Question 1

Full Question: Let $A$ be a square matrix of order 3 such that $\text{det}(A) = -2$ and $\text{det}(3 \text{ adj}(-6 \text{ adj}(3A))) = 2^{m+n} \cdot 3^{mn}$, $m > n$. Then $4m + 2n$ is equal to.

Options: (Integer Type)

Correct Answer: 34

Year: JEE Main 2025 (Online) 22nd January Morning Shift

Solution: As $A \text{ adj } A = |A|I$ and $\text{det}(\lambda A) = \lambda^n \text{det } A$. $\text{det}(3 \text{ adj}(-6 \text{ adj}(3A))) = 3^3 \text{det}(\text{adj}(-6 \text{ adj}(3A))) = 3^3 (-6 \text{ adj}(3A))^2 = 3^3 (-6)^6 |3A|^4 = 3^9 2^6 \cdot 3^{12} \cdot (-2)^4 = 3^{21} \cdot 2^{10}$. Comparing: $m+n=10, mn=21 \Rightarrow m=7, n=3$. $4m+2n=34$.

Step Solution:

1.  Apply property $|\lambda \text{adj}(B)| = \lambda^n |B|^{n-1}$ for $n=3$: $D = 3^3 |\text{adj}(-6 \text{ adj}(3A))| = 27 \cdot |-6 \text{ adj}(3A)|^2$.

2.  Use $|kC| = k^3 |C|$: $D = 27 \cdot ((-6)^3 |\text{adj}(3A)|)^2 = 3^3 \cdot 6^6 \cdot |\text{adj}(3A)|^2$.

3.  Apply $|\text{adj}(3A)| = |3A|^2 = (3^3 |A|)^2 = 3^6 |A|^2$.

4.  Substitute and simplify: $D = 3^3 \cdot 2^6 \cdot 3^6 \cdot (3^6 |A|^2)^2 = 3^{21} \cdot 2^6 \cdot |A|^4$.

5.  With $|A| = -2$: $D = 3^{21} \cdot 2^6 \cdot 16 = 3^{21} \cdot 2^{10}$. Solve $m+n=10$ and $mn=21$ for $m=7, n=3$.

Difficulty Level: Hard

Concept Name: Properties of Adjoint and Determinants

Short cut solution: Use the general formula $|\text{adj}(k \text{ adj}(mA))| = k^{n-1} m^{n(n-1)^2} |A|^{(n-1)^2}$ to directly compute powers of 2 and 3.

 Question 5

Full Question: Let $I$ be the identity matrix of order $3 \times 3$ and for the matrix $A = \begin{bmatrix} \lambda & 2 & 3 \\ 4 & 5 & 6 \\ 7 & -1 & 2 \end{bmatrix}$, $|A| = -1$. Let $B$ be the inverse of the matrix $\text{adj}(A \text{ adj}(A^2))$. Then $|(\lambda B + I)|$ is equal to.

Options: (Integer Type)

Correct Answer: 38

Year: JEE Main 2025 (Online) 3rd April Evening Shift

Solution: $B = [\text{adj}(A \text{ adj}(A^2))]^{-1}$. Since $A \text{ adj}(A^2) = A (\text{adj } A)^2 = (A \text{ adj } A) \text{adj } A = |A| \text{ adj } A = A^{-1}$. Thus $B = (\text{adj}(A^{-1}))^{-1} = (|A^{-1}| A)^{-1} = \frac{A^{-1}}{-1} = -A^{-1}$. $|A| = -1 \Rightarrow \lambda = 3$. $|3B+I| = |I - 3A^{-1}| = \frac{|A-3I|}{|A|} = 38$.

Step Solution:

1.  Solve $|A| = -1$ for $\lambda$: $\lambda(10+6) - 2(8-42) + 3(-4-35) = -1 \Rightarrow 16\lambda - 49 = -1 \Rightarrow \lambda = 3$.

2.  Simplify inner matrix: $A \text{ adj}(A^2) = A (\text{adj } A)^2 = |A| \text{ adj } A = -1 \cdot \text{adj } A$.

3.  Simplify $B$: $B = [\text{adj}(-\text{adj } A)]^{-1} = [(-1)^{3-1} \text{adj}(\text{adj } A)]^{-1} = (|A| A)^{-1} = -A^{-1}$.

