(A) Use the coding matrix $A=\left[\begin{smallmatrix}1&3\\5&-3\end{smallmatrix}\right]$ to encode the word jump
(B) Using it's Inverse $A^{-1}= \left[\begin{smallmatrix}\frac{1}{6}&-\frac{1}{6}\\\frac{5}{18}&\frac{1}{18}\end{smallmatrix}\right]$ to decode $\left[\begin{smallmatrix}26\\ -40 \\ 53\\23\end{smallmatrix}\right]$
I'm confused about how to encode and decode these two problems can anyone help??? My teacher tried to explain to me but his accent is heavy and I couldn't understand him. This is frustrating!
Presumably this coding matrix works on only two letters at a time and treats the standard basis as a=1, b=2, c=3, d=4, etc...
So, two letters at a time try to encode a word. As an indirect example, suppose we wanted to encode the word "apples." Originally, I would think of "apples" as the string (1, 16, 16, 12, 5, 19) since it is the first letter, 16th letter, 16th letter, 12th letter, etc...
Let us see how the first two letters change when I encode it: $A=\left[\begin{smallmatrix}1&3\\5&-3\end{smallmatrix}\right]\cdot \left[\begin{smallmatrix}1\\16\end{smallmatrix}\right]$. The first entry will be $1\cdot 1 + 3\cdot 16=49$, and the second entry will be $5\cdot 1 + (-3)\cdot 16=-43$. For more detail on how this operation occurred, look up the definition of matrix multiplication.
Continuing, you see that the string (1,16,16,12,5,19) gets encoded as (49,-43,52,44,62,-32).
Decode the message in a similar fashion, noting that if $Ax=b$ you have $x=A^{-1}b$ (assuming $A$ is invertible, which in this example it is).
If we were to apply $A^{-1}$ to the letters two at a time from (49,-43,52,44,62,-32), we will get an output of (1,16,16,12,5,19). We can now interpret these numbers as letters A,P,P,L,E,S.
The benefit to encoding words like this is that if you see a number in your coded message, it doesn't always refer to the same letter in the uncoded message, as opposed to a normal caesarian cipher.
So, for your problems:
What is "jump" in terms of numbers? J is the 10th letter, U is the 21st letter, etc...
two at a time use $A$ to encode the message, $\left[\begin{smallmatrix}1&3\\5&-3\end{smallmatrix}\right]\cdot \left[\begin{smallmatrix}10\\21\end{smallmatrix}\right]=?$,...
Use $A^{-1}$ to decode the word (20,-40,53,23) two numbers at a time,
Reinterpret the resulting string as letters.