Answer:
Step-by-step explanation:
root(23) - 8 < root(23) - 3 -3 is greater than -8
3 root(62) > 3 root(59) 62 > 59 Since the numbers are positive the roots of the larger number is going to be larger than the roots of the smaller number
-root(50) < - 3
The root of 50 is about 7. - 7 is smaller than - 3. Think money. Would you rather pay 3 dollars for a cup of coffee than 7 dollars for a cup of coffee?
root 15 + 2 < root(65) - 1
root(15) is roughly 4 when 2 is added to it, the result is about 6
root (65) is a touch over 8. When 1 is subtracted the result is about 7.
That's larger than the left side.
Color 2/8 of grid in green, 1/8 grid in red and 1/8 grid in blue
You could use flash lights, or lots of candles. Candles in the dark is truly beautiful.
I wrote my decision on a piece of paper.
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.
