Q. 123.9( 7 Votes )

Find the matrix A

Answer :

Here we have been given a matrix equation,


We need to find the matrix A.


Let matrix A be of order 2 × 2, and can be represented as



Then, we get



Take L.H.S:


So, first let us calculate


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.


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


(2, 1).(a, c) = (2 × a) + (1 × c)


(2, 1).(a, c) = 2a + c



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


(2, 1).(b, d) = (2 × b) + (1 × d)


(2, 1).(b, d) = 2b + d



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


(3, 2).(a, c) = (3 × a) + (2 × c)


(3, 2).(a, c) = 3a + 2c



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


(3, 2).(b, d) = (3 × b) + (2 × d)


(3, 2).(b, d) = 3b + 2d



Let X.Y = Z


Now, we need to find .


That is,



Where, let .


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


(2a + c, 2b + d).(-3, 5) = ((2a + c) × -3) + ((2b + d) × 5)


(2a + c, 2b + d).(-3, 5) = -6a – 3c + 10b + 5d


(2a + c, 2b + d).(-3, 5) = -6a + 10b – 3c + 5d



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


(2a + c, 2b + d).(2, -3) = ((2a + c) × 2) + ((2b + d) × -3)


(2a + c, 2b + d).(2, -3) = 4a + 2c – 6b – 3d


(2a + c, 2b + d).(2, -3) = 4a – 6b + 2c – 3d



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


(3a + 2c, 3b + 2d).(-3, 5) = ((3a + 2c) × -3) + ((3b + 2d) × 5)


(3a + 2c, 3b + 2d).(-3, 5) = -9a – 6c + 15b + 10d


(3a + 2c, 3b + 2d).(-3, 5) = -9a + 15b – 6c + 10d



Multiply 2nd row of matrix Z by matching members of 2nd column of matrix Q, then sum them up.


(3a + 2c, 3b + 2d).(2, -3) = ((3a + 2c) × 2) + ((3b + 2d) × -3)


(3a + 2c, 3b + 2d).(2, -3) = 6a + 4c – 9b – 6d


(3a + 2c, 3b + 2d).(2, -3) = 6a – 9b + 4c – 6d



So, we have



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



For matrices having same order, we can write as


-6a + 10b – 3c + 5d = 1 …(i)


4a – 6b + 2c – 3d = 0 …(ii)


-9a + 15b – 6c + 10d = 0 …(iii)


6a – 9b + 4c – 6d = 1 …(iv)


We have 4 variables to find, namely, a, b, c and d; and 4 equations.


So, on adding equations (i) and (iv), we get


(-6a + 10b – 3c + 5d) + (6a – 9b + 4c – 6d) = 1 + 1


-6a + 6a + 10b – 9b – 3c + 4c + 5d – 6d = 2


0 + b + c – d = 2


d = b + c – 2 …(v)


Now, adding equations (ii) and (iii), we get


(4a – 6b + 2c – 3d) + (-9a + 15b – 6c + 10d) = 0 + 0


4a – 9a – 6b + 15b + 2c – 6c – 3d + 10d = 0


-5a + 9b – 4c + 7d = 0 …(vi)


On adding equations (iv) and (vi), we get


(6a – 9b + 4c – 6d) + (-5a + 9b – 4c + 7d) = 1 + 0


6a – 5a – 9b + 9b + 4c – 4c – 6d + 7d = 1


a + 0 + 0 + d = 1


d = 1 – a …(vii)


Putting the value of d from equation (vii) in (v), we get


(1 – a) = b + c – 2


b + c – 2 – 1 = -a


b + c – 3 = -a


a = 3 – b – c …(viii)


Now, putting values of a and d from equations (vii) and (viii) in equation (iii), we get


-9(3 – b – c) + 15b – 6c + 10(1 – a) = 0


-9(3 – b – c) + 15b – 6c + 10(1 – (3 – b – c)) = 0 [ a = 3 – b – c]


-27 + 9b + 9c + 15b – 6c + 10(1 – 3 + b + c) = 0


-27 + 9b + 9c + 15b – 6c + 10(-2 + b + c) = 0


-27 + 9b + 9c + 15b – 6c – 20 + 10b + 10c = 0


9b + 15b + 10b + 9c – 6c + 10c – 27 – 20 = 0


34b + 13c – 47 = 0


34b + 13c = 47 …(ix)


Also, putting values of a and d from equations (vii) and (viii) in equation (ii), we get


4(3 – b – c) – 6b + 2c – 3(1 – a) = 0


12 – 4b – 4c – 6b + 2c – 3(1 – (3 – b – c)) = 0


12 – 4b – 4c – 6b + 2c – 3(1 – 3 + b + c) = 0


12 – 4b – 4c – 6b + 2c – 3(-2 + b + c) = 0


12 – 4b – 4c – 6b + 2c + 6 – 3b – 3c = 0


-4b – 6b – 3b – 4c + 2c – 3c + 12 + 6 = 0


-13b – 5c + 18 = 0


13b + 5c = 18 …(x)


On multiplying equation (ix) by 5 and equation (x) by 13, we get


(ix) 5(34b + 13c) = 5 × 47


170b + 65c = 235 …(xi)


(x) 13(13b + 5c) = 13 × 18


169b + 65c = 234 …(xii)


Subtracting equations (xi) and (xii), we get


(170b + 65c) – (169b + 65c) = 235 – 234


170b – 169b + 65c – 65c = 1


b = 1


Putting b = 1 in equation (x), we get


13(1) + 5c = 18


13 + 5c = 18


5c = 18 – 13


5c = 5



c = 1


Putting b = 1 and c = 1 in equation (viii), we get


a = 3 – b – c


a = 3 – 1 – 1


a = 3 – 2


a = 1


Putting a = 1 in equation (vii), we get


d = 1 – a


d = 1 – 1


d = 0


Thus, the matrix A is



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