = B^{-1}(BY) = (B^{-1}B)Y = IY = Y$

from this, it follows that:

$Y^{-1} = (B^{-1}A)^{-1} = A^{-1}(B^{-1})^{-1} = A^{-1}B$.

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

IF, however, the original equation is:

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

$A^{-1}(Ax) = A^{-1}b$
$Iy = A^{-1}b$
$y = A^{-1}b$

8. ## Re: Matrices

Originally Posted by tmoria
$Y^{-1} = \dfrac {1}{Y}$

Should be easy enough from there

btw, if you have given the correct $AY^{-1}=B$, the answer in your book is wrong
this makes no sense, for vectors OR matrices, division is not defined.

9. ## Re: Matrices

I maybe reading worngly, but is it that you got $Y^{-1}=A^{-1}B?$ The answer is different, states that it is $Y=A^{-1}B$

Nontheless, thank you so so much as usual!

10. ## Re: Matrices

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

11. ## Re: Matrices

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.

12. ## Re: Matrices

The question:

Make Y the subject of
$AY^{-1} = B$

Hopefully I have not typed wrongly!

13. ## Re: Matrices

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..

14. ## Re: Matrices

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

$A = \begin{bmatrix}0&1\\0&0 \end{bmatrix};\ Y^{-1} = \begin{bmatrix}1&0\\0&1 \end{bmatrix};\ B = A$

then we certainly have:

$AY^{-1} = B$

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

15. ## Re: Matrices

an interesting point from my earlier error though...

treating as ordinary algebra

$AY^{-1}=B$

$Y^{-1}=\dfrac {B}{A}$

since $\dfrac{1}{A}=A^{-1}$

we have $Y^{-1}=A^{-1}B$

which agrees with your matrix solution.

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

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