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Let A, B, C, D be real matrices (not necessarily square) such that AT = BCD, BT = CDA, CT = DAB and DT = ABC, where AT represents transpose of A. Then for the matrix S = ABCD
- a)S9 = S
- b)S = S4
- c)S2 = S
- d)S3 = S4
Correct answer is option 'A'. Can you explain this answer?
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Let A, B, C, D be real matrices (not necessarily square) such that AT ...
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Let A, B, C, D be real matrices (not necessarily square) such that AT ...
To find the value of S, we can substitute the given equations into each other and simplify.
1. Substituting AT = BCD into DT = ABC, we get:
DT = A(BCD) = (AB)CD
2. Substituting BT = CDA into CT = DAB, we get:
CT = D(BT) = D(CDA) = (CD)A
3. Substituting CT = (CD)A into DT = (AB)CD, we get:
(CD)A = (AB)(CD)
Multiplying both sides by C^-1 from the left, we have:
C^-1(CD)A = C^-1(AB)(CD)
Simplifying, we get:
DA = (C^-1AB)(CD)
4. Substituting AT = BCD into DA = (C^-1AB)(CD), we get:
(BCD)A = (C^-1AB)(CD)
Expanding the left side, we have:
BA = (C^-1AB)(CD)
Multiplying both sides by A^-1 from the right, we get:
BA(A^-1) = (C^-1AB)(CD)(A^-1)
Simplifying, we get:
B = (C^-1AB)(CD)(A^-1)
Thus, we have found an expression for B in terms of A, C, and D.
5. Substituting BT = CDA into BA = (C^-1AB)(CD)(A^-1), we get:
(CDA)(A^-1) = (C^-1AB)(CD)(A^-1)
Simplifying, we get:
C = (C^-1AB)(CD)(A^-1)
Multiplying both sides by D^-1 from the right, we have:
CD^-1 = (C^-1AB)(CD)(A^-1)(D^-1)
Simplifying, we get:
CD^-1 = (C^-1AB)(A^-1)
Multiplying both sides by C^-1 from the left, we have:
C^-1CD^-1 = (C^-1AB)(A^-1)C^-1
Simplifying, we get:
D^-1 = (C^-1AB)(A^-1)C^-1
Thus, we have found an expression for D^-1 in terms of A, B, and C.
6. Substituting D^-1 = (C^-1AB)(A^-1)C^-1 into DT = ABC, we get:
(C^-1AB)(A^-1)C^-1T = ABC
Simplifying, we get:
(C^-1AB)(A^-1) = AB(C^-1AB)(A^-1)C^-1
Multiplying both sides by C from the left, we have:
C(C^-1AB)(A^-1) = AB(C^-1AB)(A^-1)C^-1C
Simplifying, we get:
(C^-1AB)(A^-1) = AB
Thus, we have found an expression for (C^-1AB)(A^-1) in terms of A and B.
7. Substituting (C^-1AB)(A^-1) = AB into D^-1 = (C^-1AB)(A^-1)C^-1, we get:
D^-1
1. Substituting AT = BCD into DT = ABC, we get:
DT = A(BCD) = (AB)CD
2. Substituting BT = CDA into CT = DAB, we get:
CT = D(BT) = D(CDA) = (CD)A
3. Substituting CT = (CD)A into DT = (AB)CD, we get:
(CD)A = (AB)(CD)
Multiplying both sides by C^-1 from the left, we have:
C^-1(CD)A = C^-1(AB)(CD)
Simplifying, we get:
DA = (C^-1AB)(CD)
4. Substituting AT = BCD into DA = (C^-1AB)(CD), we get:
(BCD)A = (C^-1AB)(CD)
Expanding the left side, we have:
BA = (C^-1AB)(CD)
Multiplying both sides by A^-1 from the right, we get:
BA(A^-1) = (C^-1AB)(CD)(A^-1)
Simplifying, we get:
B = (C^-1AB)(CD)(A^-1)
Thus, we have found an expression for B in terms of A, C, and D.
5. Substituting BT = CDA into BA = (C^-1AB)(CD)(A^-1), we get:
(CDA)(A^-1) = (C^-1AB)(CD)(A^-1)
Simplifying, we get:
C = (C^-1AB)(CD)(A^-1)
Multiplying both sides by D^-1 from the right, we have:
CD^-1 = (C^-1AB)(CD)(A^-1)(D^-1)
Simplifying, we get:
CD^-1 = (C^-1AB)(A^-1)
Multiplying both sides by C^-1 from the left, we have:
C^-1CD^-1 = (C^-1AB)(A^-1)C^-1
Simplifying, we get:
D^-1 = (C^-1AB)(A^-1)C^-1
Thus, we have found an expression for D^-1 in terms of A, B, and C.
6. Substituting D^-1 = (C^-1AB)(A^-1)C^-1 into DT = ABC, we get:
(C^-1AB)(A^-1)C^-1T = ABC
Simplifying, we get:
(C^-1AB)(A^-1) = AB(C^-1AB)(A^-1)C^-1
Multiplying both sides by C from the left, we have:
C(C^-1AB)(A^-1) = AB(C^-1AB)(A^-1)C^-1C
Simplifying, we get:
(C^-1AB)(A^-1) = AB
Thus, we have found an expression for (C^-1AB)(A^-1) in terms of A and B.
7. Substituting (C^-1AB)(A^-1) = AB into D^-1 = (C^-1AB)(A^-1)C^-1, we get:
D^-1
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Let A, B, C, D be real matrices (not necessarily square) such that AT = BCD, BT = CDA, CT = DAB and DT = ABC, where AT represents transpose of A. Then for the matrix S = ABCDa)S9 = Sb)S = S4c)S2 = Sd)S3 = S4Correct answer is option 'A'. Can you explain this answer?
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