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Matrices and Determinants question

2016 · 10 Apr · Shift 1 · Q25
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Matrices and Determinants question

2016 · 10 Apr · Shift 1 · Q25

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let A be a 3 ×\times× 3 matrix such that A2 −-− 5A + 7I = 0 Statement - I : A −-− 1 = 17{1 \over 7}71​(5I −-− A). Statement - II : The polynomial A3 −-− 2A2 −-− 3A + I can be reduced to 5(A −-− 4I). Then :
  1. A
    Statement-I is true, but Statement-II is false.
  2. B
    Statement-I is false, but Statement-II is true.
  3. C
    Both the statements are true.
  4. D
    Both the statements are false
View written solutionFree

Correct answer: C

  1. Given matrix equation

We are given A2−5A+7I=0A^2-5A+7I=0A2−5A+7I=0 so A2=5A−7I.A^2=5A-7I.A2=5A−7I.

We will use this to test both statements.

  1. Check Statement I

Statement I says: A−1=17(5I−A).A^{-1}=\frac{1}{7}(5I-A).A−1=71​(5I−A).

From A2−5A+7I=0,A^2-5A+7I=0,A2−5A+7I=0, rewrite as 5A−A2=7I.5A-A^2=7I.5A−A2=7I. Factor out AAA: A(5I−A)=7I.A(5I-A)=7I.A(5I−A)=7I.

Since AAA is a matrix polynomial in AAA, it commutes with 5I−A5I-A5I−A, so also (5I−A)A=7I.(5I-A)A=7I.(5I−A)A=7I.

Hence 17(5I−A)\frac{1}{7}(5I-A)71​(5I−A) is both a left and right inverse of AAA. Therefore, A−1=17(5I−A).A^{-1}=\frac{1}{7}(5I-A).A−1=71​(5I−A).

So Statement I is true.

  1. Check Statement II

We need to simplify A3−2A2−3A+I.A^3-2A^2-3A+I.A3−2A2−3A+I.

Using A2=5A−7I,A^2=5A-7I,A2=5A−7I, first compute A3A^3A3: A3=A⋅A2=A(5A−7I)=5A2−7A.A^3=A\cdot A^2=A(5A-7I)=5A^2-7A.A3=A⋅A2=A(5A−7I)=5A2−7A. Again substitute A2=5A−7IA^2=5A-7IA2=5A−7I: A3=5(5A−7I)−7A=25A−35I−7A=18A−35I.A^3=5(5A-7I)-7A=25A-35I-7A=18A-35I.A3=5(5A−7I)−7A=25A−35I−7A=18A−35I.

Now, A3−2A2−3A+I=(18A−35I)−2(5A−7I)−3A+I.A^3-2A^2-3A+I=(18A-35I)-2(5A-7I)-3A+I.A3−2A2−3A+I=(18A−35I)−2(5A−7I)−3A+I. Simplify: =18A−35I−10A+14I−3A+I=18A-35I-10A+14I-3A+I=18A−35I−10A+14I−3A+I =(18A−10A−3A)+(−35I+14I+I)=(18A-10A-3A)+(-35I+14I+I)=(18A−10A−3A)+(−35I+14I+I) =5A−20I=5A-20I=5A−20I =5(A−4I).=5(A-4I).=5(A−4I).

So Statement II is true.

  1. Conclusion

Both Statement I and Statement II are true.

Therefore, the correct option is C.\boxed{\text{C}}.C​.

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