Multiplication of Two Matrices

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 = \(\begin{bmatrix} a & b\\ c & d \end{bmatrix}\) and B = \(\begin{bmatrix} x & y & z\\ l & m & n \end{bmatrix}\)

A is a 2 × 2 matrix and B is a 2 × 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 = \(\begin{bmatrix} a & b\\ c & d \end{bmatrix}\) \(\begin{bmatrix} x & y & z\\ l & m & n \end{bmatrix}\)

   = \(\begin{bmatrix} a(x) + b(l) & a(y) + b(m) & a(z) + b(n)\\c(x) +d(l) & c(y) + d(m) & c(z) + d(n) \end{bmatrix}\)

Product of Two Matrices
Multiplication of Two Matrices

Clearly, the product BA is not possible because the number of columns in B(=3) ≠ 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 = \(\begin{bmatrix} 2 & 5\\ -1 & 3 \end{bmatrix}\) and B = \(\begin{bmatrix} 2 & 5\\ -1 & 3 \end{bmatrix}\). Find AB and BA. Is AB = BA?

Solution:

Here, A is of the order 2 × 2 and B is of the order 2 × 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 = \(\begin{bmatrix} 2 & 5\\ -1 & 3 \end{bmatrix}\) \(\begin{bmatrix} 2 & 5\\ -1 & 3 \end{bmatrix}\)

     = \(\begin{bmatrix} 2 × 1 + 5 × 4 & 2 × 1 + 5 × (-2)\\ (-1) × 1 + 3 × 4 & (-1) × 1 + 3 × (-2) \end{bmatrix}\) 

     = \(\begin{bmatrix} 22 & -8\\ 11 & -7 \end{bmatrix}\)

BA = \(\begin{bmatrix} 2 & 5\\ -1 & 3 \end{bmatrix}\) \(\begin{bmatrix} 2 & 5\\ -1 & 3 \end{bmatrix}\)

     = \(\begin{bmatrix} 1 × 2 + 1 × (-1) & 1 × 5 + 1 × 3\\ 4 × 2 + (-2) × (-1) & 4 × 5 + (-2) × 3 \end{bmatrix}\) 

     = \(\begin{bmatrix} 1 & 8\\ 10 & 14 \end{bmatrix}\).


Clearly, \(\begin{bmatrix} 22 & -8\\ 11 & -7 \end{bmatrix}\) ≠ \(\begin{bmatrix} 1 & 8\\ 10 & 14 \end{bmatrix}\).

Therefore, AB ≠ BA.


2. Let X = \(\begin{bmatrix} 11 & 4\\ -5 & 2 \end{bmatrix}\) and I = \(\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\). Prove that XI = IX = A.

Solution:

XI = \(\begin{bmatrix} 11 & 4\\ -5 & 2 \end{bmatrix}\) \(\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\)

    = \(\begin{bmatrix} 11 × 1 + 4 × 0 & 11 × 0 + 4 × 1\\ -5 × 1 + 2 × 0 & -5 × 0 + 2 × 1 \end{bmatrix}\) 

    = \(\begin{bmatrix} 11 & 4\\ -5 & 2 \end{bmatrix}\) = X

IX = \(\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\)\(\begin{bmatrix} 11 & 4\\ -5 & 2 \end{bmatrix}\) 

    = \(\begin{bmatrix} 1 × 11 + 0 × (-5) & 1 × 4 + 0 × 2\\ 0 × 11 + 1 × (-5) & 0 × 4 + 1 × 2 \end{bmatrix}\) 

    = \(\begin{bmatrix} 11 & 4\\ -5 & 2 \end{bmatrix}\) = X


Therefore, AI = IA =A. (Proved)





10th Grade Math

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