Question

Any help ? Pls
Ex 4 , 2)
http://i.imgur.com/MHhs4yS.jpg

Answer 1

OK. Find the new matrix inside and then take the determinat. So start with getting \(A^2\), then multiply that by 3. Get the Transpsose of that, multiply by 2. Then find the inverse of that answer, and multiply by 4. Finally take the determinant.

Answer 2

Which step are you getting stuck at? If you can do exercise 1, I do not see most of this as an issue. However, it may be a process or terminology problem.

Answer 3

Ex 4
2)

Answer 4

Yes, how far are you getting with it? Which step fo you get stuck at?

Answer 5

Actually In the first step

Answer 6

\[\left[
\bengin{array}{ccc}
1 & 2 & -1 \\
2 & 1 & 3 \\
0 & 3 & -2
\end{array}
\right]^2
\Rightarrow \\
\bengin{array}{ccc}
1 & 2 & -1 \\
2 & 1 & 3 \\
0 & 3 & -2
\end{array}
\right]
\bengin{array}{ccc}
1 & 2 & -1 \\
2 & 1 & 3 \\
0 & 3 & -2
\end{array}
\right]
\]

Answer 7

Argh. Having issues with the left nad right bracets on the matrix....

Answer 8

\[\left[\begin{matrix}
1 & 2 & -1 \\
2 & 1 & 3 \\
0 & 3 & -2
\end{matrix}\right]^2
\Rightarrow \\
\left[\begin{matrix}
1 & 2 & -1 \\
2 & 1 & 3 \\
0 & 3 & -2
\end{matrix}\right]
\left[\begin{matrix}
1 & 2 & -1 \\
2 & 1 & 3 \\
0 & 3 & -2
\end{matrix}\right]
\]

Answer 9

The exponent rules still apply. It just indicates multiplication. Have you done 3x3 multiplications? It is similar to 2x2.

Answer 10

Ok . I will try
Sorry suffer from this mitrix

Answer 11

np.

Answer 12

Thank you

Answer 13

I got:
\[
\left[\begin{matrix}
1+4 & 2+2+3 & -1+6+2 \\
2+2 & 4+1+9 & -2+3+6 \\
6 & 3-6 & 9+4
\end{matrix}\right]
\Rightarrow \\
\left[\begin{matrix}
5 & 1 & 7 \\
4 & 14 & 7 \\
6 & -3 & 13
\end{matrix}\right]
\]
Unless I made a math mistake, that is the first step.

Answer 14

Error 😰

Answer 15

With?

Answer 16

You side first a over 2 or get T ? I don't know how can I start , do you have example ? 😥 Pls

Answer 17

Is ths still the same problem?

Answer 18

I made a small muistake n the square. It is supposed to be a -2+3-6, not -2+3+6. So that spot ends as -5:
\[
\left[\begin{matrix}
5 & 1 & 7 \\
4 & 14 & -5 \\
6 & -3 & 13
\end{matrix}\right]\]

Answer 19

http://i.imgur.com/gzjTNIR.jpg

Answer 20

AH! OK. The problem is in the \(A^2\) This means matrix multiplied by itself.
\[A^2=AA
\Rightarrow \\
\left[\begin{matrix}
1 & 2 & -1 \\
2 & 1 & 3 \\
0 & 3 & -2
\end{matrix}\right]^2
=
\left[\begin{matrix}
1 & 2 & -1 \\
2 & 1 & 3 \\
0 & 3 & -2
\end{matrix}\right]
\left[\begin{matrix}
1 & 2 & -1 \\
2 & 1 & 3 \\
0 & 3 & -2
\end{matrix}\right]
\]

Answer 21

Oooh , i do not get it , sorry

Answer 22

Wait a minute. Doing some work on paper.

Answer 23

I went too fast and made a mistake there too. LOL. It is supposed to be 5.

Answer 24

Let me see if I can fimd a good reference or make one real fast.

