I want to check is some number is larger than the width of a matrix. How do I access one dimension of a matrix variable instead of both dimensions?

```
dim([A]->L1
L1(1
L1(2
```

Weird. It told me there was a Data Type error when I tried that earlier. Thanks anyway.

Really? Weird. I wonder if theres a certain way you have to do it. Will have to test

Edit:As long as you are only retrieving one of the dimensions and not the other:

```
:dim([A]
:Ans(1
or
:Ans(2
```

The solution to a complex problem is often a simple answer.

Ok, that seems to work. I must have typed it in wrong earlier.

Is there any way to do it with one line of code though? I tried

`dim([A])(1`

but it multiplied the matrix by 1 instead of accessing the first element of the dimensions list.

I would like to make sure the program user correctly input a 2x2 matrix using:

**If dim([A])≠ {2,2}**

however the calculator is giving me a Data Type error ("*Wrong value or variable type entered. Ex: Attempted to store a matrix to a list*."

Alternatively I could do:

**dim([A])->L _{1}
If not(L_{1}(1)=2 and L_{1}(2)=2)**

but that seems kind of inelegant.

Instead of storing the dimension to a list, you can use the If() statement for checking the dimension of matrix A like this:

`If 4≠sum(dim([A]`

This method I mentioned above takes the sum( of the list dimension of [A], it combines the numbers of the list. If 4 is not equal to the sum of {2,2} then the code under the If statement is executed.

`Hewwo, my name is Achak Claw. I was formerly BioHazard.`

Not really ideal, since the matrix could be of the size `3,1` or `1,3` and pass.

The missing piece is the `min(` command:

`If min(dim([A])={2,2`

See, when comparing to lists, you are comparing each element to its counterpart, returning a list of 0's and 1's of the same size as the lists based on whether each pair of elements were equal. Taking the minimum of this list tells you if *any* of these values are 0, which would imply that at least one pair of elements doesn't match.

The solution to a complex problem is often a simple answer.