4.  Setup final determinant: $|\lambda B + I| = |3(-A^{-1}) + I| = |I - 3A^{-1}|$.

5.  Calculate: $|I - 3A^{-1}| = \frac{|A-3I|}{|A|}$. Calculate $|A-3I|$ using $\lambda=3$ to get $38$, then divide by $|-1|$.

Difficulty Level: Hard

Concept Name: Matrix Inversion and Adjoint Properties

Short cut solution: Recognize that for $n=3$, $\text{adj}(A \text{ adj}(A^2))$ simplifies to $A^{-1}$ because $|A|=-1$, meaning $B$ is simply $(-A^{-1})^{-1}$, which is $-A$.

 Question 8

Full Question: For a $3 \times 3$ matrix $M$, let trace $(M)$ denote the sum of all the diagonal elements of $M$. Let $A$ be a $3 \times 3$ matrix such that $|A| = \frac{1}{2}$ and trace $(A) = 3$. If $B = \text{adj}(\text{adj}(2A))$, then the value of $|B| + \text{trace}(B)$ equals:

Options: A. 56, B. 132, C. 174, D. 280

Correct Answer: D

Year: JEE Main 2025 (Online) 22nd January Evening Shift

Solution: $B = \text{adj}(\text{adj}(2A)) = \text{det}(2A) \cdot (2A)$. Since $|2A| = 2^3|A| = 8(1/2) = 4$, $B = 4(2A) = 8A$. $|B| = |8A| = 8^3|A| = 512(1/2) = 256$. $\text{trace}(B) = \text{trace}(8A) = 8 \cdot \text{trace}(A) = 8(3) = 24$. Sum $= 256+24 = 280$.

Step Solution:

1.  Use property $\text{adj}(\text{adj } M) = |M|^{n-2} M$. For $n=3$, $B = |2A| \cdot 2A$.

2.  Calculate $|2A|$: $|2A| = 2^3 |A| = 8 \times \frac{1}{2} = 4$.

3.  Define $B$ in terms of $A$: $B = 4 \times 2A = 8A$.

4.  Calculate $|B|$: $|B| = |8A| = 8^3 |A| = 512 \times \frac{1}{2} = 256$.

5.  Calculate $\text{trace}(B)$: $\text{trace}(8A) = 8 \times \text{trace}(A) = 8 \times 3 = 24$. Total $= 256 + 24 = 280$.

Difficulty Level: Medium

Concept Name: Trace of a Matrix and Adjoint Properties

Short cut solution: Since $\text{adj}(\text{adj } M) = |M|M$ for $n=3$, the matrix $B$ is always a scalar multiple of $A$. Scaling the trace and determinant by that scalar saves calculation time.

Question 16

Full Question: For some $a, b$ let, $f(x) = \begin{vmatrix} a + \frac{\sin x}{x} & 1 & b \\ a & 1 + \frac{\sin x}{x} & b \\ a & 1 & b + \frac{\sin x}{x} \end{vmatrix}, x \neq 0$. If $\lim_{x \to 0} f(x) = \lambda + \mu a + \nu b$, then $(\lambda + \mu + \nu)^2$ is equal to:

Options: A. 25, B. 16, C. 9, D. 36

Correct Answer: B

Year: JEE Main 2025 (Online) 24th January Evening Shift

Solution: $\lim_{x \to 0} f(x) = \begin{vmatrix} a + 1 & 1 & b \\ a & 1 + 1 & b \\ a & 1 & b + 1 \end{vmatrix} = (a+1)(2(b+1)-b) + 1(ab-a(b+1)) + ba = (a+1)(b+2) - a + ab = b+a+2 = \lambda + \mu a + \nu b$. $\lambda = 2, \mu = 1, \nu = 1 \Rightarrow (\lambda + \mu + \nu)^2 = 16$.

Step Solution:

1.  Apply the limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$ to each element of the determinant.

2.  The resulting determinant is $D = \begin{vmatrix} a+1 & 1 & b \\ a & 2 & b \\ a & 1 & b+1 \end{vmatrix}$.

3.  Expand the determinant along the first row: $D = (a+1)[2(b+1) - b] - 1[a(b+1) - ab] + b[a - 2a]$.

4.  Simplify the expression: $D = (a+1)(b+2) - a - ab = ab + 2a + b + 2 - a - ab = a + b + 2$.

5.  Compare $a + b + 2$ with $\lambda + \mu a + \nu b$ to get $\lambda = 2, \mu = 1, \nu = 1$. Calculate $(2+1+1)^2 = 16$.

