Let A = bmatrix 2 & 1 & 0 1 & 2 & -1 0 & -1 & 2 bmatrix. If | adj( adj( adj(2A))) | = (16)^n, then n is equal to:

Given: A = bmatrix 2 & 1 & 0 1 & 2 & -1 0 & -1 & 2 bmatrix and | adj( adj( adj(2A))) | = 16^n. Find: The value of n. First compute the determinant of A: |A| = 2 vmatrix2 & -1 -1 & 2 vmatrix - 1 vmatrix1 & -1 0 & 2 vmatrix = 2(4-1) - (2-0) = 6 - 2 = 4. Now A is of order 3, so for any 3 3 matrix M, | adj(M)| = |M|^3-1 = |M|^2. Also, |2A| = 2^3|A| = 8 4 = 32. Apply the adjugate determinant property repeatedly: | adj(2A)| = |2A|^2 = 32^2 | adj( adj(2A))| = (32^2)^2 = 32^4 | adj( adj( adj(2A)))| = (32^4)^2 = 32^8. Therefore, 32^8 = (2^5)^8 = 2^40. Since 16^n = (2^4)^n = 2^4n, we get 2^40 = 2^4n 4n = 40 n = 10. Therefore, the value of n is 10. The correct option is D.

Let A = bmatrix 2 & 1 & 0 1 & 2 & -1 0 & -1 & 2 bmatrix. If | adj… — Solution | JEE Mathematics

MCQMediumJEE 2023Inverse & Adjoint of a Matrix

Let $A = \begin{bmatrix} 2 & 1 & 0 \\ 1 & 2 & -1 \\ 0 & -1 & 2 \end{bmatrix}.$ If $\left| \operatorname{adj}(\operatorname{adj}(\operatorname{adj}(2A))) \right| = (16)^n$ , then $n$ is equal to:

A
$8$
B
$9$
C
$12$
D
$10$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: $A = \begin{bmatrix} 2 & 1 & 0 \\ 1 & 2 & -1 \\ 0 & -1 & 2 \end{bmatrix}$ and $\left|\operatorname{adj}(\operatorname{adj}(\operatorname{adj}(2A)))\right| = 16^n$ .

Find: The value of $n$ .

First compute the determinant of $A$ :

|A| = 2\begin{vmatrix}2 & -1 \\ -1 & 2\end{vmatrix} - 1\begin{vmatrix}1 & -1 \\ 0 & 2\end{vmatrix}

= 2(4-1) - (2-0) = 6 - 2 = 4.

Now $A$ is of order $3$ , so for any $3 \times 3$ matrix $M$ ,

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

Also,

|2A| = 2^3|A| = 8 \cdot 4 = 32.

Apply the adjugate determinant property repeatedly:

|\operatorname{adj}(2A)| = |2A|^2 = 32^2

|\operatorname{adj}(\operatorname{adj}(2A))| = (32^2)^2 = 32^4

|\operatorname{adj}(\operatorname{adj}(\operatorname{adj}(2A)))| = (32^4)^2 = 32^8.

Therefore,

32^8 = (2^5)^8 = 2^{40}.

Since

16^n = (2^4)^n = 2^{4n},

we get

2^{40} = 2^{4n} \Rightarrow 4n = 40 \Rightarrow n = 10.

Therefore, the value of $n$ is $10$ . The correct option is D.

Common mistakes

Using $|\operatorname{adj}(M)| = |M|$ is incorrect for a $3 \times 3$ matrix. The correct relation is $|\operatorname{adj}(M)| = |M|^{2}$ . Always use $|M|^{r-1}$ for a matrix of order $r$ .
Forgetting that $|2A| \neq 2|A|$ for a $3 \times 3$ matrix is a common error. Since the scalar multiplies each row, $|2A| = 2^3|A|$ .
Applying the adjugate property only once is wrong here because the adjugate is taken three times. Repeat the determinant transformation at each stage in sequence.

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