Strassen’s matrix multiplication method is based on a divide & conquer rule. We have implemented a simple formula for you to find the Strassen’s matrix multiplication of the 4×4 matrix. In this post, I will try to explain the concept of Strassen’s 4×4 matrix multiplication with an example.

**Overview of Strassen’s algorithm**

- Strassen’s matrix multiplication will take less time to multiply two matrices using programming as compare to normal matrix multiplication.
- It is applicable to the matrices of order (n*n).
- If we have a 4×4 matrix, then we need to divide it into 4 equal parts.

## Formula of Strassen’s matrix multiplication

Suppose we have two matrices A & B then the applicable formula is,

**P1=**a * ( f – h )**P2 =**h * ( a + b )**P3 =**e * ( c + d )**P4 =**d * ( g – e )**P5 =**( a + d ) * ( e + h )**P6 =**( b – d ) * ( g + h )**P7 =**( a – c ) * ( e + f )

So the final matrix is,

### Example of 4×4 Strassen’s matrix

Consider a matrix which is of order 4×4, we need to find its multiplication.

To solve & multiply these matrices we need to follow some steps,

**Step 1: **Firstly divide A & B into 4 sub equivalent parts.

**Step 2:**** **Write this matrix in the form of four 2×2 matrices and name them accordingly like A1, A2, A3, A4, B1, B2, B3, B4

**Step 3: **Finally we need to find the multiplications of these matrices one by one in such a way that

- C1 = A1.B1
- C2 = A2.B2
- C3 = A3.B3
- C4 = A4.B4

So for C1, take consider the matrix A1 & B1

We know C1 = A1.B1

So,

**P1=**a*(f – h) = 3*(6-3) = 9**P2 =**h*(a + b) = 3*(3+2) = 15**P3 =**e*(c + d) = 5*(5+6) = 55**P4 =**d*(g – e) = 6*(1-5) = -24**P5 =**(a + d)*(e + h)= (3+6)*(5+3) = 72**P6 =**(b – d)*(g + h)= (2-6)*(1+3) = -16**P7 =**(a – c)*(e + f)= (3-5)+(5+6) = -22

We know,

So here,

Now for C2, take consider the matrix A2 & B2

We know C2 = A2.B2

Similarly here we have,

**P1=**a * ( f – h ) = 1**P2 =**h * ( a + b ) = 0**P3 =**e * ( c + d ) = 63**P4 =**d * ( g – e ) = -35**P5 =**( a + d ) * ( e + h ) = 56**P6 =**( b – d ) * ( g + h ) = -14**P7 =**( a – c ) * ( e + f ) = -8

Similarly,

So, C2 becomes,

For C3, We have to consider A3 & B3 and implement it in the form of C3=A3.B3.

Now we have to find P1, P2…..P7

**P1=**a * ( f – h ) = 21**P2 =**h * ( a + b ) = 39**P3 =**e * ( c + d ) = 15**P4 =**d * ( g – e ) = -6**P5 =**( a + d ) * ( e + h ) = 72**P6 =**( b – d ) * ( g + h ) = 20**P7 =**( a – c ) * ( e + f ) = 66

Now we have to find C,

So, C3 becomes

Finally we need to find C4 which will be equivalent to C4=A4.B4

So consider A4 & B4

Finally find P1, P2….P7

**P1=**a * ( f – h ) = 5**P2 =**h * ( a + b ) = 0**P3 =**e * ( c + d ) = 56**P4 =**d * ( g – e ) = 10**P5 =**( a + d ) * ( e + h ) = 70**P6 =**( b – d ) * ( g + h ) = -27**P7 =**( a – c ) * ( e + f ) = 16

Now we have find C,

Finally C4 will be equal to,

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-> Algorithm ( A Complete Introduction )

-> Getch() in C

C1 = A1*B1 + A1*B3

Got the wrong answer when I run into 4×4 matrix multiplication calculation.