Difficulty Level: Medium

Concept Name: Limits and Determinant Expansion

Short cut solution: Use row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_1$ on the limit matrix to get $\begin{vmatrix} a+1 & 1 & b \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{vmatrix}$, which simplifies instantly to $a+b+2$.

 Question 18

Full Question: Let $M$ and $m$ respectively be the maximum and the minimum values of $f(x) = \begin{vmatrix} 1 + \sin^2 x & \cos^2 x & 4\sin 4x \\ \sin^2 x & 1 + \cos^2 x & 4\sin 4x \\ \sin^2 x & \cos^2 x & 1 + 4\sin 4x \end{vmatrix}, x \in R$. Then $M^4 - m^4$ is equal to:

Options: A. 1280, B. 1040, C. 1215, D. (Option not provided in source)

Correct Answer: A

Year: JEE Main 2025 (Online) 29th January Morning Shift

Solution: Apply $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$. $f(x) = \begin{vmatrix} 1 + \sin^2 x & \cos^2 x & 4\sin 4x \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{vmatrix}$. Expand about $R_1$: $f(x) = 2 + 4\sin 4x$. $M = \max f(x) = 6$, $m = \min f(x) = -2$. $M^4 - m^4 = 1280$.

Step Solution:

1.  Perform row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$ to simplify the determinant.

2.  Expand the resulting determinant $\begin{vmatrix} 1 + \sin^2 x & \cos^2 x & 4\sin 4x \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{vmatrix}$ along $R_3$ or $R_1$.

3.  The expansion yields $f(x) = (1 + \sin^2 x)(1) - \cos^2 x(-1) + 4\sin 4x(1) = 1 + \sin^2 x + \cos^2 x + 4\sin 4x = 2 + 4\sin 4x$.

4.  Determine extrema: Since $-1 \leq \sin 4x \leq 1$, the max value $M = 2 + 4(1) = 6$ and min value $m = 2 + 4(-1) = -2$.

5.  Calculate the final value: $M^4 - m^4 = 6^4 - (-2)^4 = 1296 - 16 = 1280$.

Difficulty Level: Medium

Concept Name: Properties of Determinants (Row Operations) and Trigonometric Extrema

Short cut solution: Notice that subtracting $R_1$ from other rows makes two rows highly sparse (containing mostly -1, 1, and 0), making the determinant calculation trivial.

Question 19

Full Question: Let $A = [a_{ij}] = \begin{bmatrix} \log_5 128 & \log_4 5 \\ \log_5 8 & \log_4 25 \end{bmatrix}$. If $A_{ij}$ is the cofactor of $a_{ij}$ and $C_{ij} = \sum_{k=1}^2 a_{ik}A_{jk}, 1 \leq i, j \leq 2$, and $C = [C_{ij}]$, then $8|C|$ is:

Options: A. 288, B. 262, (C and D not explicitly listed but calculation leads to 242)

Correct Answer: 242 (Option D)

Year: JEE Main 2025 (Online) 29th January Morning Shift

Solution: $|A| = (\log_5 128)(\log_4 25) - (\log_4 5)(\log_5 8) = \frac{7 \log 2}{\log 5} \cdot \frac{2 \log 5}{2 \log 2} - \frac{\log 5}{2 \log 2} \cdot \frac{3 \log 2}{\log 5} = 7 - 1.5 = 5.5 = \frac{11}{2}$. Matrix $C$ elements are defined such that $C_{11} = |A|, C_{12} = 0, C_{21} = 0, C_{22} = |A|$. Thus $|C| = |A|^2 = \frac{121}{4}$. $8|C| = 242$.

Step Solution:

1.  Calculate $|A|$ using the change of base formula: $|A| = (\log_5 2^7)(\log_{2^2} 5^2) - (\log_{2^2} 5)(\log_5 2^3)$.

2.  Simplify terms: $|A| = (7 \log_5 2)(1 \log_2 5) - (\frac{1}{2} \log_2 5)(3 \log_5 2) = 7(1) - 1.5(1) = 5.5 = \frac{11}{2}$.

