{"id":4897,"date":"2021-08-31T17:52:51","date_gmt":"2021-08-31T17:52:51","guid":{"rendered":"https:\/\/mathemerize.com\/?p=4897"},"modified":"2022-01-16T17:01:22","modified_gmt":"2022-01-16T11:31:22","slug":"multiplication-of-matrices","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/multiplication-of-matrices\/","title":{"rendered":"Multiplication of Matrices – Examples and Definition"},"content":{"rendered":"
Here you will learn multiplication of matrices with definition and examples.<\/p>\n
Let’s begin –<\/p>\n
Definition : <\/strong>Two matrices A and B are conformable for the product AB if the number of columns in A is same as the number of row in B.<\/p>\n Let matrix A is of order \\(m\\times n\\) then m is the number of rows and n is the number of coumns in A<\/p>\n and matrix B is of order \\(n\\times p\\) then n is the number of rows and p is the number of columns in B.<\/p>\n Thus, if A = \\([a_{ij}]_{m\\times n}\\) and B = \\([b_{ij}]_{n\\times p}\\) are two matrices of order \\(m\\times n\\) and \\(n\\times p\\) respectively, then their product AB is conformable and of order \\(m\\times p\\) and is defined as<\/p>\n \\((AB)_{ij}\\) = ( \\(i^{th}\\) row of A) (( \\(j^{th}\\) column of B) for all i = 1, 2, ….., m and j = 1, 2, ……. , p.<\/p>\n \\(\\implies\\) \\((AB)_{ij}\\) = \\(\\begin{bmatrix} a_{i1} & a_{i2} & …… & a_{in} \\end{bmatrix}\\) \\(\\begin{bmatrix} b_{1j} \\\\ b_{2j} \\\\ . \\\\ . \\\\ . \\\\ b_{nj} \\end{bmatrix}\\)<\/p>\n Note<\/strong> : If A and B are two matrices such that AB exists, then BA may or may not exist.<\/p>\n If A = \\(\\begin{bmatrix} 2 & 1 & 3 \\\\ 3 & -2 & 1 \\\\ -1 & 0 & 1 \\end{bmatrix}\\) and B = \\(\\begin{bmatrix} 1 & -2 \\\\ 2 & 1 \\\\ 4 & -3 \\end{bmatrix}\\), then A is a \\(3\\times 3\\) matrix and B is a \\(3\\times 2\\) matrix. Therefore, A and b are conformable for the product AB and it is of order \\(3\\times 2\\) such that<\/p>\n \\((AB)_{11}\\) = (first row of A)(first column of B)<\/p>\n \\(\\implies\\) \\((AB)_{11}\\) = \\(\\begin{bmatrix} 2 & 1 & 3 \\end{bmatrix}\\) \\(\\begin{bmatrix} 1 \\\\ 2 \\\\ 4 \\end{bmatrix}\\)<\/p>\n = 2*1 + 1*2 + 3*4 = 16<\/p>\n \\((AB)_{12}\\) = (first row of A)(second column of B)<\/p>\n \\(\\implies\\) \\((AB)_{12}\\) = \\(\\begin{bmatrix} 2 & 1 & 3 \\end{bmatrix}\\) \\(\\begin{bmatrix} -2 \\\\ 1 \\\\ 3 \\end{bmatrix}\\)<\/p>\n = 2*(-2) + 1*1 + 3*(-3) = -12<\/p>\n \\((AB)_{21}\\) = (second row of A)(first column of B)<\/p>\n \\(\\implies\\) \\((AB)_{21}\\) = \\(\\begin{bmatrix} 3 & -2 & 1 \\end{bmatrix}\\) \\(\\begin{bmatrix} 1 \\\\ 2 \\\\ 4 \\end{bmatrix}\\)<\/p>\n = 3*1 + (-2)*2 + 1*(4) = 3<\/p>\n Similarly, we obtain <\/p>\n \\((AB)_{22}\\) = -11 , \\((AB)_{31}\\) = 3 and \\((AB)_{32}\\) = -1<\/p>\n \\(\\therefore\\) AB = \\(\\begin{bmatrix} 16 & -12 \\\\ 3 & -11 \\\\ 3 & -1 \\end{bmatrix}\\) <\/p>\n Note : <\/strong>In this case BA does not exist, because the number of columns in B is not same as the number of rows in A.<\/p>\n\n\nExample<\/span><\/strong> :<\/h2>\n