# the function MINVERSE

Posted by Sacha on December 09, 2001 4:34 AM

I try MINVERSE unsuccessfully. Can you help?

Posted by Tom Urtis on December 09, 2001 6:03 AM

Care to give any details about what difficulty you are having? Just as a guess though, because that function returns an inverse matrix for the matrix stored in an array, you need to be sure of 2 things:

(1) Your original array must have an equal number of rows and columns (example, A1:E5), and...
(2) Because you want to produce an array from an array, you need to enter the MINVERSE function as if it were an array formula (Ctrl+Shift+Enter).

Any help?

Tom Urtis

Posted by Aladin Akyurek on December 09, 2001 6:34 AM

Sacha --

Not sure what you're looking for, but hope the following will be of some help (my matrix algebra is pretty rusty; a good into is one by the mathematician/philosopher Kemeny and others: Kemeny, J. G., Snell, J. L., & Thompson, G. L. (1974). Introduction to finite mathematics. Prentice-Hall.)

Lets say that A1:C3 houses the following square matrix

={4,0,3;0,1,0;5,-6,4}

Note 1. Copy this including the =-sign. Activate A1 in a worksheet, paste it, and hit enter. Activate A1, select the range A1:C3, go to the Formula Bar, and hit CONTROL+SHIFT+ENTER at the same time. Edit|Copy the range then activate just A1 and do Edit|Paste Special -> Values.

In D1 enter: =MINVERSE(A1:C3)

You'll just see 4 in D1.

However, activate D1, go to the Formula Bar, select the MINVERSE formula and hit F9. What you see is the inverse of the matrix in A1:C3. Hit Escape to get the original formula.

If you'd want to have this inverse matrix in a range instead of as a constant array, activate D1, select D1:F3, go the Formula Bar, and hit CONTROL+SHIFT+ENTER at the same time.

In H1 enter: =MMULT(A1:C3,D1:F3)

and now activate H1, select the range H1:J3, go to the Formula Bar, and hit CONTROL+SHIFT+ENTER at the same time. Activate immediately Format H1:J3 such that decimals are set to 0.

H1:J3 is the identity matrix, showing that the original matrix in A1:C3 has indeed an inverse.

Posted by Aladin Akyurek on December 09, 2001 6:47 AM

Tom --

I didn't see your reply; Was busy composing my longish reply (it seems it took an half hour!).

My apologies.