3.  Identify matrix $C$: By the property of cofactors, $\sum a_{ik}A_{jk}$ is $|A|$ if $i=j$ and $0$ if $i \neq j$. Thus $C = \begin{bmatrix} |A| & 0 \\ 0 & |A| \end{bmatrix}$.

4.  Calculate $|C|$: $|C| = |A|^2 = (\frac{11}{2})^2 = \frac{121}{4}$.

5.  Find the final value: $8|C| = 8 \times \frac{121}{4} = 2 \times 121 = 242$.

Difficulty Level: Hard

Concept Name: Logarithmic Properties and Matrix-Cofactor Expansion Property ($\sum a_{ik}A_{jk} = \delta_{ij}|A|$)

Short cut solution: Recognize that the definition of $C_{ij}$ is simply the matrix product $A \cdot (\text{adj } A)^T$, which is always $|A|I$ for any square matrix. Thus $|C| = | |A|I | = |A|^n$. For $n=2$, $|C| = |A|^2$. This avoids calculating individual cofactors.

 Question 24

Full Question: Let $a \in R$ and $A$ be a matrix of order $3 \times 3$ such that $\text{det}(A) = -4$ and $A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix}$, where $I$ is the identity matrix of order $3 \times 3$. If $\text{det}((a + 1)\text{ adj}((a - 1)A))$ is $2^m 3^n, m, n \in \{0, 1, 2, \dots, 20\}$, then $m + n$ is equal to:

Options: A. 14, B. 17, C. 15, D. 16

Correct Answer: D

Year: JEE Main 2025 (Online) 2nd April Morning Shift

Solution: $A = (A+I) - I = \begin{bmatrix} 0 & a & 1 \\ 2 & 0 & 0 \\ a & 1 & 1 \end{bmatrix}$. $\text{det}(A) = -a(2) + 1(2) = -2a + 2 = -4 \Rightarrow a=3$. Now find $\text{det}(4 \text{ adj}(2A)) = 4^3 \text{ det}(\text{adj}(2A)) = 4^3 |2A|^2 = 4^3 (2^3 |A|)^2 = 2^6 (2^3 \cdot 2^2)^2 = 2^6 \cdot 2^{10} = 2^{16} \cdot 3^0$. Thus $m=16, n=0 \Rightarrow m+n=16$.

Step Solution:

1.  Extract Matrix A: Subtract the Identity matrix from $A+I$ to get $A = \begin{bmatrix} 0 & a & 1 \\ 2 & 0 & 0 \\ a & 1 & 1 \end{bmatrix}$.

2.  Solve for $a$: Calculate $|A| = 0(0) - a(2-0) + 1(2-0) = -2a + 2$. Set $-2a + 2 = -4$, resulting in $a=3$.

3.  Setup Target: Substitute $a=3$ into the required expression: $\text{det}((3 + 1)\text{ adj}((3 - 1)A)) = \text{det}(4 \text{ adj}(2A))$.

4.  Apply Determinant Properties: Use $|\lambda \text{adj } B| = \lambda^3 |B|^2$ and $|kB| = k^3 |B|$. Thus, $4^3 |\text{adj}(2A)| = 4^3 |2A|^2 = 4^3 (2^3 |A|)^2$.

5.  Calculate Final Powers: $2^6 \cdot (2^3 \cdot |-4|)^2 = 2^6 \cdot (2^3 \cdot 2^2)^2 = 2^6 \cdot 2^{10} = 2^{16}$. This matches $2^m 3^n$ with $m=16, n=0$. $m+n=16$.

Difficulty Level: Hard

Concept Name: Determinant and Adjoint Properties

Short cut solution: Recognize that since $|A| = -2^2$, the final result will only contain powers of 2. Focus calculation solely on the exponents of 2 to save time.

