Described high mining link mph log no bugs bugs
2.48 rounded to the nearest 10th would be 2.5<span />
Answer:
1. The matrix A isn't the inverse of matrix B.
2. |B|=12, |A|=12
Step-by-step explanation:
1. We want to know if matrix A is the inverse of matrix B, this means that if you do the product between B and A you have to obtain the identity matrix.
We have:
![A=\left[\begin{array}{cc}4&-2\\-1&3\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-2%5C%5C-1%263%5Cend%7Barray%7D%5Cright%5D)
and
![B=\left[\begin{array}{cc}3&2\\1&4\end{array}\right]](https://tex.z-dn.net/?f=B%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%262%5C%5C1%264%5Cend%7Barray%7D%5Cright%5D)
A and B are 2×2 matrices (2 rows and 2 columns), if you multiply them you have to obtain a 2×2 matrix.
Then if A is the inverse of B:

Where,
![I=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]](https://tex.z-dn.net/?f=I%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D)
Observation:
If you have two matrices:
![A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\and\\B=\left[\begin{array}{cc}e&f\\g&h\end{array}\right]\\\\\\A.B=\left[\begin{array}{cc}(a.e+b.g)&(a.f+b.h)\\(c.e+d.g)&(c.f+d.h)\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%5C%5Cand%5C%5CB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7De%26f%5C%5Cg%26h%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CA.B%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%28a.e%2Bb.g%29%26%28a.f%2Bb.h%29%5C%5C%28c.e%2Bd.g%29%26%28c.f%2Bd.h%29%5Cend%7Barray%7D%5Cright%5D)
Now:
![B.A=\left[\begin{array}{cc}3&2\\1&4\end{array}\right].\left[\begin{array}{cc}4&-2\\-1&3\end{array}\right]\\\\\\B.A=\left[\begin{array}{cc}4.3+(-2).1&4.2+(-2).4\\(-1).3+3.1&(-1).2+3.4\end{array}\right]\\\\\\B.A=\left[\begin{array}{cc}12-2&8-8\\-3+3&-2+12\end{array}\right]\\\\\\B.A=\left[\begin{array}{cc}10&0\\0&10\end{array}\right]](https://tex.z-dn.net/?f=B.A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%262%5C%5C1%264%5Cend%7Barray%7D%5Cright%5D.%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-2%5C%5C-1%263%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CB.A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4.3%2B%28-2%29.1%264.2%2B%28-2%29.4%5C%5C%28-1%29.3%2B3.1%26%28-1%29.2%2B3.4%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CB.A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D12-2%268-8%5C%5C-3%2B3%26-2%2B12%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CB.A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D10%260%5C%5C0%2610%5Cend%7Barray%7D%5Cright%5D)
![B.A=\left[\begin{array}{cc}10&0\\0&10\end{array}\right]\neq \left[\begin{array}{cc}1&0\\0&1\end{array}\right]=I\\\\\\B.A\neq I](https://tex.z-dn.net/?f=B.A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D10%260%5C%5C0%2610%5Cend%7Barray%7D%5Cright%5D%5Cneq%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%3DI%5C%5C%5C%5C%5C%5CB.A%5Cneq%20I)
Then, the matrix A isn't the inverse of matrix B.
2. If you have a matrix A:
![A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D)
The determinant of the matrix is:

Then the determinant of B is:
![B=\left[\begin{array}{cc}3&2\\1&4\end{array}\right]](https://tex.z-dn.net/?f=B%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%262%5C%5C1%264%5Cend%7Barray%7D%5Cright%5D)


The determinant of A is:
![A=\left[\begin{array}{cc}4&-2\\-1&3\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-2%5C%5C-1%263%5Cend%7Barray%7D%5Cright%5D)


Part (a)
The angles would stay the same. This is because similar figures keep the same shape, but the size will shrink or grow.
A real world application is to simply move your head closer to the page and the figure will appear to get larger. The angles themselves stay the same but the side lengths will get larger.
====================================================
Part (b)
While the angles stay the same under a dilation operation, the side lengths do not. As the instructions state: "The side lengths are 3 times longer" (paraphrased).
So we multiply each side length in the diagram by 3
- DE = 3*JK = 3*4.5 = 13.5
- EF = 3*KL = 3*3 = 9
- FG = 3*LM = 3*2 = 6
- GD = 3*MJ = 3*2.5 = 7.5
Refer to the diagram below. I've used the resize feature to enlarge the image to make figure DEFG
Answer:
<h2>Hence the second option is the correct answer</h2>
Step-by-step explanation:
In this problem we are required to factor out the given terms as stated in the expression, what is obtainable is that we look for terms(greatest terms) that are common to all the terms in the expression and factor it out.
here is a way to do it
given the following expression

From the expression above we can see that 3 is common to all the terms of the expression so that we have

hence the second option is the correct answer