三个矩阵乘法的效率及通用表达式推导问题咨询

Hey there! I totally get where you're coming from—trying to nail down the general expression for multiplying three matrices and wrap your head around the nuances can feel frustrating when you're stuck. Let me break this down clearly for you.

通用表达式推导 #

First, let's define our matrices properly to avoid confusion:

• Let A be an ( m \times n ) matrix
• Let B be an ( n \times p ) matrix
• Let C be an ( p \times q ) matrix

Matrix multiplication is associative, meaning ( (AB)C = A(BC) )—the end result will be the same regardless of which pair you multiply first. The general element-wise expression for the resulting ( m \times q ) matrix D is:
[
D[i][j] = \sum_{k=1}^{p} \left( \sum_{l=1}^{n} A[i][l] \times B[l][k] \right) \times C[k][j]
]
Or, since multiplication is associative, you can rearrange the order of summation to:
[
D[i][j] = \sum_{l=1}^{n} \sum_{k=1}^{p} A[i][l] \times B[l][k] \times C[k][j]
]
In plain terms: each element in the final matrix is the sum of the products of corresponding elements traced through the three matrices. For row ( i ) of A, column ( k ) of B, and column ( j ) of C, you multiply the three values together and sum all those products.

关于计算效率的实用技巧 #

While the mathematical result is the same, the number of arithmetic operations (multiplications and additions) can differ drastically based on the order you choose to multiply the matrices:

• If you compute ( AB ) first: This takes ( m \times n \times p ) multiplications, then multiplying the result by C takes ( m \times p \times q ) multiplications. Total operations: ( mnp + mpq = mp(n + q) )
• If you compute ( BC ) first: This takes ( n \times p \times q ) multiplications, then multiplying by A takes ( m \times n \times q ) multiplications. Total operations: ( npq + mnq = nq(m + p) )

You’ll want to pick the order with the smaller total number of operations. For example, if ( m=10 ), ( n=100 ), ( p=20 ), ( q=50 ):

• ( AB ) first: ( 1010020 + 102050 = 20000 + 10000 = 30000 ) operations
• ( BC ) first: ( 1002050 + 1010050 = 100000 + 50000 = 150000 ) operations

Clearly, multiplying ( AB ) first is way more efficient here!

Also, I noticed you mentioned an image in your original question—if you can share the specific details from that image (like particular matrix dimensions, a problem example, or any constraints you're working with), I can refine this explanation even further to fit your exact scenario.

备注：内容来源于stack exchange，提问作者Araf Raihan