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Matrices and Determinants question

2025 · 22 Jan · Shift 2 · Q34
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Matrices and Determinants question

2025 · 22 Jan · Shift 2 · Q34

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
For a 3×33 \times 33×3 matrix MMM, let trace (M)(M)(M) denote the sum of all the diagonal elements of MMM. Let AAA be a 3×33 \times 33×3 matrix such that ∣A∣=12|A|=\frac{1}{2}∣A∣=21​ and trace (A)=3(A)=3(A)=3. If B=adj⁡(adj⁡(2A))B=\operatorname{adj}(\operatorname{adj}(2 A))B=adj(adj(2A)), then the value of ∣B∣+|B|+∣B∣+ trace (B)(B)(B) equals :
  1. A
    56
  2. B
    132
  3. C
    174
  4. D
    280
View written solutionFree

Correct answer: D

  1. Use the adjugate identity for an n×nn\times nn×n matrix

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

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

Also,

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

Since here the matrix size is 3×33\times 33×3, we get

adj⁡(adj⁡(M))=∣M∣ M.\operatorname{adj}(\operatorname{adj}(M))=|M|\,M.adj(adj(M))=∣M∣M.
  1. Apply this to M=2AM=2AM=2A

Given

B=adj⁡(adj⁡(2A)).B=\operatorname{adj}(\operatorname{adj}(2A)).B=adj(adj(2A)).

Therefore,

B=∣2A∣(2A).B=|2A|(2A).B=∣2A∣(2A).

Now,

∣2A∣=23∣A∣=8⋅12=4.|2A|=2^3|A|=8\cdot \frac12=4.∣2A∣=23∣A∣=8⋅21​=4.

Hence,

B=4(2A)=8A.B=4(2A)=8A.B=4(2A)=8A.
  1. Find ∣B∣|B|∣B∣

Since B=8AB=8AB=8A and AAA is 3×33\times 33×3,

∣B∣=∣8A∣=83∣A∣=512⋅12=256.|B|=|8A|=8^3|A|=512\cdot \frac12=256.∣B∣=∣8A∣=83∣A∣=512⋅21​=256.
  1. Find trace⁡(B)\operatorname{trace}(B)trace(B)

Trace is linear, so

trace⁡(B)=trace⁡(8A)=8trace⁡(A)=8⋅3=24.\operatorname{trace}(B)=\operatorname{trace}(8A)=8\operatorname{trace}(A)=8\cdot 3=24.trace(B)=trace(8A)=8trace(A)=8⋅3=24.
  1. Compute the required value
∣B∣+trace⁡(B)=256+24=280.|B|+\operatorname{trace}(B)=256+24=280.∣B∣+trace(B)=256+24=280.
  1. Check options

The correct option is

280\boxed{280}280​

which is Option D.

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