Notes On Multiplication of Matrices - CBSE Class 12 Maths
In multiplication we multiply the elements in a row of the first matrix by the elements in a column of the second matrix. We have two matrices X and Y. X = $\left[\begin{array}{ccc}{\text{x}}_{\text{11}}& {\text{x}}_{\text{12}}& {\text{x}}_{\text{13}}\\ {\text{x}}_{\text{21}}& {\text{x}}_{\text{22}}& {\text{x}}_{\text{23}}\end{array}\right]$ Y = $\left[\begin{array}{cc}{\text{y}}_{\text{11}}& {\text{y}}_{\text{12}}\\ {\text{y}}_{\text{21}}& {\text{y}}_{\text{22}}\\ {\text{y}}_{\text{31}}& {\text{y}}_{\text{32}}\end{array}\right]$ We will first multiply the elements in the first row of X to the corresponding elements in the first column of Y.And we multiply the elements in the first row X to the elements in the second column of Y. We will write the sum of the products of the elements. In similar manner, we multiply the elements in the other rows and columns. Finally, we get the matrix as shown. = $\left[\begin{array}{cc}{{\text{x}}_{\text{11}}\text{y}}_{\text{11}}\text{+}{{\text{x}}_{\text{12}}\text{y}}_{\text{21}}\text{+}{\text{x}}_{\text{13}}{\text{y}}_{\text{31}}& {{\text{x}}_{\text{11}}\text{y}}_{\text{12}}\text{+}{{\text{x}}_{\text{12}}\text{y}}_{\text{22}}\text{+}{\text{x}}_{\text{13}}{\text{y}}_{\text{32}}\\ {\text{x}}_{\text{11}}{\text{y}}_{\text{11}}\text{+}{{\text{x}}_{\text{22}}\text{y}}_{\text{21}}\text{+}{{\text{x}}_{\text{23}}\text{y}}_{\text{31}}& {\text{x}}_{\text{21}}{\text{y}}_{\text{12}}\text{+}{{\text{x}}_{\text{22}}\text{y}}_{\text{22}}\text{+}{\text{x}}_{\text{23}}{\text{y}}_{\text{32}}\end{array}\right]$ The multiplication operation is possible only when the number of columns of the first matrix, X, is equal to the number of rows of the second matrix, Y. If we have two matrices, X and Y, with orders m x p and p x n, respectively, then their product is of the order m x n. X = [xik]mxp = $\left[\begin{array}{ccccccc}{\text{x}}_{\text{11}}& {\text{x}}_{\text{12}}& {\text{x}}_{\text{13}}& \cdots & {\text{x}}_{\text{1k}}& \cdots & {\text{x}}_{\text{1p}}\\ {\text{x}}_{\text{21}}& {\text{x}}_{\text{22}}& {\text{x}}_{\text{23}}& \cdots & {\text{x}}_{\text{2k}}& \cdots & {\text{x}}_{\text{2p}}\\ ⋮& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}\\ {\text{x}}_{\text{i1}}& {\text{x}}_{\text{i2}}& {\text{x}}_{\text{i3}}& \text{⋯}& {\text{x}}_{\text{ik}}& \text{⋯}& {\text{x}}_{\text{ip}}\\ \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}\\ {\text{x}}_{\text{m1}}& {\text{x}}_{\text{m2}}& {\text{x}}_{\text{m3}}& \text{⋯}& {\text{x}}_{\text{mk}}& \text{⋯}& {\text{x}}_{\text{mp}}\end{array}\right]$ 1 ≤ i ≤ m; 1 ≤ k ≤ p Y =     [ykj]pxn    = $\left[\begin{array}{ccccccc}{\text{y}}_{\text{11}}& {\text{y}}_{\text{12}}& {\text{y}}_{\text{13}}& \cdots & {\text{y}}_{\text{1j}}& \cdots & {\text{y}}_{\text{1n}}\\ {\text{y}}_{\text{21}}& {\text{y}}_{\text{22}}& {\text{y}}_{\text{23}}& \cdots & {\text{y}}_{\text{2j}}& \cdots & {\text{y}}_{\text{2n}}\\ ⋮& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}\\ {\text{y}}_{\text{k1}}& {\text{y}}_{\text{k2}}& {\text{y}}_{\text{k3}}& \text{⋯}& {\text{y}}_{\text{kj}}& \text{⋯}& {\text{y}}_{\text{kn}}\\ \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}\\ {\text{y}}_{\text{p1}}& {\text{y}}_{\text{p2}}& {\text{y}}_{\text{p3}}& \text{⋯}& {\text{y}}_{\text{pj}}& \text{⋯}& {\text{y}}_{\text{pn}}\end{array}\right]$  1 ≤ k ≤ p; 1 ≤ j ≤ n The product of these matrices be equal to matrix Z, which is of the order m by n. Z = [zij]mxn  =  $\left[\begin{array}{ccccccc}{\text{z}}_{\text{11}}& {\text{z}}_{\text{12}}& {\text{z}}_{\text{13}}& \cdots & {\text{z}}_{\text{1j}}& \cdots & {\text{z}}_{\text{1n}}\\ {\text{z}}_{\text{21}}& {\text{z}}_{\text{22}}& {\text{z}}_{\text{23}}& \cdots & {\text{z}}_{\text{2j}}& \cdots & {\text{z}}_{\text{2n}}\\ ⋮& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}\\ {\text{z}}_{\text{i1}}& {\text{z}}_{\text{i2}}& {\text{z}}_{\text{i3}}& \text{⋯}& {\text{z}}_{\text{ij}}& \text{⋯}& {\text{z}}_{\text{in}}\\ \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}\\ {\text{z}}_{\text{m1}}& {\text{z}}_{\text{m2}}& {\text{z}}_{\text{m3}}& \text{⋯}& {\text{z}}_{\text{mj}}& \text{⋯}& {\text{z}}_{\text{mn}}\end{array}\right]$      1 ≤ i ≤ m; 1 ≤ j ≤ n zij = xi1y1j + xi2y2j + xi3y3j + ..... + xikykj + .... + xipypj = $\sum _{\text{k=1}}^{\text{p}}\text{}$  xikykj In the expression, Zij represents the element in the ith row and jth column of matrix Z. The multiplication operation over two matrices does not satisfy the commutative property. X x Y ≠ Y x X (Not always) Case 1: XY defined, YX not defined. X = ${\left[\begin{array}{ccc}\text{1}& \text{2}& \text{3}\\ \text{4}& \text{5}& \text{6}\end{array}\right]}_{\text{2x3}}$ Y = ${\left[\begin{array}{ccc}\text{1}& \text{2}& \text{7}\\ \text{3}& \text{4}& \text{8}\\ \text{5}& \text{6}& \text{9}\end{array}\right]}_{\text{3x3}}$ No. of columns of X =3=No. of rows of Y No. of columns of Y = 3 ≠ 2 = No. of rows of Y Product YX is not possible. Case 2: Sometimes, product XY and YX are both defined, but XY does not necessarily equal YX. XY : Defined, YX: Defined but XY ≠ YX X = ${\left[\begin{array}{ccc}\text{1}& \text{1}& \text{2}\\ \text{3}& \text{1}& \text{5}\\ \text{1}& \text{3}& \text{2}\end{array}\right]}_{\text{3x3}}$ Y = ${\left[\begin{array}{ccc}\text{2}& \text{1}& \text{5}\\ \text{1}& \text{4}& \text{2}\\ \text{3}& \text{4}& \text{1}\end{array}\right]}_{\text{3x3}}$ XY = $\left[\begin{array}{ccc}\text{2+1+6}& \text{1+4+8}& \text{5+2+2}\\ \text{6+1+15}& \text{3+4+20}& \text{15+2+5}\\ \text{2+3+6}& \text{1+12+8}& \text{5+6+2}\end{array}\right]$ =$\left[\begin{array}{ccc}\text{9}& \text{13}& \text{9}\\ \text{22}& \text{27}& \text{22}\\ \text{11}& \text{21}& \text{13}\end{array}\right]$ YX = $\left[\begin{array}{ccc}\text{2+3+5}& \text{2+1+15}& \text{4+5+10}\\ \text{1+12+2}& \text{1+4+6}& \text{2+20+4}\\ \text{3+12+1}& \text{3+4+3}& \text{6+20+2}\end{array}\right]$ = $\left[\begin{array}{ccc}\text{10}& \text{18}& \text{19}\\ \text{15}& \text{11}& \text{26}\\ \text{16}& \text{10}& \text{28}\end{array}\right]$ XY ≠ YX Case 3: Multiplication of diagonal matrices of same order will commute A = ${\left[\begin{array}{ccc}\text{1}& \text{0}& \text{0}\\ \text{0}& \text{2}& \text{0}\\ \text{0}& \text{0}& \text{3}\end{array}\right]}_{\text{3x3}}$ B = ${\left[\begin{array}{ccc}\text{2}& \text{0}& \text{0}\\ \text{0}& \text{3}& \text{0}\\ \text{0}& \text{0}& \text{5}\end{array}\right]}_{\text{3x3}}$ AB = ${\left[\begin{array}{ccc}\text{2}& \text{0}& \text{0}\\ \text{0}& \text{6}& \text{0}\\ \text{0}& \text{0}& \text{15}\end{array}\right]}_{\text{3x3}}$ BA = ${\left[\begin{array}{ccc}\text{2}& \text{0}& \text{0}\\ \text{0}& \text{6}& \text{0}\\ \text{0}& \text{0}& \text{15}\end{array}\right]}_{\text{3x3}}$ Case 4: XY = O (null matrix) may not always imply that either X = O or Y = O Ex: X =         Y =      XY =          x           =        The multiplication property over matrices satisfies the associative law and the distributive law. For any matrices X, Y and Z, (X x Y) x Z = X x (Y x Z) Ex: A =           B = $\left[\begin{array}{cc}\text{7}& \text{4}\\ \text{2}& \text{5}\\ \text{1}& \text{-3}\end{array}\right]$      C =   $\left[\begin{array}{ccc}\text{2}& \text{-3}& \text{1}\\ \text{4}& \text{3}& \text{2}\end{array}\right]$                              (A x B) x C = A x (B x C) Now L.H.S = (A x B) x C A x B =    x $\left[\begin{array}{cc}\text{7}& \text{4}\\ \text{2}& \text{5}\\ \text{1}& \text{-3}\end{array}\right]$  =                                                              A x B =  (A x B) x C = x     =   =                            R.H.S = A x (B x C)  B x C  =   x     B x C  =   x    =                         B x C = A x (B x C)   =       x                                       = $\left[\begin{array}{ccc}\text{60+96+10}& \text{-18+36+12}& \text{30+48+5}\\ \text{90+24-50}& \text{-27+9-60}& \text{45+12-25}\\ \text{180+48-10}& \text{-54+18-12}& \text{90+24-5}\end{array}\right]$ = (A x B) x C = A x (B x C) Both the matrices are equal. We can see that the multiplication satisfies the associative law. According to the distributive law for any matrices X, Y and Z, X x (Y + Z) = X x Y + X x Z (X + Y) x Z = X x Z + Y x Z Ex: (A x B) x C = A x C + B x C A =      B =   C = $\left[\begin{array}{c}\text{5}\\ \text{1}\\ \text{4}\end{array}\right]$ (A + B) x C A + B =   +            (A + B) x C =     x  $\left[\begin{array}{c}\text{5}\\ \text{1}\\ \text{4}\end{array}\right]$                      =   =  A x C + B x C A x C =     x   =  =                                      B x C =   +  =  =                  A x C + B x C = +  =                The distributive law is applicable to the multiplication of matrices. Multiplicative identity exists for any square matrix. If X is a square matrix, there exists an identity matrix, I, of similar order such that X = [xij]nxn , I = [iij]nxn X x I = I x X = X .

