# Easy way to remember Strassen’s Matrix Equation

Strassen’s matrix is a Divide and Conquer method that helps us to multiply two matrices(of size n X n).

You can refer to the link, for having the knowledge about Strassen’s Matrix first :

Divide and Conquer | Set 5 (Strassen’s Matrix Multiplication)

But this method needs to cram few equations, so I’ll tell you the simplest way to remember those :

You just need to remember 4 Rules :

- AHED (Learn it as ‘Ahead’)
- Diagonal
- Last CR
- First CR

Also, consider X as (Row +) and Y as (Column -) matrix

Follow the Steps :

- Write P1 = A; P2 = H; P3 = E; P4 = D
- For P5 we will use Diagonal Rule i.e.

(Sum the Diagonal Elements Of Matrix X ) * (Sum the Diagonal Elements Of Matrix Y ), we get

P5 = (A + D)* (E + H) - For P6 we will use Last CR Rule i.e. Last Column of X and Last Row of Y and remember that Row+ and Column- so i.e. (B – D) * (G + H), we get

P6 = (B – D) * (G + H) - For P7 we will use First CR Rule i.e. First Column of X and First Row of Y and remember that Row+ and Column- so i.e. (A – C) * (E + F), we get

P7 = (A – C) * (E + F)

P1 = A

P2= H

P3= E

P4= D

P5= ( A + D ) * ( E + H )

P6= ( B – D ) * ( G + H)

P7= ( A – C ) * ( E + F) - Come Back to P1 : we have A there and it’s adjacent element in Y Matrix is E, since Y is Column Matrix so we select a column in Y such that E won’t come, we find F H Column, so multiply A with (F – H)

So, finally P1 = A * (F – H)

P1 = A * ( F – H)

P2= H

P3= E

P4= D

P5= ( A + D ) * ( E + H )

P6= ( B – D ) * ( G + H)

P7= ( A – C ) * ( E + F) - Come Back to P2 : we have H there and it’s adjacent element in X Matrix is D, since X is Row Matrix so we select a Row in X such that D won’t come, we find A B Column, so multiply H with (A + B)

So, finally P2 = (A + B) * H - Come Back to P3 : we have E there and it’s adjacent element in X Matrix is A, since X is Row Matrix so we select a Row in X such that A won’t come, we find C D Column, so multiply E with (C + D)

So, finally P3 = (C + D) * E

P1= A * ( F – H )

P2= H * ( A + B )

P3= E * ( C + D )

P4= D

P5= ( A + D ) * ( E + H )

P6= ( B – D ) * ( G + H)

P7= ( A – C ) * ( E + F) - Come Back to P4 : we have D there and it’s adjacent element in Y Matrix is H, since Y is Column Matrix so we select a column in Y such that H won’t come, we find G E Column, so multiply D with (G – E)

So, finally P4 = D * (G – E)We are done with P1 – P7 equations, so now we move to C1 – C4 equations in Final Matrix C :

- Remember Counting : Write P1 + P2 at C2
- Write P3 + P4 at its diagonal Position i.e. at C3
- Write P4 + P5 + P6 at 1st position and subtract P2 i.e. C1 = P4 + P5 + P6 – P2
- Write odd values at last Position with alternating – and + sign i.e. P1 P3 P5 P7 becomes

C4 = P1 – P3 + P5 – P7

This article is contributed by **Mohit Gupta 🙂**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**

In case you wish to attend **live classes **with experts, please refer **DSA Live Classes for Working Professionals **and **Competitive Programming Live for Students**.