Q. 225.0( 1 Vote )

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Answer :

We have matrices A, B and C, such that




In order to multiply two matrices, A and B, the number of columns in A must equal the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.


(i). We need to verify: (AB)C = A(BC)


Take L.H.S = (AB)C


First, compute AB.



Multiply 1st row of matrix A by matching members of 1st column of matrix B, then sum them up.


(1, 2)(2, 3) = (1 × 2) + (2 × 3)


(1, 2)(2, 3) = 2 + 6


(1, 2)(2, 3) = 8



Multiply 1st row of matrix A by matching members of 2nd column of matrix B, then sum them up.


(1, 2)(3, -4) = (1 × 3) + (2 × -4)


(1, 2)(3, -4) = 3 – 8


(1, 2)(3, -4) = -5



Similarly, let us fill for the rest of elements.





Let .


Now, compute for DC. [ (AB)C = DC]



Multiply 1st row of matrix D by matching members of 1st column of matrix C, then sum them up.


(8, -5)(1, -1) = (8 × 1) + (-5 × -1)


(8, -5)(1, -1) = 8 + 5


(8, -5)(1, -1) = 13



Multiply 1st row of matrix D by matching members of 2nd column of matrix C, then sum them up.


(8, -5)(0, 0) = (8 × 0) + (-5 × 0)


(8, -5)(0, 0) = 0 + 0


(8, -5)(0, 0) = 0



Similarly, let us fill for the rest of elements.





So,



Take R.H.S: A(BC)


First, compute BC.



Multiply 1st row of matrix B by matching members of 1st column of matrix C, then sum them up.


(2, 3)(1, -1) = (2 × 1) + (3 × -1)


(2, 3)(1, -1) = 2 – 3


(2, 3)(1, -1) = -1



Multiply 1st row of matrix B by matching members of 2nd column of matrix C, then sum them up.


(2, 3)(0, 0) = (2 × 0) + (3 × 0)


(2, 3)(0, 0) = 0 + 0


(2, 3)(0, 0) = 0



Similarly, let us fill for the rest of the elements.





Let .


Now, compute for AE.



Multiply 1st row of matrix A by matching members of 1st column of matrix E, then sum them up.


(1, 2)(-1, 7) = (1 × -1) + (2 × 7)


(1, 2)(-1, 7) = -1 + 14


(1, 2)(-1, 7) = 13



Multiply 1st row of matrix A by matching members of 2nd column of matrix E, then sum them up.


(1, 2)(0, 0) = (1 × 0) + (2 × 0)


(1, 2)(0, 0) = 0 + 0


(1, 2)(0, 0) = 0



Similarly, repeat the step for the other elements.





So,



Thus, (AB)C = A(BC).


(ii). We need to verify: A(B + C) = AB + AC


Take L.H.S: A(B + C)


Add B + C.





Let B + C = F, such that



Now, multiply A and F.



Multiply 1st row of matrix A by matching members of 1st column of matrix F, then sum them up.


(1, 2)(3, 2) = (1 × 3) + (2 × 2)


(1, 2)(3, 2) = 3 + 4


(1, 2)(3, 2) = 7



Multiply 1st row of matrix A by matching members of 2nd column of matrix F, then sum them up.


(1, 2)(3, -4) = (1 × 3) + (2 × -4)


(1, 2)(3, -4) = 3 – 8


(1, 2)(3, -4) = -5



Similarly, repeat the steps for the other elements.





So,



Now, take R.H.S: AB + AC


Compute AB.



Multiply 1st row of matrix A by matching members of 1st column of matrix B, then sum them up.


(1, 2)(2, 3) = (1 × 2) + (2 × 3)


(1, 2)(2, 3) = 2 + 6


(1, 2)(2, 3) = 8



Multiply 1st row of matrix A by matching members of 2nd column of matrix B, then sum them up.


(1, 2)(3, -4) = (1 × 3) + (2 × -4)


(1, 2)(3, -4) = 3 – 8


(1, 2)(3, -4) = -5



Similarly, repeat the steps for the other elements.





So,



Now, compute AC.



Multiply 1st row of matrix A by matching members of 1st column of matrix C, then sum them up.


(1, 2)(1, -1) = (1 × 1) + (2 × -1)


(1, 2)(1, -1) = 1 – 2


(1, 2)(1, -1) = -1



Multiply 1st row of matrix A by matching members of 2nd column of matrix C, then sum them up.


(1, 2)(0, 0) = (1 × 0) + (2 × 0)


(1, 2)(0, 0) = 0 + 0


(1, 2)(0, 0) = 0



Similarly, repeat the steps for the other elements.





So,



Adding AB + AC.



Matrices of same order can be added or subtracted.




So, clearly L.H.S = R.H.S.


Thus, A(B + C) = AB + AC.


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