#### Summary

In multiplication we multiply the elements in a row of the first matrix by the elements in a column of the second matrix. We have two matrices X and Y. X = $\left[\begin{array}{ccc}{\text{x}}_{\text{11}}& {\text{x}}_{\text{12}}& {\text{x}}_{\text{13}}\\ {\text{x}}_{\text{21}}& {\text{x}}_{\text{22}}& {\text{x}}_{\text{23}}\end{array}\right]$ Y = $\left[\begin{array}{cc}{\text{y}}_{\text{11}}& {\text{y}}_{\text{12}}\\ {\text{y}}_{\text{21}}& {\text{y}}_{\text{22}}\\ {\text{y}}_{\text{31}}& {\text{y}}_{\text{32}}\end{array}\right]$ We will first multiply the elements in the first row of X to the corresponding elements in the first column of Y.And we multiply the elements in the first row X to the elements in the second column of Y. We will write the sum of the products of the elements. In similar manner, we multiply the elements in the other rows and columns. Finally, we get the matrix as shown. = $\left[\begin{array}{cc}{{\text{x}}_{\text{11}}\text{y}}_{\text{11}}\text{+}{{\text{x}}_{\text{12}}\text{y}}_{\text{21}}\text{+}{\text{x}}_{\text{13}}{\text{y}}_{\text{31}}& {{\text{x}}_{\text{11}}\text{y}}_{\text{12}}\text{+}{{\text{x}}_{\text{12}}\text{y}}_{\text{22}}\text{+}{\text{x}}_{\text{13}}{\text{y}}_{\text{32}}\\ {\text{x}}_{\text{11}}{\text{y}}_{\text{11}}\text{+}{{\text{x}}_{\text{22}}\text{y}}_{\text{21}}\text{+}{{\text{x}}_{\text{23}}\text{y}}_{\text{31}}& {\text{x}}_{\text{21}}{\text{y}}_{\text{12}}\text{+}{{\text{x}}_{\text{22}}\text{y}}_{\text{22}}\text{+}{\text{x}}_{\text{23}}{\text{y}}_{\text{32}}\end{array}\right]$ The multiplication operation is possible only when the number of columns of the first matrix, X, is equal to the number of rows of the second matrix, Y. If we have two matrices, X and Y, with orders m x p and p x n, respectively, then their product is of the order m x n. X = [xik]mxp = $\left[\begin{array}{ccccccc}{\text{x}}_{\text{11}}& {\text{x}}_{\text{12}}& {\text{x}}_{\text{13}}& \cdots & {\text{x}}_{\text{1k}}& \cdots & {\text{x}}_{\text{1p}}\\ {\text{x}}_{\text{21}}& {\text{x}}_{\text{22}}& {\text{x}}_{\text{23}}& \cdots & {\text{x}}_{\text{2k}}& \cdots & {\text{x}}_{\text{2p}}\\ ⋮& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}\\ {\text{x}}_{\text{i1}}& {\text{x}}_{\text{i2}}& {\text{x}}_{\text{i3}}& \text{⋯}& {\text{x}}_{\text{ik}}& \text{⋯}& {\text{x}}_{\text{ip}}\\ \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}\\ {\text{x}}_{\text{m1}}& {\text{x}}_{\text{m2}}& {\text{x}}_{\text{m3}}& \text{⋯}& {\text{x}}_{\text{mk}}& \text{⋯}& {\text{x}}_{\text{mp}}\end{array}\right]$ 1 ≤ i ≤ m; 1 ≤ k ≤ p Y =     [ykj]pxn    = $\left[\begin{array}{ccccccc}{\text{y}}_{\text{11}}& {\text{y}}_{\text{12}}& {\text{y}}_{\text{13}}& \cdots & {\text{y}}_{\text{1j}}& \cdots & {\text{y}}_{\text{1n}}\\ {\text{y}}_{\text{21}}& {\text{y}}_{\text{22}}& {\text{y}}_{\text{23}}& \cdots & {\text{y}}_{\text{2j}}& \cdots & {\text{y}}_{\text{2n}}\\ ⋮& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}\\ {\text{y}}_{\text{k1}}& {\text{y}}_{\text{k2}}& {\text{y}}_{\text{k3}}& \text{⋯}& {\text{y}}_{\text{kj}}& \text{⋯}& {\text{y}}_{\text{kn}}\\ \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}\\ {\text{y}}_{\text{p1}}& {\text{y}}_{\text{p2}}& {\text{y}}_{\text{p3}}& \text{⋯}& {\text{y}}_{\text{pj}}& \text{⋯}& {\text{y}}_{\text{pn}}\end{array}\right]$  1 ≤ k ≤ p; 1 ≤ j ≤ n The product of these matrices be equal to matrix Z, which is of the order m by n. Z = [zij]mxn  =  $\left[\begin{array}{ccccccc}{\text{z}}_{\text{11}}& {\text{z}}_{\text{12}}& {\text{z}}_{\text{13}}& \cdots & {\text{z}}_{\text{1j}}& \cdots & {\text{z}}_{\text{1n}}\\ {\text{z}}_{\text{21}}& {\text{z}}_{\text{22}}& {\text{z}}_{\text{23}}& \cdots & {\text{z}}_{\text{2j}}& \cdots & {\text{z}}_{\text{2n}}\\ ⋮& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}\\ {\text{z}}_{\text{i1}}& {\text{z}}_{\text{i2}}& {\text{z}}_{\text{i3}}& \text{⋯}& {\text{z}}_{\text{ij}}& \text{⋯}& {\text{z}}_{\text{in}}\\ \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}& \text{⋮}\\ {\text{z}}_{\text{m1}}& {\text{z}}_{\text{m2}}& {\text{z}}_{\text{m3}}& \text{⋯}& {\text{z}}_{\text{mj}}& \text{⋯}& {\text{z}}_{\text{mn}}\end{array}\right]$      1 ≤ i ≤ m; 1 ≤ j ≤ n zij = xi1y1j + xi2y2j + xi3y3j + ..... + xikykj + .... + xipypj = $\sum _{\text{k=1}}^{\text{p}}\text{}$  xikykj In the expression, Zij represents the element in the ith row and jth column of matrix Z. The multiplication operation over two matrices does not satisfy the commutative property. X x Y ≠ Y x X (Not always) Case 1: XY defined, YX not defined. X = ${\left[\begin{array}{ccc}\text{1}& \text{2}& \text{3}\\ \text{4}& \text{5}& \text{6}\end{array}\right]}_{\text{2x3}}$ Y = ${\left[\begin{array}{ccc}\text{1}& \text{2}& \text{7}\\ \text{3}& \text{4}& \text{8}\\ \text{5}& \text{6}& \text{9}\end{array}\right]}_{\text{3x3}}$ No. of columns of X =3=No. of rows of Y No. of columns of Y = 3 ≠ 2 = No. of rows of Y Product YX is not possible. Case 2: Sometimes, product XY and YX are both defined, but XY does not necessarily equal YX. XY : Defined, YX: Defined but XY ≠ YX X = ${\left[\begin{array}{ccc}\text{1}& \text{1}& \text{2}\\ \text{3}& \text{1}& \text{5}\\ \text{1}& \text{3}& \text{2}\end{array}\right]}_{\text{3x3}}$ Y = ${\left[\begin{array}{ccc}\text{2}& \text{1}& \text{5}\\ \text{1}& \text{4}& \text{2}\\ \text{3}& \text{4}& \text{1}\end{array}\right]}_{\text{3x3}}$ XY = $\left[\begin{array}{ccc}\text{2+1+6}& \text{1+4+8}& \text{5+2+2}\\ \text{6+1+15}& \text{3+4+20}& \text{15+2+5}\\ \text{2+3+6}& \text{1+12+8}& \text{5+6+2}\end{array}\right]$ =$\left[\begin{array}{ccc}\text{9}& \text{13}& \text{9}\\ \text{22}& \text{27}& \text{22}\\ \text{11}& \text{21}& \text{13}\end{array}\right]$ YX = $\left[\begin{array}{ccc}\text{2+3+5}& \text{2+1+15}& \text{4+5+10}\\ \text{1+12+2}& \text{1+4+6}& \text{2+20+4}\\ \text{3+12+1}& \text{3+4+3}& \text{6+20+2}\end{array}\right]$ = $\left[\begin{array}{ccc}\text{10}& \text{18}& \text{19}\\ \text{15}& \text{11}& \text{26}\\ \text{16}& \text{10}& \text{28}\end{array}\right]$ XY ≠ YX Case 3: Multiplication of diagonal matrices of same order will commute A = ${\left[\begin{array}{ccc}\text{1}& \text{0}& \text{0}\\ \text{0}& \text{2}& \text{0}\\ \text{0}& \text{0}& \text{3}\end{array}\right]}_{\text{3x3}}$ B = ${\left[\begin{array}{ccc}\text{2}& \text{0}& \text{0}\\ \text{0}& \text{3}& \text{0}\\ \text{0}& \text{0}& \text{5}\end{array}\right]}_{\text{3x3}}$ AB = ${\left[\begin{array}{ccc}\text{2}& \text{0}& \text{0}\\ \text{0}& \text{6}& \text{0}\\ \text{0}& \text{0}& \text{15}\end{array}\right]}_{\text{3x3}}$ BA = ${\left[\begin{array}{ccc}\text{2}& \text{0}& \text{0}\\ \text{0}& \text{6}& \text{0}\\ \text{0}& \text{0}& \text{15}\end{array}\right]}_{\text{3x3}}$ Case 4: XY = O (null matrix) may not always imply that either X = O or Y = O Ex: X =         Y =      XY =          x           =        The multiplication property over matrices satisfies the associative law and the distributive law. For any matrices X, Y and Z, (X x Y) x Z = X x (Y x Z) Ex: A =           B = $\left[\begin{array}{cc}\text{7}& \text{4}\\ \text{2}& \text{5}\\ \text{1}& \text{-3}\end{array}\right]$      C =   $\left[\begin{array}{ccc}\text{2}& \text{-3}& \text{1}\\ \text{4}& \text{3}& \text{2}\end{array}\right]$                              (A x B) x C = A x (B x C) Now L.H.S = (A x B) x C A x B =    x $\left[\begin{array}{cc}\text{7}& \text{4}\\ \text{2}& \text{5}\\ \text{1}& \text{-3}\end{array}\right]$  =                                                              A x B =  (A x B) x C = x     =   =                            R.H.S = A x (B x C)  B x C  =   x     B x C  =   x    =                         B x C = A x (B x C)   =       x                                       = $\left[\begin{array}{ccc}\text{60+96+10}& \text{-18+36+12}& \text{30+48+5}\\ \text{90+24-50}& \text{-27+9-60}& \text{45+12-25}\\ \text{180+48-10}& \text{-54+18-12}& \text{90+24-5}\end{array}\right]$ = (A x B) x C = A x (B x C) Both the matrices are equal. We can see that the multiplication satisfies the associative law. According to the distributive law for any matrices X, Y and Z, X x (Y + Z) = X x Y + X x Z (X + Y) x Z = X x Z + Y x Z Ex: (A x B) x C = A x C + B x C A =      B =   C = $\left[\begin{array}{c}\text{5}\\ \text{1}\\ \text{4}\end{array}\right]$ (A + B) x C A + B =   +            (A + B) x C =     x  $\left[\begin{array}{c}\text{5}\\ \text{1}\\ \text{4}\end{array}\right]$                      =   =  A x C + B x C A x C =     x   =  =                                      B x C =   +  =  =                  A x C + B x C = +  =                The distributive law is applicable to the multiplication of matrices. Multiplicative identity exists for any square matrix. If X is a square matrix, there exists an identity matrix, I, of similar order such that X = [xij]nxn , I = [iij]nxn X x I = I x X = X .

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