 Question 28

Full Question: Let $A$ be a matrix of order $3 \times 3$ and $|A| = 5$. If $|2 \text{ adj}(3A \text{ adj}(2A))| = 2^\alpha \cdot 3^\beta \cdot 5^\gamma, \alpha, \beta, \gamma \in N$, then $\alpha + \beta + \gamma$ is equal to:

Options: A. 26, B. 27, C. 25, D. 28

Correct Answer: B

Year: JEE Main 2025 (Online) 3rd April Morning Shift

Solution: $|2 \text{ adj}(3A \text{ adj}(2A))| = 2^3 |3A \text{ adj}(2A)|^2 = 2^3 (3^3 |A| |\text{adj } 2A|)^2 = 2^3 \cdot 3^6 \cdot |A|^2 \cdot (|2A|^2)^2 = 2^3 \cdot 3^6 \cdot 5^2 \cdot (2^3 |A|)^4 = 2^{15} \cdot 3^6 \cdot 5^6$. Sum $= 15+6+6=27$.

Step Solution:

1.  Remove outer scalar: Using $|kB| = k^3|B|$, the expression becomes $2^3 |\text{adj}(3A \text{ adj}(2A))|$.

2.  Remove outer Adjoint: Using $|\text{adj } B| = |B|^2$, we get $2^3 |3A \text{ adj}(2A)|^2$.

3.  Decompose product: Apply determinant product property: $2^3 (3^3 |A| |\text{adj } 2A|)^2 = 2^3 \cdot 3^6 \cdot |A|^2 \cdot |\text{adj } 2A|^2$.

4.  Simplify inner Adjoint: $|\text{adj } 2A|^2 = (|2A|^2)^2 = |2A|^4 = (2^3 |A|)^4 = 2^{12} |A|^4$.

5.  Combine and calculate sum: Result is $2^3 \cdot 3^6 \cdot |A|^2 \cdot 2^{12} \cdot |A|^4 = 2^{15} \cdot 3^6 \cdot 5^6$. Sum $\alpha+\beta+\gamma = 15+6+6=27$.

Difficulty Level: Medium

Concept Name: Successive Adjoint Properties

Short cut solution: Use the direct formula $|\text{adj}(k A \text{ adj } m A)| = k^{n(n-1)} m^{n(n-1)^2} |A|^{n(n-1)}$. For $n=3$, this is $k^6 m^{12} |A|^6$. Multiplying by the initial $2^3$ gives the result.

 Question 31

Full Question: Let $A$ be a $3 \times 3$ matrix such that $|\text{adj}(\text{adj}(\text{adj } A))| = 81$. If $S = \{n \in \mathbb{Z} : (|\text{adj}(\text{adj } A)|)^{\frac{(n-1)^2}{2}} = |A|^{(3n^2 - 5n - 4)}\}$, then $\sum_{n \in S} |A^{(n^2 + n)}|$ is equal to:

Options: A. 820, B. 866, C. 750, D. 732

Correct Answer: D

Year: JEE Main 2025 (Online) 7th April Morning Shift

Solution: $|A|^{(2^3)} = 3^4 \Rightarrow |A| = 3^{1/2}$. Then $(|A|^4)^{\frac{(n-1)^2}{2}} = |A|^{3n^2-5n-4} \Rightarrow 2(n-1)^2 = 3n^2-5n-4 \Rightarrow n^2-n-6=0 \Rightarrow n = 3, -2$. Sum $= |A|^6 + |A|^{12} = 3^3 + 3^6 = 27 + 729 = 732$.

Step Solution:

1.  Find $|A|$: Use property $|\text{adj}^k A| = |A|^{(n-1)^k}$. Here $|A|^{(2^3)} = 81 \Rightarrow |A|^8 = 3^4 \Rightarrow |A| = 3^{1/2} = \sqrt{3}$.

2.  Simplify exponential equation: Left side is $(|A|^4)^{\frac{(n-1)^2}{2}} = |A|^{2(n-1)^2}$.

3.  Equate exponents: Set $2(n-1)^2 = 3n^2 - 5n - 4$, which simplifies to $2n^2 - 4n + 2 = 3n^2 - 5n - 4$.

4.  Solve for $n$: $n^2 - n - 6 = 0 \Rightarrow (n-3)(n+2) = 0$. The set $S = \{3, -2\}$.

5.  Final Summation: Calculate $|A|^{3^2+3} + |A|^{(-2)^2-2} = |A|^{12} + |A|^2 = (\sqrt{3})^{12} + (\sqrt{3})^2 = 729 + 3 = 732$.

