What is this question asking me? Help appreciated! I was wondering if anyone would be able to help me understand what this question is asking me. How would I go about working these out on Wolfram Alpha? I'm not too sure how to input them. Any help is appreciated!
(Text transcribed from image here.)

This question will use the following matrices:
  $$A =\left[\begin{array}{ccc}
1 & 1 & 1 \\
2 & 1 & 2 \\
1 & 1 & 2 
\end{array}\right] \qquad
C = \left[\begin{array}{cc}
1 & 1 \\
2 & 0 \\
2 & 1
\end{array}\right]$$ Perform teh following matrix operations, either by hand or using a computer program such as Matlab, Excel, or Wolfram Alpha (http://wolframalpha.com): $$(a) \; |A^4| \qquad (b) \; A^{-1} \qquad (c)\; D = A^{-1} C \qquad (d) \; AD - C$$ What can you say about your answer to part $(d)$, given the equation from part $(c)$?

 A: The nice thing about Wolfram Alpha is that it attempts to interpret human language. I am sure after this you will use it often. You can ask the questions however you please and modify your question based on results to get the answer you are looking for. I just put your first few in Wolfram, and I think this will get you rolling. The learning moment here is how to type a matrix into Wolfram Alpha, and how you can ask your question in a variety of ways. Try typing these phrases:
a) Determinant of {{1,1,1},{2,1,2},{1,1,2}} 
then try
Determinant of {{1,1,1},{2,1,2},{1,1,2}}^4
b) Inverse of {{1,1,1},{2,1,2},{1,1,2}}
c) Inverse of {{1,1,1},{2,1,2},{1,1,2}} times {{1,1},{2,0},{2,1}}
I will leave the rest to you, just follow this lead. When you type a matrix in, the rows go in curl brackets {}, and all of those rows go into a main bracket like {{},{},{}}. Have fun with Wolfram Alpha, and take the time to read about the philosophy behind this great Wolfram gift.
You might also consider developing some proficiency at performing these sorts of matrix operations with a pencil and paper. It really is quite liberating.
A: So, matrix multiplication works like this:
$\begin{bmatrix}
 a_1 & b_1 & c_1 \\
 a_2 & b_2 & c_2
\end{bmatrix}
\begin{bmatrix}
 x_1 & y_1 \\
 x_2 & y_2 \\
 x_3 & y_3
\end{bmatrix}
= \begin{bmatrix}
a_1x_1+b_1x_2+c_1x_3 & a_1y_1+b_1y_2+c_1y_3 \\
a_2x_1+b_2x_2+c_2x_3 & a_2y_1+b_2y_2+c_2y_3 \\
\end{bmatrix}$
The identity matrix looks like this:
$I_1 = \begin{bmatrix}1\end{bmatrix},
I_2 = \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix},
I_3 = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
And if there's a square matrix $A$ and you want its inverse, then you multiply $A$ with a square matrix filled with symbols $a, b, c, ...$ which will give you a matrix of formulas. If you set the formula in the topleft to 1, and solve for the symbols there, and went on to make the product matrix the identity matrix, you would get the inverse $A^{-1}$ of $A$. But that's not usually how you find the inverse. You usually use Gaussian elimination, which you can find information about using to get an inverse on Wikipedia and YouTube.
When you take a 2x3 matrix (meaning 2 rows and 3 columns - notice in the multiplication formula the rows change with those of the LEFThand matrix), and you multiply it with a 3-dimensional column vector on the right (vector-matrix multiplication is another thing), you find it gives you a 2-dimensional vector. So, as a function, there's no way this can have an inverse. But importantly, it shows that you're taking a map from 3D space to 2D space. When you perform Gaussian elimination, you're actually just changing the basis of the output space, literally looking at it from a different perspective, where the coordinates look nicer. If you wrote the matrix like this:
$\begin{array}
& x \\ y
\end{array}
\begin{bmatrix}
 1 & 2 & 3 \\
 4 & 5 & 6
\end{bmatrix}$
A row reduction operation would give you this:
$\begin{array}
& x-y \\ y
\end{array}
\begin{bmatrix}
 -3 & -3 & -3 \\
 4 & 5 & 6
\end{bmatrix}$
and the expressions on the left of the matrix will still span the x and y axes.
