Question 1204571
<font color=black size=3>
Part (a)


Let's ignore the Z7 notation for a moment.


For now we're just going to worry about finding the inverse.


The given matrix is
{{{A = (matrix(3,3,3,2,4,4,3,5,2,1,2))}}}
I'll place the 9 items in separate boxes
<table border = "1" cellpadding = "5"><tr><td>3</td><td>2</td><td>4</td></tr><tr><td>4</td><td>3</td><td>5</td></tr><tr><td>2</td><td>1</td><td>2</td></tr></table>


Append the 3x3 identity matrix to the right hand side like so
<table border = "1" cellpadding = "5"><tr><td>3</td><td>2</td><td>4</td><td>1</td><td>0</td><td>0</td></tr><tr><td>4</td><td>3</td><td>5</td><td>0</td><td>1</td><td>0</td></tr><tr><td>2</td><td>1</td><td>2</td><td>0</td><td>0</td><td>1</td></tr></table>
The goal is to use matrix row reduction (i.e. Gauss-Jordan Elimination) to transform the left 3x3 block into the identity matrix. 
The resulting right hand side will be the inverse.


<table border = "1" cellpadding = "5"><tr><td>3</td><td>2</td><td>4</td><td>1</td><td>0</td><td>0</td></tr><tr><td>4</td><td>3</td><td>5</td><td>0</td><td>1</td><td>0</td></tr><tr><td>2</td><td>1</td><td>2</td><td>0</td><td>0</td><td>1</td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>2</td><td>1</td><td>2</td><td>0</td><td>0</td><td>1</td><td>R1 <--> R3</td></tr><tr><td>4</td><td>3</td><td>5</td><td>0</td><td>1</td><td>0</td></tr><tr><td>3</td><td>2</td><td>4</td><td>1</td><td>0</td><td>0</td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>1/2</td><td>1</td><td>0</td><td>0</td><td>1/2</td><td>(1/2)*R1 --> R1</td></tr><tr><td>4</td><td>3</td><td>5</td><td>0</td><td>1</td><td>0</td></tr><tr><td>3</td><td>2</td><td>4</td><td>1</td><td>0</td><td>0</td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>1/2</td><td>1</td><td>0</td><td>0</td><td>1/2</td></tr><tr><td>0</td><td>1</td><td>1</td><td>0</td><td>1</td><td>-2</td><td>R2  -  4*R1 --> R2</td></tr><tr><td>3</td><td>2</td><td>4</td><td>1</td><td>0</td><td>0</td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>1/2</td><td>1</td><td>0</td><td>0</td><td>1/2</td></tr><tr><td>0</td><td>1</td><td>1</td><td>0</td><td>1</td><td>-2</td></tr><tr><td>0</td><td>1/2</td><td>1</td><td>1</td><td>0</td><td>-3/2</td><td>R3  -  3*R1 --> R3</td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>1/2</td><td>1</td><td>0</td><td>0</td><td>1/2</td></tr><tr><td>0</td><td>1</td><td>1</td><td>0</td><td>1</td><td>-2</td></tr><tr><td>0</td><td>0</td><td>1/2</td><td>1</td><td>-1/2</td><td>-1/2</td><td>R3  -  (1/2)*R2 --> R3</td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>1/2</td><td>1</td><td>0</td><td>0</td><td>1/2</td></tr><tr><td>0</td><td>1</td><td>1</td><td>0</td><td>1</td><td>-2</td></tr><tr><td>0</td><td>0</td><td>1</td><td>2</td><td>-1</td><td>-1</td><td>2*R3 --> R3</td></tr></table>

