Disclaimer: This is for the language associated with the TI-83+ through TI-84+SE models. I'm unfamiliar with the syntax you're using.

Let's say I have a matrix with five rows and three columns…

```
[A]
[[4 6 5]
[3 9 5]
[4 1 2]
[7 5 3]
[8 1 6]]
```

…and I use Matr►list( to transform the second column into a list:

```
Matr►list([A],2,
L₁
Done
L₁
{6 9 1 5 1}
```

(Lists are more flexible, and should be the default if you're working with arrangements of numbers that don't require multiple rows.)

Suppose I want to find out which element position has a five in it. Performing the test outright…

…returns a list where each element has been compared. The fourth slot returned "true," so it evidently holds the five.

If you were to now multiply this result by a new list where each element identifies with the position that it occupies…

```
Ans*{1,2,3,4,5
{0 0 0 4 0}
```

…just the four gets multiplied by the "1" you saw earlier. You can now take the max( of this result to obtain your answer.

A seq( command will produce the ordered list shown above, but there is a different approach: assuming that you have a list prepared in Ans, you can multiply Ans by zero, not( all of the elements inside it to turn it into a list of ones, and then take the cumSum( of that to get the numbers in increasing order. For example:

```
L₁
{6 9 1 5 1}
cumSum(not(0Ans
{1 2 3 4 5}
```

Proceed to multiply that by the binary list generated from a comparison, and then pull out the position number using max(.

```
max((L₁=5)cumSum
(not(0L₁
4
```

Addendum: You can use a For( loop to test each column of a multicolumnar matrix.

Layperson's technique (for the simple-minded):

0→A

0→B

dim([A]→L1

For(R,1,L1(1

For(C,1,L1(2

If 5=[A](R,C

Then

R→A

C→B

End

End

End

Disp "ROW:",A,"COLUMN:",B