Find index of specific ordered pair in layered matrix

I have an ordered pair matrix, A=[x y], and a layered matrix, N, with three layers. What I want to do is search layer 2 for the correct x value and then layer 3 for the corresponding y value. Layer 2 represents the x axis and layer 3 represents the y axis. So in essence, I want to scan layer 2 and highlight the column that contains the x value and then scan the 3rd layer, highlight the row that contains the y value, and return the index at which layer 2 and layer 3 intersect. Any ideas on how to do this? I showed my setup of the layered matrix N and the values in matrix A. Layer 1 of matrix N are just random values and are irrelevant for this problem. Printing out layer 2 and 3 of matrix N would help visualize what I am talking about. Thanks!

cn = 3;
N11 = randi([0,5],cn,cn);
N12 = randi([1,4],cn,cn);
N13 = randi([0,7],cn,cn);
N14 = randi([0,6],cn,cn);
N21 = randi([1,3],cn,cn);
N22 = randi([1,7],cn,cn);
N23 = randi([4,6],cn,cn);
N24 = randi([2,7],cn,cn);
N31 = randi([1,4],cn,cn);
N32 = randi([6,7],cn,cn);
N33 = randi([1,7],cn,cn);
N34 = randi([3,4],cn,cn);
N41 = randi([2,5],cn,cn);
N42 = randi([0,3],cn,cn);
N43 = randi([5,6],cn,cn);
N44 = randi([0,4],cn,cn); 
N = [N11  N12  N13  N14;
     N21  N22  N23  N24;
     N31  N32  N33  N34;
     N41  N42  N43  N44];

[x, y] = size(N);
H1 = -x/2:1:x/2;
H2 = H1(find(H1~=0));
V1 = -y/2:1:y/2;
V2 = V1';
V3 = V2(find(V2~=0));
N(:,:,2) = H2(ones(x,1),:);
N(:,:,3) = V3(:,ones(y,1)) % finish making layered matrix, N 

A = [-6 -1; -3 4; 2 2; 5 -1];

Answer 1

[Kevin Xia]

Suppose that you want to find the row and column index where the second layer is -6 (A(1,1)) and where the third layer is -1 (A(1,2)). To go about doing so, use:

[row col]=find(N(:,:,2)==A(1,1) & N(:,:,3)==A(1,2));

This will give you the row and column where the layers intersect. To understand why this is the case, it is necessary to break down the boolean expression into three parts:

1.) N(:,:,2)==A(1,1)
2.) N(:,:,3)==A(1,2)
3.) &

Part 1 generates a logical matrix. If a value of N(:,:,2) is -6, then value of the logical matrix at that index is 1. Otherwise it is 0. The following logical matrix is thus generated:

   1   0   0   0   0   0   0   0   0   0   0   0
   1   0   0   0   0   0   0   0   0   0   0   0
   1   0   0   0   0   0   0   0   0   0   0   0
   1   0   0   0   0   0   0   0   0   0   0   0
   1   0   0   0   0   0   0   0   0   0   0   0
   1   0   0   0   0   0   0   0   0   0   0   0
   1   0   0   0   0   0   0   0   0   0   0   0
   1   0   0   0   0   0   0   0   0   0   0   0
   1   0   0   0   0   0   0   0   0   0   0   0
   1   0   0   0   0   0   0   0   0   0   0   0
   1   0   0   0   0   0   0   0   0   0   0   0
   1   0   0   0   0   0   0   0   0   0   0   0

Part 2 also generates a similar logical matrix, only for the condition where N(:,:,3) is -1:

   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0
   1   1   1   1   1   1   1   1   1   1   1   1
   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0

In part 3, the elementwise AND operator is used. The reason why the standard AND operator (&&) is not used is because && is used for scalar binary values. The elementwise AND (&) operator takes the corresponding element of both logical matrices and applies the AND operation. Thus, the following logical matrix is generated:

[Kevin Xia]

   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0
   1   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0

Find just finds the row index and column index of that 1 in this logical matrix.

[Andrew Poissant]

Thank you, this was very helpful and similar to my previous efforts. Is there a way to generalize this to do this for the entire matrix of A? Would I need to use two for loops to iterate over the matrix?

[Kevin Xia]

You would need one for loop to iterate over the rows of matrix A.

[Andrew Poissant]

Why would I not need to iterate over the columns? Does matlab automatically pair up the two numbers in the same row? What would that look like?

[Kevin Xia]

Here's a hint:

[row col]=find(N(:,:,2)==A(1,1) & N(:,:,3)==A(1,2));

The above code works for the first row of A. How would you get the above code to work for the second row of A?