Answer:
![A^{-1}=\left[\begin{array}{cc}-3&-4\\1&1\end{array}\right]](https://tex.z-dn.net/?f=A%5E%7B-1%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-3%26-4%5C%5C1%261%5Cend%7Barray%7D%5Cright%5D)
message SHOW_ME_THE_MONEY_
Step-by-step explanation:
The matrix
![A=\left[\begin{array}{cc}1&4\\-1&-3\end{array}\right]\rightarrow |A|=(1 \times -3)-(-1\times 4)=1\\\rightarrow A^{-1}=\left[\begin{array}{cc}-3&-4\\1&1\end{array}\right] \\](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%264%5C%5C-1%26-3%5Cend%7Barray%7D%5Cright%5D%5Crightarrow%20%7CA%7C%3D%281%20%5Ctimes%20-3%29-%28-1%5Ctimes%204%29%3D1%5C%5C%5Crightarrow%20A%5E%7B-1%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-3%26-4%5C%5C1%261%5Cend%7Barray%7D%5Cright%5D%20%5C%5C)
We can check that in fact A*A^⁻1=I_2 the identity matrix of size 2 x 2.
Now the message was divided in 1 x 2 matrices, then we have that the sequence given is the result of multiplying m by A, so to get m again we multiply now by A^⁻1. and we get the next table
Encoded message Decoded message message in letters by association
11 52 19 8 S H
-8 -9 15 23 O W
-13 -39 0 13 _ M
5 20 5 0 E _
12 56 20 8 T H
5 20 5 0 E _
-2 7 13 15 M O
9 41 14 5 N E
25 100 25 0 Y _
Then the message decoded is SHOW_ME_THE_MONEY_
I don’t know the depth but I can tell you that you need to multiply everything using the formula v=l*w*h. So v=3.5*0.5*x
X= weight of TV
y= weight of CD
3x+5y= 62.5
3x+ 2y= 52
You can either solve for a variable, or eliminate a variable by subtracting the second equation from the first:
3x + 5y = 62.5
-(3x+2y = 52)
0 + 3y = 10.5, so y = 10.5/3 or 3.5.
Now plug y= 3.5 back into any equation to get x:
3x + 5(3.5) = 62.5
3x + 17.5 = 62.5
3x = 45
x = 15
Always check work. Both equations should be true for x=15, y= 3.5
Thye have to have like denominators.
So it becomes, 6/15+5/15 which is equal to 11/15
