Multiplication of Matrices

A matrix is said to be as ordered rectangular array of number. The operation on matrices that is the multiplication of a matrix generally falls into two categories

Scalar Multiplication: In the matrix, a real number is called a scalar in which a single number is being multiplied by all the elements present in the matrix.
Multiplication of the matrix with another entire matrix.

Scalar Multiplication

Multiplication of scalar means, multiplying a matrix by a number i.e. a real number. In general, we may define multiplication of a matrix by a scalar as follows: If $A=[a_{ij}]_{m*n}$ is a matrix and k is a scalar, then kA is another matrix which is obtained by multiplying each element of A by the scalar number k. In other words, $kA=k[a_{ij}]_{m*n}$ that is, (i,j) element of kA is $[k_{ij}]$ for all possible values of i and j.

Example 1: Suppose Mohan has a factory which produces clothes for boys and girls in three different price tags. The quantity produced by a factory is represented as matrices given below : $Boys\begin {bmatrix} 50\\40 \\60 \end {bmatrix}$ $Girls\begin{bmatrix} 80\\30\\70 \end{bmatrix}$ . Now Mohan wants to doubled the production for all three different prices tags clothes. Represent it in a matrix form.

Solution: From the above information the matrix can be represented by doubling each price categories as follows: $Boys=\begin {bmatrix} 2*50\\2*40 \\2*60 \end {bmatrix}$ and $Girls=\begin{bmatrix} 2*80\\2*30\\2*70 \end{bmatrix}$

Total Production = $\begin {bmatrix} 100&160\\80&60 \\120&140 \end {bmatrix}$

Example 2: If $A=\begin {bmatrix} 2 & 4\\ 6 & 8\end {bmatrix}$ . Find 4A.

Solution : $4A=4\begin {bmatrix} 2 & 4\\ 6 & 8\end {bmatrix}$ = $\begin {bmatrix} 8 & 16\\ 24 & 22\end {bmatrix}$

Multiplication between two matrices

Multiplication between two matrices is feasible if the number of columns of the first matrix is same as the matrix of rows in another matrix then matrix multiplication can be done. In general, Let $A=[a_{ij}]$ be an m*n matrix and $B=[b_{jk}]$ 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 $[c_{jk}]$ of the matrix C, we take the $i^{th}$ row of A and $k^{th}$ column of B, multiply then element wise and take sum of all these products. In other words, If $A=[a_{ij}]_{m*n}$ , $B=[b_{ij}]_{n*p}$ then the $i^{th}$ row of A is $\begin {bmatrix} a_{i1} & a_{i2} &... a_{in} \end {bmatrix}$ and the $k^{th}$ column of B is $\begin {bmatrix} b_{1k}\\ b_{2k}\\...\\ b_{nk} \end {bmatrix}$ , then $[c_{jk}]$ = $a_{i1} b_{1k}+a_{i2} b_{2k}+ a_{i3} b_{3k} + .... a_{in} b_{nk}$ . The matrix $C=[c_{ik}]_{m*p}$ is the product of A and B.

Properties of multiplication of matrices

The associative law: For any three matrices A, B and C. We have (AB)C =A(BC), whenever both sides of the equality are defined.
The distributed law: For three matrices A, B and C. (i) A (B+C) = AB + AC and (ii) (A+B)C = AC + BC, whenever both sides of equality are defined.
The existence of multiplicative identity: For every square matrix A, there exists an identity matrix of same order such that IA = AI = A.

Example 3: If $A=\begin {bmatrix} 1 & -1 & 2\\ 0 & 3 & 4 \end {bmatrix}$ and $B=\begin {bmatrix} 2 & 7\\ -1 &1 \\ 5&-4 \end {bmatrix}$ . Find product of AB.

Solution: This is a 2*2 matrix in which each entry in the sum of the product across some row of the A with the corresponding entries down some columns of B. These four computations is:

Entry in first row first column $\begin {bmatrix} \textbf{1}& \mathbf{-1} & \textbf{2}\\ 0 & 3 & 4 \end {bmatrix} \begin {bmatrix} \textbf{2} & 7\\ \textbf{-1} &1 \\ \textbf{5}&-4 \end {bmatrix}$ = $\begin {bmatrix} (1)(2)+(-1)(-1)+(2)(5)&.. \\ .. &.. \end {bmatrix}$

Entry in first row second column $\begin {bmatrix} \textbf{1}& \mathbf{-1} & \textbf{2}\\ 0 & 3 & 4 \end {bmatrix} \begin {bmatrix} 2 & \textbf{7}\\ -1 &\textbf{1} \\ 5&\textbf{-4} \end {bmatrix}$ = $\begin {bmatrix} 13&(1)(7)+(-1)(1)+2(-4) \\ .. &.. \end {bmatrix}$

Entry in second row first column $\begin {bmatrix} 1 & -1 & 2\\ \textbf{0} & \textbf{3} & \textbf{4} \end {bmatrix} \begin {bmatrix} \textbf{2} & 7\\ \textbf{-1} &1 \\ \textbf{5}&-4 \end {bmatrix}$ = $\begin {bmatrix} 13&-2 \\ 0(2)+3(-1)+4(5) &.. \end {bmatrix}$

Entry in second row second column $\begin {bmatrix} 1 & -1 & 2\\ \textbf{0} & \textbf{3} & \textbf{4} \end {bmatrix} \begin {bmatrix} 2 & \textbf{7}\\ -1 &\textbf{1} \\ 5&\textbf{-4} \end {bmatrix}$ = $\begin {bmatrix} 13&-2 \\ 17 &0(7)+3(1)+4(-4) \end {bmatrix}$

Thus $CD=\begin {bmatrix} 13&-2 \\ 17 &-13 \end {bmatrix}$

Exercise

If $A=\begin {bmatrix} -1&2 \\ 5&1 \end {bmatrix}$ , Find 2A
Let $A=\begin {bmatrix} 1&1 \\ 0&0 \end {bmatrix}$ and $B=\begin {bmatrix} 1&0 \\ -1&0 \end {bmatrix}$ . Find AB and BA.
If $A=\begin {bmatrix} 7 & 3\\ 2 &8 \\ 6&5 \end {bmatrix}$ and $B=\begin {bmatrix} 8 &14 &0 \\ 7& 11& 5\\ 8&-4 &19 \end {bmatrix}$ . Find AB.
Show that the matrix $A=\begin {bmatrix} 1&2 \\ 1&3 \end {bmatrix}$ satisfies the equation $A^{2}-4A+I=0$ .
If $A=\begin {bmatrix}-5\\6 \\ 0 \end {bmatrix}$ and $B=\begin {bmatrix} 3 &-1 \end {bmatrix}$ . Find AB.

Scalar Multiplication

Multiplication between two matrices

Properties of multiplication of matrices

Exercise

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