MathematicsSolution of System of Linear Equations (Matrix Method and Cramer's Rule)JEE Advanced 2009Moderate
Visualized Solution (Hindi)
Structure of Symmetric Matrix A
- A symmetric 3×3 matrix has the form A=adedbfefc
- Total entries = 9. Given: five 1s and four 0s.
- Symmetry implies off-diagonal entries (d,e,f) appear in pairs.
- Let k be the number of 1s on the diagonal. Then 5−k must be even.
- Possible values for k: 1 or 3.
Case 1: One 1 on the Diagonal
- Case 1: k=1 (1 on diagonal, 2 zeros on diagonal)
- Ways to choose diagonal: (13)=3
- Remaining 1s = 4, which forms 2 off-diagonal pairs.
- Ways to choose off-diagonal pairs: (23)=3
- Total matrices for Case 1: 3×3=9
Case 2: Three 1s on the Diagonal
- Case 2: k=3 (all 3 on diagonal are 1s)
- Ways to choose diagonal: (33)=1
- Remaining 1s = 2, which forms 1 off-diagonal pair.
- Ways to choose off-diagonal pairs: (13)=3
- Total matrices for Case 2: 1×3=3
- Total matrices in A = 9+3=12
Condition for Unique Solution
- System Ax=b has a unique solution if ∣A∣=0.
- Determinant ∣A∣=abc+2def−af2−be2−cd2
- Since entries ∈{0,1}, ∣A∣=abc+2def−af−be−cd
Counting Unique Solutions
- Checking 12 matrices:
- From Case 1 (k=1): 6 matrices have ∣A∣=−1 (e.g., a=1,e=1,f=1,b=c=d=0)
- From Case 2 (k=3): All 3 matrices have ∣A∣=0
- Total matrices with unique solution = 6
- This fits the range: at least 4 but less than 7
Condition for Inconsistency
- System is inconsistent if ∣A∣=0 and at least one of Dx,Dy,Dz=0.
- We check the 6 matrices where ∣A∣=0.
- Example: A=110110001 gives x+y=1 and x+y=0, which is inconsistent.
Final Count for Inconsistency
- Out of 6 matrices with ∣A∣=0:
- Inconsistent cases: 4
- Consistent (Infinite solutions) cases: 2
- Total inconsistent matrices = 4
- Answer: more than 2
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