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Here we will learn the process of Multiplication of two matrices.
Two matrices A and B are conformable (compatible) for multiplication
(i) AB if the number of columns in A = the number of rows in B
(ii) BA if the number of columns in B = the number of rows in A.
To find the product AB when A and B are conformable for multiplication AB
Let A = [abcd] and B = [xyzlmn]
A is a 2 Γ 2 matrix and B is a 2 Γ 3 matrix.
Therefore, the number of columns in A = the number of rows in B = 2.
Therefore, AB can be found because A, B are conformable for multiplication AB.
The product AB is defined as
AB = [abcd] [xyzlmn]
= [a(x)+b(l)a(y)+b(m)a(z)+b(n)c(x)+d(l)c(y)+d(m)c(z)+d(n)]
Clearly, the product BA is not possible because the number of columns in B(=3) β the number of rows in A(=2).
Note: Given two matrices A and B, AB may be found but BA may not be found. It is also possible that neither AB nor BA can be found, or both AB and BA can be found.
Solved Example on Multiplication of Two Matrices:
1. Let A = [25β13] and B = [25β13]. Find AB and BA. Is AB = BA?
Solution:
Here, A is of the order 2 Γ 2 and B is of the order 2 Γ 2.
So, the number of columns in A = the number of rows in B. Hence, AB can be found. Also, the number of columns in B = the number of rows in A. Hence, BA can also found.
Now,
AB = [25β13] [25β13]
= [2Γ1+5Γ42Γ1+5Γ(β2)(β1)Γ1+3Γ4(β1)Γ1+3Γ(β2)]
= [22β811β7]
BA = [25β13] [25β13]
= [1Γ2+1Γ(β1)1Γ5+1Γ34Γ2+(β2)Γ(β1)4Γ5+(β2)Γ3]
= [181014].
Clearly, [22β811β7] β [181014].
Therefore, AB β BA.
2. Let X = [114β52] and I = [1001]. Prove that XI = IX = A.
Solution:
XI = [114β52] [1001]
= [11Γ1+4Γ011Γ0+4Γ1β5Γ1+2Γ0β5Γ0+2Γ1]
= [114β52] = X
IX = [1001][114β52]
= [1Γ11+0Γ(β5)1Γ4+0Γ20Γ11+1Γ(β5)0Γ4+1Γ2]
= [114β52] = X
Therefore, AI = IA =A. (Proved)
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