Difficulty Level: Medium

Concept Name: Exponential Matrix Equations and Adjoint Properties

Short cut solution: Since the question asks for a sum of determinants, and determinants are numbers, recognize $|A^k| = |A|^k$. This allows you to treat the entire problem as a standard algebraic exponent problem once $|A|$ is found.

Question 33

Full Question: Let $A = \begin{bmatrix} 2 & 2 + p & 2 + p + q \\ 4 & 6 + 2 p & 8 + 3 p + 2 q \\ 6 & 12 + 3 p & 20 + 6 p + 3 q \end{bmatrix}$. If $\det(\text{adj}(\text{adj}(3 A))) = 2^{m} \cdot 3^{n}, m, n \in \mathbb{N}$, then $m + n$ is equal to:

Options: A. 22, B. 20, C. 24, D. 26

Correct Answer: C

Year: JEE Main 2025 (Online) 8th April Evening Shift

Solution: $|A| = \begin{vmatrix} 2 & 2 + p & 2 + p + q \\ 4 & 6 + 2 p & 8 + 3 p + 2 q \\ 6 & 12 + 3 p & 20 + 6 p + 3 q \end{vmatrix}$. Applying row operations $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$, we get $|A| = \begin{vmatrix} 2 & 2 + p & 2 + p + q \\ 0 & 2 & 4 + p \\ 0 & 6 & 14 + 3 p \end{vmatrix}$. Further operation $R_3 \to R_3 - 3R_2$ gives $|A| = \begin{vmatrix} 2 & 2 + p & 2 + p + q \\ 0 & 2 & 4 + p \\ 0 & 0 & 2 \end{vmatrix} = 8 = 2^3$. $|\text{adj}(\text{adj}(3 A))| = |3 A|^{(3-1)^2} = |3 A|^4 = (3^3 |A|)^4 = (3^3 \times 2^3)^4 = 2^{12} \times 3^{12}$. $\Rightarrow m + n = 24$.

Step Solution:

1.  Evaluate $|A|$: Simplify the determinant using row operations $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$ to get a simpler matrix where $|A| = \begin{vmatrix} 2 & 2+p & 2+p+q \\ 0 & 2 & 4+p \\ 0 & 6 & 14+3p \end{vmatrix}$.

2.  Continue row operations: Perform $R_3 \to R_3 - 3R_2$. The matrix becomes upper triangular: $\begin{bmatrix} 2 & 2+p & 2+p+q \\ 0 & 2 & 4+p \\ 0 & 0 & 2 \end{bmatrix}$. Thus, $|A| = 2 \times 2 \times 2 = 8 = 2^3$.

3.  Apply Adjoint properties: For a $3 \times 3$ matrix, $|\text{adj}(\text{adj } M)| = |M|^{(n-1)^2} = |M|^4$. Let $M = 3A$.

4.  Compute $|3A|^4$: Use $|kA| = k^3|A|$. So, $|3A|^4 = (3^3 \cdot |A|)^4 = (3^3 \cdot 2^3)^4$.

5.  Find $m$ and $n$: Simplify to powers of 2 and 3: $3^{12} \cdot 2^{12}$. Comparing with $2^m \cdot 3^n$, we get $m=12, n=12$. Sum $m+n = 24$.

Difficulty Level: Medium

Concept Name: Properties of Adjoint and Determinant Evaluation

Short cut solution: Applying $R_2 \to R_2 - 2R_1$ and $R_3 \to R_2 - 1.5R_2$ immediately removes all $p$ and $q$ variables, leaving a simple triangular determinant.

 Question 37

Full Question: Let $\mathcal{A} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha \end{bmatrix}$. If $|\text{det}(2A)|^3 = 2^{21}$ where $\alpha, \beta \in Z$, then $\alpha$ is:

Options: A. 3, B. 5, C. 17, D. 9

Correct Answer: B

Year: JEE Main 2024, 29th January Shift 1

Solution: $|A| = 1(\alpha^2 - \beta^2)$. Given $|2A|^3 = 2^{21} \Rightarrow |2A| = 2^7$. As $A$ is a $3 \times 3$ matrix, $|2A| = 2^3 |A|$. So $2^3 |A| = 2^7 \Rightarrow |A| = 2^4 = 16$. Thus $\alpha^2 - \beta^2 = 16 \Rightarrow (\alpha + \beta)(\alpha - \beta) = 16$. Testing options, $\alpha = 5$ gives $25 - \beta^2 = 16 \Rightarrow \beta^2 = 9$, which works for integers.

