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MCQMediumJEE 2023Inverse & Adjoint of a Matrix

JEE Mathematics 2023 Question with Solution

Let x,y,z>1x, y, z > 1 and

A=[1log⁡xylog⁡xzlog⁡yx2log⁡yzlog⁡zxlog⁡zy3].A = \begin{bmatrix} 1 & \log_x y & \log_x z \\ \log_y x & 2 & \log_y z \\ \log_z x & \log_z y & 3 \end{bmatrix}.

Then adj(adj A2)adj(adj\, A^2) is equal to:

  • A

    646^4

  • B

    282^8

  • C

    484^8

  • D

    242^4

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given:

A=[1log⁡xylog⁡xzlog⁡yx2log⁡yzlog⁡zxlog⁡zy3]A = \begin{bmatrix} 1 & \log_x y & \log_x z \\ \log_y x & 2 & \log_y z \\ \log_z x & \log_z y & 3 \end{bmatrix}

with x,y,z>1x,y,z>1.

Find: adj(adj A2)adj(adj\, A^2).

Using logarithmic relations from the solution,

log⁡yx=1log⁡xy,log⁡zx=1log⁡xz,log⁡zy=log⁡xylog⁡xz\log_y x = \frac{1}{\log_x y}, \qquad \log_z x = \frac{1}{\log_x z}, \qquad \log_z y = \frac{\log_x y}{\log_x z}

and the determinant simplifies to

∣A∣=2.|A| = 2.

Now use the adjoint property for a 3×33 \times 3 matrix:

∣adj⁡(M)∣=∣M∣n−1=∣M∣2|\operatorname{adj}(M)| = |M|^{n-1} = |M|^2

where n=3n=3.

For M=A2M=A^2,

∣A2∣=∣A∣2=22=4.|A^2| = |A|^2 = 2^2 = 4.

Hence,

∣adj⁡(A2)∣=∣A2∣2=42=24.|\operatorname{adj}(A^2)| = |A^2|^2 = 4^2 = 2^4.

Applying the same determinant-of-adjoint property again to adj⁡(A2)\operatorname{adj}(A^2),

∣adj⁡(adj⁡(A2))∣=∣adj⁡(A2)∣2=(24)2=28.\left|\operatorname{adj}(\operatorname{adj}(A^2))\right| = \left|\operatorname{adj}(A^2)\right|^2 = (2^4)^2 = 2^8.

Therefore, the correct option is B.

Determinant and adjoint property chain

Given: the matrix AA has determinant ∣A∣=2|A|=2 as concluded in the provided solution.

Find: the value associated with adj(adj A2)adj(adj\, A^2).

First compute the determinant of A2A^2:

∣A2∣=∣A∣2=22=4.|A^2| = |A|^2 = 2^2 = 4.

For any non-singular 3×33 \times 3 matrix MM,

∣adj⁡(M)∣=∣M∣3−1=∣M∣2.|\operatorname{adj}(M)| = |M|^{3-1} = |M|^2.

So with M=A2M=A^2,

∣adj⁡(A2)∣=∣A2∣2=42=24.|\operatorname{adj}(A^2)| = |A^2|^2 = 4^2 = 2^4.

Now apply the same rule once more to the matrix adj⁡(A2)\operatorname{adj}(A^2):

∣adj⁡(adj⁡(A2))∣=(∣adj⁡(A2)∣)2=(24)2=28.\left|\operatorname{adj}(\operatorname{adj}(A^2))\right| = \left(|\operatorname{adj}(A^2)|\right)^2 = (2^4)^2 = 2^8.

Thus the final answer is 282^8.

Common mistakes

  • Using ∣adj⁡(M)∣=∣M∣|\operatorname{adj}(M)| = |M| is incorrect. For an n×nn \times n matrix, the correct relation is ∣adj⁡(M)∣=∣M∣n−1|\operatorname{adj}(M)| = |M|^{n-1}. Here n=3n=3, so the exponent must be 22.

  • Forgetting that ∣A2∣=∣A∣2|A^2| = |A|^2 leads to a wrong base before applying the adjoint formula. First compute ∣A2∣|A^2| correctly, then use the adjoint determinant property.

  • Treating adj(adj A2)adj(adj\, A^2) as if one adjoint operation cancels the other is wrong. The expression requires applying the adjoint-related determinant rule twice, not simplifying by cancellation.

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