Q. 274.0( 6 Votes )

Find A, if<

Answer :

We have,


We need to find the matrix A.


Let us see what the order of the matrices given are.


We know what order of matrix is,


If a matrix has M rows and N columns, the order of matrix is M × N.


Order of .


Number of rows = 3


M = 3


Number of column = 1


N = 1


Then, order of matrix X = M × N


Order of matrix X = 3 × 1


Order of .


Number of rows = 3


M = 3


Number of columns = 3


N = 3


Then, order of matrix Y = M × N


Order of matrix Y = 3 × 3


We must understand that, when a matrix of order 1 × 3 is multiplied to the matrix X, only then matrix Y is produced.


Let matrix A be of order 1 × 3, and can be represented as



Then, we get



Take L.H.S:


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.


So, we have



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


(4)(a) = 4a



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


(4)(b) = 4b



Multiply 1st row of matrix X by matching member of 3rd column of matrix A, then sum them up.


(4)(c) = 4c



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


(1)(a) = a



Multiply 2nd row of matrix X by matching member of 2nd column of matrix A, then sum them up.


(1)(b) = b



Multiply 2nd row of matrix X by matching member of 3rd column of matrix A, then sum them up.


(1)(c) = c



Multiply 3rd row of matrix X by matching member of 1st column of matrix A, then sum them up.


(3)(a) = 3a



Multiply 3rd row of matrix X by matching member of 2nd column of matrix A, then sum them up.


(3)(b) = 3b



Multiply 3rd row of matrix X by matching member of 3rd column of matrix A, then sum them up.


(3)(c) = 3c



Now, L.H.S = R.H.S




Since, the matrices on either sides are of same order, we can say that


4a = -4 …(i)


4b = 8 …(ii)


4c = 4 …(iii)


a = -1 …(iv)


b = 2 …(v)


c = 1 …(vi)


3a = -3 …(vii)


3b = 6 …(viii)


3c = 3 …(ix)


From equation (i), we can find the value of a,


4a = -4



a = -1


From equation (ii), we can find the value of b,


4b = 8



b = 2


From equation (iii), we can find the value of c,


4c = 4



c = 1


And it will satisfy other equations (iv), (v), (vi), (vii), (viii) and (ix) too.


Thus, the matrix A is



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