Answer 25

Ok , take your time , i'm trying too 😅
http://i.imgur.com/T1OoDOt.jpg

Answer 26

This is \(3\times 3\) matrix multiplication:

\[\left[\begin{matrix}
a & b & c \\
d & e & f \\
h & i & j
\end{matrix}\right]
\left[\begin{matrix}
k & l & m \\
n & o & p \\
q & r & s
\end{matrix}\right]
=\\

\left[\begin{matrix}
ak+bm+cq & al+bo+cr & am+bp+cs \\
dk+em+fq & dl+eo+fr & dm+ep+fs \\
hk+im+jq & hl+io+jr & hm+ip+js
\end{matrix}\right]
\]

Answer 27

OK. I checked the math this timeL

\[
\left[\begin{matrix}
1 & 2 & -1 \\
2 & 1 & 3 \\
0 & 3 & -2
\end{matrix}\right]
\left[\begin{matrix}
1 & 2 & -1 \\
2 & 1 & 3 \\
0 & 3 & -2
\end{matrix}\right]
\Rightarrow \\
\left[\begin{matrix}
1+4+0 & 2+2-3 & -1+6+2 \\
2+2+0 & 4+1+9 & -2+3-6 \\
0+6+0 & 0+3-6 & 0+9+4
\end{matrix}\right]
\Rightarrow \\
\left[\begin{matrix}
5 & 1 & 7 \\
4 & 14 & -5 \\
6 & -3 & 13
\end{matrix}\right]
\]
That is how I got \(A^2\) Does that make more sense now?

Answer 28

The next step is multiply by a constant of 3. I got the same thing you did there.\[
3
\left[\begin{matrix}
5 & 1 & 7 \\
4 & 14 & -5 \\
6 & -3 & 13
\end{matrix}\right]
\Rightarrow
\left[\begin{matrix}
15 & 3 & 21 \\
12 & 42 & -15 \\
18 & -9 & 39
\end{matrix}\right]
\]

Answer 29

Now transpose all that. Again, same answer:
\[
\left[\begin{matrix}
15 & 12 & 18 \\
3 & 42 & -9 \\
21 & -15 & 39
\end{matrix}\right]
\]

Answer 30

Another constant multiple. 2 this time.
\[
\left[\begin{matrix}
30 & 24 & 36 \\
6 & 84 & -18 \\
42 & -30 & 78
\end{matrix}\right]
\]

Answer 31

Ok , I'm so happy 😭, next take invers or *2 ?

Answer 32

2 then inverse. Some of these do not matter in the order, but I am following the ( ) to make it a logical sequence.

Answer 33

http://i.imgur.com/PpAQGl3.jpg
It is true ?
I hope 😁

Answer 34

How did you get that many zeros in the inverse? I am still working on the inverse. This one uses large numbers, so it is taking a bit. But you should get fractions in all spots.

Answer 35

Output fractures
Followed method 3 * 3

Answer 36

For the inverse?

Answer 37

Yes 😁 , Do I have mistakes?

Answer 38

This has a \(3\times 3\) lower down the page. There are also a few other ways to do this.
http://www.mathportal.org/linear-algebra/matrices/gauss-jordan.php

Answer 39

Oh, that was a lot of work! Let me write up the inverse I got.

Answer 40

Inverse:

\[
\left[\begin{matrix}
\frac{167}{486} & -\frac{41}{243} & -\frac{16}{81} \\
-\frac{17}{243} & \frac{23}{486} & \frac{7}{162} \\
-\frac{103}{486} & \frac{53}{486} & \frac{11}{81}
\end{matrix}\right]
\]

Answer 41

I can watch a lot of hard work
Finally ended this matrix
I need to make a lot of effort in the future
Thank you very much for your patience and effort

Answer 42

Yes, the inverse is not always easy. I am almost done with the next steps.

Answer 43

That times 4:
\[
\left[\begin{matrix}
\frac{334}{243} & -\frac{164}{243} & -\frac{64}{81} \\
-\frac{68}{243} & \frac{46}{243} & \frac{14}{81} \\
-\frac{260}{243} & \frac{106}{243} & \frac{44}{81}
\end{matrix}\right]
\]

Answer 44

Link that you putting explains how the solution
I could not very well so far, I need some time until perfected

Answer 45

Add finally, the determinant of that is last! Yes, it takes practice. Let me give you another link that also shows a different way to do the same thing. It uses the Adjoint and Determinant.
http://www.mathwords.com/i/inverse_of_a_matrix.htm
With a \(3\times 3) like this, the Adjoint and Determinant might be easier.

Answer 46

This is another good one for learning this:
http://www.purplemath.com/modules/mtrxinvr.htm
Sometimes it helos me to read what many people say about it. I learn a little form each one and finally I understand it all.