Step Solution:

1.  Find $|A|$: Expand along the first row to get $|A| = 1 \cdot (\alpha \cdot \alpha - \beta \cdot \beta) = \alpha^2 - \beta^2$.

2.  Simplify given equation: Take the cube root of both sides of $|2A|^3 = 2^{21}$ to find $|2A| = 2^7$.

3.  Use scalar property: Since $A$ is $3 \times 3$, $|2A| = 2^3 |A|$. Equating gives $2^3 |A| = 128$, so $|A| = 16$.

4.  Set up integer equation: $\alpha^2 - \beta^2 = 16$. Factorize as $(\alpha - \beta)(\alpha + \beta) = 16$.

5.  Test for $\alpha$: Substitute the option $\alpha = 5$ into the equation: $25 - \beta^2 = 16 \Rightarrow \beta^2 = 9 \Rightarrow \beta = \pm 3$. Since $\beta$ is an integer, $\alpha = 5$ is correct.

Difficulty Level: Easy

Concept Name: Properties of Determinants (Scalar Multiplication)

Short cut solution: Immediately recognize that $|A|$ must be $2^4=16$. Quickly check which option for $\alpha$ allows $\alpha^2 - 16$ to be a perfect square ($\beta^2$). Only $5^2 - 16 = 9$ (a perfect square) works.

 Question 39

Full Question: Let $A = \begin{bmatrix} 2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2 \end{bmatrix}$ and $P = \begin{bmatrix} 1 & 2 & 0 \\ 5 & 0 & 2 \\ 7 & 1 & 5 \end{bmatrix}$. If prime factors of $|P^{-1}AP - 2I|$ are $f_1, f_2, \dots$, then their sum is:

Options: A. 26, B. 27, C. 66, D. 23

Correct Answer: A

Year: JEE Main 2024, 29th January Shift 2

Solution: $|P^{-1}AP - 2I| = |P^{-1}AP - 2P^{-1}IP| = |P^{-1}(A-2I)P| = |P^{-1}| \cdot |A-2I| \cdot |P| = |A-2I|$. Calculating $|A-2I| = \begin{vmatrix} 0 & 1 & 2 \\ 6 & 0 & 11 \\ 3 & 3 & 0 \end{vmatrix} = 0 - 1(0-33) + 2(18-0) = 33 + 36 = 69$. Prime factors of 69 are 3 and 23. Sum $= 3 + 23 = 26$.

Step Solution:

1.  Use similarity property: $|P^{-1}AP - 2I| = |P^{-1}(A-2I)P|$. By the multiplicative property of determinants, this equals $|P^{-1}| \cdot |A-2I| \cdot |P| = |A-2I|$.

2.  Construct $A-2I$: Subtract 2 from the diagonal elements of $A$: $M = \begin{bmatrix} 2-2 & 1 & 2 \\ 6 & 2-2 & 11 \\ 3 & 3 & 2-2 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 2 \\ 6 & 0 & 11 \\ 3 & 3 & 0 \end{bmatrix}$.

3.  Evaluate $|M|$: Expand along the first row: $0(0 - 33) - 1(0 - 33) + 2(18 - 0)$.

4.  Calculate determinant: $0 + 33 + 36 = 69$.

5.  Sum prime factors: Find the prime factorization of 69, which is $3 \times 23$. The sum of these factors is $3 + 23 = 26$.

Difficulty Level: Medium

Concept Name: Similarity Transformation Property of Determinants

Short cut solution: Similar matrices have the same characteristic polynomial. The expression $|P^{-1}AP - \lambda I|$ is identical to $|A - \lambda I|$. Setting $\lambda = 2$ allows you to ignore matrix $P$ entirely and calculate $|A-2I|$ directly.

