Number of scalar addition required to compute P(QR) ,where P,Q,R be matrices or order $3× 5,5×7,7×3$ Firstly i don't understand what is meant by scalar additions here ? This is my main concern here ,computation comes after .If anyon can help it will be great
 A: The number of scalar additions is nothing but the addition of products of scalars of a row of a matrix to the column of another matrix in the multiplication of matrices.
      Suppose we have two matrices A of m cross n order and B of n cross p order.Let, m = 2, n = 3 and p = 4 which is defined below as:

              A = a b c  ; B = g h i j 
                  d e f        k l m n
                               o p q r

       Now we want to find their product that is conformable(i.e. can find their product as the number of columns in matrix A are equal to the number of rows of matrix B that is 3).So, their product is defined as:

            A*B = a*g+b*k+c*o   a*h+b*l+c*p   a*i+b*m+c*q   a*j+b*n+c*r
                  d*g+e*k+f*o   d*h+e*l+f*p   d*i+e*m+f*q   d*j+e*n+f*r 

        Here we can see when we multiply row 1 of matrix A to the column 1 of matrix B, we find 2 no. of scalar addition(+) at a11 position i.e. (n-1) = (3-1) = 2.We'll get same for a12, a13 and a14 position.So for 1st row total no of scalar additions are = (n-1)*p = 2*4 = 8 (i.e. p times (n-1)).Similarly, for second row at a21 position, there will be 2 no. of scalar additions, so for a22, a23 and a24. Therefore, total no. of scalar additions for the second row will be again 8.

    From here we can conclude that the total no. of scalar additions are = m*(n-1)*p = 2*2*4 = 16(can verify by 8+8 = 16)

    Now, we come to the question;

    Given; P of 3 cross 5,Q of 5 cross 7 and R of 7 cross 3 and we have to find no. of scalar additions for P(QR) i. e. first we have to find for QR.So,

 for Q*R, m = 5, n = 7, p = 3 i.e. m*(n-1)*p = 5*6*3 = 90

 Then Let Q*R = Some matrix B whose order is 5 cross 3(after multiplication of Q and R)

 Now for P*B, m = 3, n = 5,p = 3 i.e. m*(n-1)*p = 3*4*3 = 36

Hence total no. of scalar additions are = 90+36 = 126

