Answer:
<u>The correct answer is D. 2x, -4y, and 8.</u>
Step-by-step explanation:
The concept of the terms of an equation is used for the monomials of each member participating in the equation. In Mathematics we distinguish between two types of terms:
1. The terms in x: when the terms are accompanied by a literal.
2. The independent terms: when the terms are constituted by numerical elements.
Answer: I think the answer is 2b+9
Step-by-step explanation:
Answer:
Acc to the exponential rule
{[A^m]} ^2
A ^m*n
Hence A raise to 3 whole raise to 2 is equal to
A raise to 3*2=6
That is A raise to 6
Your answer is 2^12 check out this video that explains it
https://www.youtube.com/watch?v=ZV430MtO12s
Answer:
A) Verified
B) Proved
Step-by-step explanation:
a) Let's verify it for 2 x 2 matrix,
![A=\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D)
and ![B=\left[\begin{array}{ccc}e&f\\g&h\end{array}\right]](https://tex.z-dn.net/?f=B%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7De%26f%5C%5Cg%26h%5Cend%7Barray%7D%5Cright%5D)
![AB=\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]\left[\begin{array}{ccc}e&f\\g&h\end{array}\right]=\left[\begin{array}{ccc}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=AB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7De%26f%5C%5Cg%26h%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da.e%2Bb.g%26a.f%2Bb.h%5C%5Cc.e%2Bd.g%26c.f%2Bd.h%5Cend%7Barray%7D%5Cright%5D)
![(AB)^{-1}=\frac{1}{(a.e+b.g)(c.f+d.h)-(a.f+b.h)(c.e+d.g)}\left[\begin{array}{ccc}c.f+d.h&-(a.f+b.h)\\-(c.e+d.g)&a.e+b.g\end{array}\right]](https://tex.z-dn.net/?f=%28AB%29%5E%7B-1%7D%3D%5Cfrac%7B1%7D%7B%28a.e%2Bb.g%29%28c.f%2Bd.h%29-%28a.f%2Bb.h%29%28c.e%2Bd.g%29%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dc.f%2Bd.h%26-%28a.f%2Bb.h%29%5C%5C-%28c.e%2Bd.g%29%26a.e%2Bb.g%5Cend%7Barray%7D%5Cright%5D)
![A^{-1}=\frac{1}{a.d-b.c} \left[\begin{array}{ccc}d&-b\\-c&a\end{array}\right]](https://tex.z-dn.net/?f=A%5E%7B-1%7D%3D%5Cfrac%7B1%7D%7Ba.d-b.c%7D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dd%26-b%5C%5C-c%26a%5Cend%7Barray%7D%5Cright%5D)
![B^{-1}=\frac{1}{e.h-f.g} \left[\begin{array}{ccc}h&-f\\-g&e\end{array}\right]](https://tex.z-dn.net/?f=B%5E%7B-1%7D%3D%5Cfrac%7B1%7D%7Be.h-f.g%7D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dh%26-f%5C%5C-g%26e%5Cend%7Barray%7D%5Cright%5D)
![B^{-1}A^{-1}=\frac{1}{(a.e+b.g)(c.f+d.h)-(a.f+b.h)(c.e+d.g)}\left[\begin{array}{ccc}c.f+d.h&-(a.f+b.h)\\-(c.e+d.g)&a.e+b.g\end{array}\right]](https://tex.z-dn.net/?f=B%5E%7B-1%7DA%5E%7B-1%7D%3D%5Cfrac%7B1%7D%7B%28a.e%2Bb.g%29%28c.f%2Bd.h%29-%28a.f%2Bb.h%29%28c.e%2Bd.g%29%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dc.f%2Bd.h%26-%28a.f%2Bb.h%29%5C%5C-%28c.e%2Bd.g%29%26a.e%2Bb.g%5Cend%7Barray%7D%5Cright%5D)
So it is proved that the results are same.
b) Now, let's prove it for any n x n matrix.
