Just as two or more real numbers can be multiplied, it is possible to multiply two or more matrices too. Multiplication of matrices generally falls into two categories, Scalar Matrix Multiplication, in which a single number is multiplied with every other element of the matrix and Vector Matrix Multiplication wherein an entire matrix is multiplied by another one.
Suggested Videos
Definition
The product of two matrices A and B is defined if the number of columns of A is equal to the number of rows of B. Let A = [aij] be an m × n matrix and B = [bjk] be an n × p matrix. Then the product of the matrices A and B is the matrix C of order m × p. To get the (i, k)th element c of the matrix C, we take the ith row of A and kth column of B, multiply them element-wise and take the sum of all these products.
Browse more Topics Under Matrices
When Do We Do Multiplication?
When the number of columns of the 1st matrix must equal the number of rows of the 2nd matrix. In other words, To multiply an m×n matrix by an n×p matrix, the ns must be the same, and the result is an m×p matrix.
(m×n) × (n×p) → m×p
Matrix Multiplication
Whenever we multiply a matrix by another one we need to find out the “dot product” of rows of the first matrix and columns of the second. Now, what does that mean? Let us see an example below:
 [source: mathisfun]
The dot product is where we multiply matching members, then sum them up:
(1, 2, 3) . (7, 9, 11) = 1×7 + 2×9 + 3×11 = 58
We match the 1st members (1 and 7), multiply them, likewise for the 2nd members (2 and 9) and the 3rd members (3 and 11), and finally sum them up.
If we multiply a 2×3 matrix with a 3×1 matrix, the product matrix is 2×1
$$ \begin{bmatrix} r_{11} & r_{12} & r_{13}\\ r_{21} & r_{22} & r_{23}\end{bmatrix} × \begin{bmatrix} c_{11} \\ c_{21} \\ c_{31} \end{bmatrix} = \begin{bmatrix} M_{11}\end{bmatrix}$$
Here is how we get M11 and M12 in the product.
M12 = r21× t11 + r22× t21  + r23×t31
Properties of Matrix Multiplication
1) Associative Law
The assosiative law for any three matrices A, B and C, we have(AB) C = A (BC), whenever both sides of the equality are defined. Example: $$ (AB) C = ( \begin{bmatrix} 1 & 3 \\ 2 & 4\end{bmatrix} × \begin{bmatrix} 1\\ 2 \end{bmatrix} )× \begin{bmatrix} 3 & 4 \end{bmatrix}$$
$$  = \begin{bmatrix} 7 \\ 10\end{bmatrix} × \begin{bmatrix} 3 & 4 \end{bmatrix} $$
$$ Â = \begin{bmatrix} 21 & 28 \\ 30 & 40\end{bmatrix}$$
$$ A (BC) = \begin{bmatrix} 1 & 3 \\ 2 & 4\end{bmatrix} × ( \begin{bmatrix} 1\\ 2 \end{bmatrix} × \begin{bmatrix} 3 & 4 \end{bmatrix} )$$
$$  = \begin{bmatrix} 1 & 3 \\ 2 & 4\end{bmatrix} × \begin{bmatrix} 3 & 4 \\ 6 & 8 \end{bmatrix} $$
$$ Â = \begin{bmatrix} 21 & 28 \\ 30 & 40\end{bmatrix}$$
Therefore, LHS = RHS
2) Distributive Law
The distributive law for three matrices A, B and C.
- A (B + C) =AB + AC
- (A + B) Â C = AC + BC, whenever both sides of equality are defined.
Example: (i)$$ A (B + C) =  \begin{bmatrix} 1 & 2 \end{bmatrix} × (\begin{bmatrix} 1\\ 2 \end{bmatrix} + \begin{bmatrix} 3 \\ 4 \end{bmatrix}) $$
$$=  \begin{bmatrix} 1 & 2 \end{bmatrix} × \begin{bmatrix} 4\\ 6 \end{bmatrix} $$
$$= \begin{bmatrix}16 \end{bmatrix} $$
$$ AB + AC =  \begin{bmatrix} 1 & 2 \end{bmatrix} × \begin{bmatrix} 1\\ 2 \end{bmatrix} + \begin{bmatrix} 1 & 2 \end{bmatrix} ×\begin{bmatrix} 3 \\ 4 \end{bmatrix} $$
$$= \begin{bmatrix}5 + 11 \end{bmatrix} $$
$$= \begin{bmatrix}16 \end{bmatrix} $$
Therefore, LHS = RHS
Similarly, (ii) can also be proved.
3) Existence of Multiplicative Identity
The existence of multiplicative identity for every square matrix A, there exists an identity matrix of same order such that  IA = AI = A. Example: $$ IA =  \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}  \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} $$
$$ = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} = A$$
$$ AI = Â \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} $$
$$ = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} = A$$
Therefore, LHS = RHS = A
Solved Examples For You
Question: $$ If \ A =\begin{bmatrix} 0 & 1\\ 0 & 0\end{bmatrix},$$ I is the unit matrix of order 2 and a,b are arbitrary constants then (aI + bA)2Â is equal to
- a2I + abA
- a2I + 2abA
- I2a + b2A
- None of these
Solution:Â $$A =\begin{bmatrix} 0 & 1\\ 0 & 0\end{bmatrix}$$
$$A^2=\begin{bmatrix} 0 & 1\\ 0 & 0\end{bmatrix} \begin{bmatrix} 0 & 1\\ 0 & 0\end{bmatrix}$$
⇒ A2 = 0
Now, consider (aIÂ +Â bA)2Â =Â (aIÂ +Â bA)Â (aIÂ +Â bA)
                     =a2I + abIA + baAI + b2A2
Question. How can one multiply matrices together?
Answer. The multiplication of matrices can take place with the following steps:
- The number of columns in the first one must the number of rows in the second one.
- Now you must multiply the first matrix’s elements of each row by the elements belonging to each column of the second matrix.
- Finally, add the products.
Question. How can one solve a 3 by 3 matrix?
Answer. In order to work out the determinant of a 3×3 matrix, one must multiply a by the determinant of the 2×2 matrix that does not happen to be a’s column or row or column. Similarly, do the same for b and for c. Finally, sum them up.
Question. Is it possible to multiply a 2×3 and 2×2 matrix?
Answer. Yes, multiplication of 2×3 and 2×2 matrix is certainly possible. Also, the result would be a 2×3 matrix.
Question. What are the different types of matrices?
Answer. There are many types of matrices that exist. However, the most commonly used are rectangular matrix, square matrix, rows matrix, columns matrix, scalar matrix, diagonal matrix, identity matrix, triangular matrix, null matrix, and transpose of a matrix.
MATHEMATICS WAS TOO DIFFICULT FOR ME BUT WHEN I LEARN FROM TOPPR I FEEL MATHEMATICS IS TOO EASY I LIKE IT