Almost done. Just a few more steps.
<table border = "1" cellpadding = "5"><tr><td>1</td><td>1/2</td><td>1</td><td>0</td><td>0</td><td>1/2</td></tr><tr><td>0</td><td>1</td><td>0</td><td>-2</td><td>2</td><td>-1</td><td>R2  -  R3 --> R2</td></tr><tr><td>0</td><td>0</td><td>1</td><td>2</td><td>-1</td><td>-1</td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>1/2</td><td>0</td><td>-2</td><td>1</td><td>3/2</td><td>R1  -  R3 --> R1</td></tr><tr><td>0</td><td>1</td><td>0</td><td>-2</td><td>2</td><td>-1</td></tr><tr><td>0</td><td>0</td><td>1</td><td>2</td><td>-1</td><td>-1</td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>0</td><td>0</td><td>-1</td><td>0</td><td>2</td><td>R1  -  (1/2)R2 --> R1</td></tr><tr><td>0</td><td>1</td><td>0</td><td>-2</td><td>2</td><td>-1</td></tr><tr><td>0</td><td>0</td><td>1</td><td>2</td><td>-1</td><td>-1</td></tr></table>
Here is a very useful tool
<a href="http://www.math.odu.edu/~bogacki/lat/">http://www.math.odu.edu/~bogacki/lat/</a>
It is called "linear algebra toolkit". It is a collection of matrix solvers that show step by step solutions. I recommend it to check your work.


We now have the left hand 3x3 block as the identity matrix.
The 3x3 block on the right hand side is the inverse.


The inverse of {{{A = (matrix(3,3,3,2,4,4,3,5,2,1,2))}}} is {{{A^(-1) = (matrix(3,3,-1,0,2,-2,2,-1,2,-1,-1))}}}


This can be verified by showing that 
{{{A*A^(-1) =  (matrix(3,3,3,2,4,4,3,5,2,1,2))*(matrix(3,3,-1,0,2,-2,2,-1,2,-1,-1)) = (matrix(3,3,1,0,0,0,1,0,0,0,1))}}} 
and 
{{{A^(-1)*A =(matrix(3,3,-1,0,2,-2,2,-1,2,-1,-1))*(matrix(3,3,3,2,4,4,3,5,2,1,2)) = (matrix(3,3,1,0,0,0,1,0,0,0,1))}}}
I'll let the student do these two calculations to verify.



Now let's return to the idea of Z7.
This is the set of integers modulo 7. We look at the remainders when dividing over 7.
Those remainders are: {0,1,2,3,4,5,6}
If we have a value too large, then subtract off 7 repeatedly until landing on one of those remainders listed above.
For instance, start with 15 and subtract off two copies of 7 to get 15-7-7 = 1, which shows that 15/7 gives remainder 1.


If we have a value too small, then repeatedly add 7 until landing on one of the remainders.
Example: start with -2 and land on -2+7 = 5
It shows that -2 = 5 (mod 7)


So when we're working with Z7, then the inverse {{{A^(-1) = (matrix(3,3,-1,0,2,-2,2,-1,2,-1,-1))}}} turns into {{{A^(-1) = (matrix(3,3,6,0,2,5,2,6,2,6,6))}}}


This wraps up part (a)


---------------------------------------------------------------------------------------------


Part (b)



Use the trick mentioned in the previous part to find that
-4 = 3 (mod 7)
9 = 2 (mod 7)


So,
{{{b = (matrix(3,1,1,-4,9)) = (matrix(3,1,1,3,2))}}} when in Z7.



The matrix equation {{{Ax = b}}} turns into {{{A^(-1)*Ax = A^(-1)*b}}} when we left-multiply both sides by the inverse of matrix A.



That turns into {{{x = A^(-1)*b}}}


I'll skip steps, but you'll then perform <a href="https://www.algebra.com/algebra/homework/Matrices-and-determiminant/Matrices-and-determiminant.faq.question.441330.html">matrix multiplication</a> to get {{{x = A^(-1)*b = (matrix(3,3,6,0,2,5,2,6,2,6,6))*(matrix(3,1,1,3,2)) = (matrix(3,1,10,23,32))}}}


Then we'll need to find the remainder of each of those items mod 7
10 = 3 (mod 7)
23 = 2 (mod 7)
32 = 4 (mod 7)


The final answer to part (b) is {{{x = (matrix(3,1,3,2,4))}}}


What this means is that
{{{(matrix(3,3,3,2,4,4,3,5,2,1,2))*(matrix(3,1,3,2,4)) = (matrix(3,1,1,3,2))}}} when in Z7.
</font>