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Matrices and Determinants question

2017 · 8 Apr · Shift 1 · Q31
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Matrices and Determinants question

2017 · 8 Apr · Shift 1 · Q31

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let A be any 3 ×\times× 3 invertible matrix. Then which one of the following is not always true ?
  1. A
    adj (A) = ∣\left| \right.∣ A ∣\left| \right.∣.A −-− 1
  2. B
    adj (adj(A)) = ∣\left| \right.∣ A ∣\left| \right.∣.A
  3. C
    adj (adj(A)) = ∣\left| \right.∣ A ∣\left| \right.∣ 2.(adj(A)) −-− 1
  4. D
    adj (adj(A)) = ∣ \left| \, \right.∣ A ∣ \left| \, \right.∣.(adj(A)) −-− 1
View written solutionFree

Correct answer: D

  1. Given identities for an invertible matrix

For any invertible n×nn\times nn×n matrix AAA,

adj⁡(A)=∣A∣A−1.\operatorname{adj}(A)=|A|A^{-1}.adj(A)=∣A∣A−1.

So for a 3×33\times 33×3 invertible matrix, option A is always true.

  1. Formula for adj⁡(adj⁡(A))\operatorname{adj}(\operatorname{adj}(A))adj(adj(A))

A standard identity is:

adj⁡(adj⁡(A))=∣A∣n−2A\operatorname{adj}(\operatorname{adj}(A))=|A|^{n-2}Aadj(adj(A))=∣A∣n−2A

for any invertible n×nn\times nn×n matrix.

Here n=3n=3n=3, so

adj⁡(adj⁡(A))=∣A∣3−2A=∣A∣A.\operatorname{adj}(\operatorname{adj}(A))=|A|^{3-2}A=|A|A.adj(adj(A))=∣A∣3−2A=∣A∣A.

Hence option B is always true.

  1. Check option C

We know

adj⁡(A)=∣A∣A−1.\operatorname{adj}(A)=|A|A^{-1}.adj(A)=∣A∣A−1.

Therefore,

(adj⁡(A))−1=(∣A∣A−1)−1=1∣A∣A.(\operatorname{adj}(A))^{-1}=(|A|A^{-1})^{-1}=\frac{1}{|A|}A.(adj(A))−1=(∣A∣A−1)−1=∣A∣1​A.

So,

∣A∣2(adj⁡(A))−1=∣A∣2⋅1∣A∣A=∣A∣A.|A|^2(\operatorname{adj}(A))^{-1}=|A|^2\cdot \frac{1}{|A|}A=|A|A.∣A∣2(adj(A))−1=∣A∣2⋅∣A∣1​A=∣A∣A.

But from step 2,

adj⁡(adj⁡(A))=∣A∣A.\operatorname{adj}(\operatorname{adj}(A))=|A|A.adj(adj(A))=∣A∣A.

Hence option C is also always true.

  1. Check option D

Again,

(adj⁡(A))−1=1∣A∣A.(\operatorname{adj}(A))^{-1}=\frac{1}{|A|}A.(adj(A))−1=∣A∣1​A.

Therefore,

∣A∣(adj⁡(A))−1=∣A∣⋅1∣A∣A=A.|A|(\operatorname{adj}(A))^{-1}=|A|\cdot \frac{1}{|A|}A=A.∣A∣(adj(A))−1=∣A∣⋅∣A∣1​A=A.

But we already have

adj⁡(adj⁡(A))=∣A∣A.\operatorname{adj}(\operatorname{adj}(A))=|A|A.adj(adj(A))=∣A∣A.

So option D would require

∣A∣A=A,|A|A=A,∣A∣A=A,

which is true only if ∣A∣=1|A|=1∣A∣=1, not for every invertible matrix.

Hence option D is not always true.

  1. Conclusion
  • A: always true
  • B: always true
  • C: always true
  • D: not always true

So the correct option is D.

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