Answer 47

I'll put that in mind
Thank you very much, you give me a lot of help that I needed

Answer 48

I used a calculator and got an approximate determinant of: -0.003657979
I do not know what this is as a fraction and accurate answer, but it looks like a lot of work! That is actually a very hard problem they gave you to do!

Answer 49

I agree with you

Answer 50

I think there is an easier way if you use the rules of determinants. Look in your book for that. I know some are not that hard to use. For example:\[det(B^{-1})=\frac{1}{det(B)}\]

You might be able to find the \( det(A^2) \) and then use the rules to make this simpler. I just do not remember quite all the rules off the top of my head. And I thought this would be an easier matrix because the numbers start small, but they get big very fast and that makes it harder.

Answer 51

As we were working on your problem, I knew there was a better way to do it. I just could not remember it then. I did a little review on the properties of determinates through basic row operations and matrix operations.

Your matrix is:

\(\left[\begin{matrix}
1 & 2 & -1\\
2 & 1 & 3\\
0 & 3 & -2
\end{matrix}\right]\)

This is a \(3\times 3\) matrix.

The problem is:

Evaluate: \(det(4(2(3A^2)^T)^{-1})\)

The steps, if taken in order, are:

1) Find AA
2) Multiply by 3
3) Get the transpose
4) Multiply by 2
5) Get the inverse
6) Multiply by 4
7) Take the determinant of all of that.

However, there is an easier way! I knew there was, but I did not remember all of the rules for it.

Here are the rules:

1) The determinate of the product of two matrices is the product of the determinants.
http://www.proofwiki.org/wiki/Determinant_of_Matrix_Product

2) Multiply the matrix by a constant \(k\) changes the determinant by \(k^n\), where \(n\) is the number of rows. That is basically the following rule, but done once for each row:
http://www.proofwiki.org/wiki/Determinant_with_Row_Multiplied_by_Constant

3) The determinant of the transpose remains the same.
http://www.proofwiki.org/wiki/Determinant_of_Transpose

4) See 2.

5) The determinant of the inverse is the inverse of the determinant.
http://www.proofwiki.org/wiki/Determinant_of_Inverse

6) See 2.

7) We do this first!

If I use these rules, I can take the determinant first and then use the operations to change it!

\(det\left(\left[\begin{matrix}
1 & 2 & -1\\
2 & 1 & 3\\
0 & 3 & -2
\end{matrix}\right]\right)\Rightarrow\)

\(det(A)=(1)(1)(-2)+(2)(3)(0)+(-1)(2)(3)-(0)(1)(-1)-(3)(3)(1)-(-2)(2)(2) \Rightarrow\)

\(det(A)=-2+0-6-0-9+8 \Rightarrow det(A)=-9\)

Now that I have the determinant of A, I can use the above rules to modify the determinant and \textbf{not} do all the math we talked about. I thought the numbers were going to stay small, but they got large very fast, which is why this is a better way. For my rule examples I will use M for Matrix.

We start with \(det(A)=-9\), then:

\(det(M^2)=(det(M))(det(M))\therefore det(A^2)=(-9)(-9)\Rightarrow det(A^2)=81\)

\(det(kM)=k^ndet(M)\therefore det(3A^2)=3^3det(A^2)\Rightarrow det(3A^2)=729\)

\(det(M^T)=det(M)\therefore det((3A^2)^T)=det(3A^2)\Rightarrow det((3A^2)^T)=729\)

\(det(kM)=k^ndet(M)\therefore det(2(3A^2)^T)=2^3det((3A^2)^T)\Rightarrow det(2(3A^2)^T)=5832\)

\(det(M^{-1})=\dfrac{1}{det(M)}\therefore det((2(3A^2)^T)^{-1})=\dfrac{1}{det(2(3A^2)^T)}\Rightarrow det((2(3A^2)^T)^{-1})=\dfrac{1}{5832}\)

\(det(kM)=k^ndet(M)\therefore det(4(2(3A^2)^T)^{-1})=4^3det((2(3A^2)^T)^{-1})\Rightarrow\\ det(4(2(3A^2)^T)^{-1})=\dfrac{64}{5832}\)

\(\dfrac{64}{5832} = \dfrac{(8)(8)}{(8)(729)}\therefore det(4(2(3A^2)^T)^{-1})=\dfrac{8}{729}\)

That is a MUCH easier way to solve that problem!