11.3: Matrix Indexing

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Let’s look at how we can reference parts of a matrix.

Example 11.3.1

Consider the matrix A = [1 2 3 4;5 6 7 8]. There are actually two ways to view this matrix, either as a rectangular array of 2 rows and 4 columns, or as a list of 8 elements. Suppose we wanted to isolate the 7 in the matrix A and store it as the variable temp. First, we can think of the 7 as being located in the second row and third column. In this case, we can type:
>> A = [1 2 3 4;5 6 7 8];
>> temp = A(2,3)

with the result being:

temp =
7
Second, we can think of 7 as being one of the eight elements total. But, it is crucial to realize that we count elements in this way using a “column precedence.” This means that we count, one at a time, down the columns. This means that we can think of 7 as being located in the 6th entry, or:

>> temp = A(6)

also gives the result:

temp =
7

For completeness, in the last example, if we think of A as a matrix:

A(1,1) = 1 A(1,2) = 2 A(1,3) = 3 A(1,4) = 4
A(2,1) = 5 A(2,2) = 6 A(2,3) = 7 A(2,4) = 8

or, thinking of A as a vector:
A(1) = 1 A(3) = 2 A(5) = 3 A(7) = 4
A(2) = 5 A(4) = 6 A(6) = 7 A(8) = 8
If we wanted to store the entire first row of A in the variable firstrow, we would say that we want “all four columns of the first row.” This suggests that we can use the colon operator to shorten our work. Namely,

>> firstrow = A(1,1:4)

which gives

firstrow =
1 2 3 4

But, there’s an even shorter way to do this! If the colon doesn’t have a start and end value, it simply lists all possible values! Namely,

>> firstrow = A(1,:)

also gives

firstrow =
1 2 3 4

Ok, now what if we wanted the first row, but not the element in the first column? There are two ways to do this. First, we can use the colon as:

>> mostoffirstrow = A(1,2:4)

which gives

mostoffirstrow =
2 3 4

But, what if the matrix changes and we don’t know how big A has changed to? Those sneaky programmers at Mathworks have a work around:

>> mostoffirstrow = A(1,2:end)

also gives

mostoffirstrow =
2 3 4

Exercise \(\PageIndex{1}\) "Hilbert matrix" rows

Consider the 5 x 5
H5 = [1 1/2 1/3 1/4 1/5;
1/2 1/3 1/4 1/5 1/6;
1/3 1/4 1/5 1/6 1/7;
1/4 1/5 1/6 1/7 1/8;
1/5 1/6 1/7 1/8 1/9];

By referencing H5, create a matrix that consists of the 2nd and 3rd rows of H5 (Do not simply retype the entries).

Answer: Add texts here. Do not delete this text first.