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Matrices and Determinants question

2018 · 15 Apr · Shift 2 · Q27
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Matrices and Determinants question

2018 · 15 Apr · Shift 2 · Q27

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Suppose A is any 3 ×\times× 3 non-singular matrix and ( A −-− 3I) (A −-− 5I) = O where I = I3 and O = O3. If α\alphaα A + β\betaβ A-1 = 4I, then α\alphaα+β\betaβ is equal to :
  1. A
    8
  2. B
    7
  3. C
    13
  4. D
    12
View written solutionFree

Correct answer: A

  1. We are given
(A−3I)(A−5I)=O(A-3I)(A-5I)=O(A−3I)(A−5I)=O

for a non-singular 3×33\times 33×3 matrix AAA.

Since III commutes with every matrix, expand:

A2−8A+15I=OA^2-8A+15I=OA2−8A+15I=O

So,

A2−8A+15I=0.A^2-8A+15I=0.A2−8A+15I=0.
  1. Because AAA is non-singular, A−1A^{-1}A−1 exists. Multiply the above equation by A−1A^{-1}A−1:
A−8I+15A−1=0.A-8I+15A^{-1}=0.A−8I+15A−1=0.

Rearrange:

A+15A−1=8I.A+15A^{-1}=8I.A+15A−1=8I.
  1. Compare this with the given form
αA+βA−1=4I.\alpha A+\beta A^{-1}=4I.αA+βA−1=4I.

Divide the equation A+15A−1=8IA+15A^{-1}=8IA+15A−1=8I by 222:

12A+152A−1=4I.\frac12 A+\frac{15}{2}A^{-1}=4I.21​A+215​A−1=4I.

Thus,

α=12,β=152.\alpha=\frac12,\qquad \beta=\frac{15}{2}.α=21​,β=215​.

Hence,

α+β=12+152=162=8.\alpha+\beta=\frac12+\frac{15}{2}=\frac{16}{2}=8.α+β=21​+215​=216​=8.
  1. Checking options:
  • A: 888 ✅
  • B: 777
  • C: 131313
  • D: 121212

Therefore, the correct answer is A.

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