Question 41

Full Question: Let $A$ be a $3 \times 3$ matrix and $\text{det}(A) = 2$. If $n = \text{det}(\underbrace{\text{adj}(\text{adj}(\dots \text{adj}}_{2024 \text{ times}}(A)\dots)))$, then the remainder when $n$ is divided by 9 is equal to:

Options: (Integer Type)

Correct Answer: 7

Year: JEE Main 2024 (Online) 30th January Evening Shift

Solution: $2^{2024} = (2^2) 2^{2022} = 4(8)^{674} = 4(9-1)^{674}$. $\Rightarrow 2^{2024} \equiv 4 \pmod 9$. $\Rightarrow 2^{2024} = 9m + 4$, $m$ is even. $2^{9m+4} \equiv 16 \cdot (2^3)^{3m} \equiv 16 \pmod 9 \equiv 7 \pmod 9$.

Step Solution:

1.  Apply the property $| \text{adj}^k(A) | = |A|^{(n-1)^k}$. For order $n=3$ and $k=2024$, $n = |A|^{2^{2024}} = 2^{2^{2024}}$.

2.  To find $n \pmod 9$, we first evaluate the exponent $2^{2024} \pmod{\phi(9)}$. Since $\phi(9) = 6$, we find $2^{2024} \pmod 6$.

3.  Observe the pattern: $2^1 \equiv 2, 2^2 \equiv 4, 2^3 \equiv 2, 2^4 \equiv 4 \pmod 6$. Since 2024 is even, $2^{2024} \equiv 4 \pmod 6$.

4.  Express the exponent as $2^{2024} = 6k + 4$.

5.  Calculate $n \pmod 9$: $2^{6k+4} = (2^6)^k \cdot 2^4 = (64)^k \cdot 16$. Since $64 \equiv 1 \pmod 9$, result is $1^k \cdot 16 \equiv 16 \equiv 7 \pmod 9$.

Difficulty Level: Hard

Concept Name: Successive Adjoint Property and Euler’s Totient Theorem

Short cut solution: Use the property that $2^x \pmod 9$ has a cycle of 6. Since $2^{2024} \equiv 4 \pmod 6$, the result is simply $2^4 \pmod 9 = 16 \pmod 9 = 7$.

 Question 42

Full Question: Let $A = \begin{bmatrix} \sqrt{2} & 1 \\ -1 & \sqrt{2} \end{bmatrix}, B = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}, C = ABA^T$ and $X = A^T C^2 A$, then $\text{det}(X)$ is:

Options: A. 243, B. 729, C. 27, D. 891

Correct Answer: B

Year: JEE Main 2024 (Online) 1st February Morning Shift

Solution: $A = \begin{bmatrix} \sqrt{2} & 1 \\ -1 & \sqrt{2} \end{bmatrix} \Rightarrow \text{det}(A) = 3$. $B = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \Rightarrow \text{det}(B) = 1$. $C = ABA^T \Rightarrow \text{det}(C) = (\text{det}(A))^2 \times \text{det}(B) = 9$. $\text{det}(X) = |A^T| |C|^2 |A| = |A|^2 |C|^2 = 9 \times 81 = 729$.

Step Solution:

1.  Find $|A|$: $|A| = (\sqrt{2})(\sqrt{2}) - (1)(-1) = 2 + 1 = 3$.

2.  Find $|B|$: $|B| = (1)(1) - (0)(1) = 1$.

3.  Evaluate $|C|$: Use $|M \cdot N \cdot P| = |M| |N| |P|$. Thus, $|C| = |A| |B| |A^T| = 3 \cdot 1 \cdot 3 = 9$.

4.  Setup $|X|$: Use $|X| = |A^T \cdot C^2 \cdot A| = |A^T| |C|^2 |A|$.

5.  Calculate Final Value: Since $|A^T| = |A| = 3$ and $|C| = 9$, then $|X| = 3 \cdot 9^2 \cdot 3 = 9 \cdot 81 = 729$.

Difficulty Level: Easy

Concept Name: Multiplicative Property of Determinants

Short cut solution: Recognize that for any square matrices, $|A^T C^2 A| = |A|^2 |C|^2$. Since $C$ is a similarity-like transform of $B$ scaled by $|A|$, $|C| = |A|^2|B|$. Substituting gives $|X| = |A|^2 (|A|^2|B|)^2 = |A|^6 |B|^2$. With $|A|=3$ and $|B|=1$, $|X| = 3^6 = 729$.

 Question 46

Full Question: Let $A$ be a $3 \times 3$ matrix such that $|\text{adj}(\text{adj}(\text{adj } A))| = 12^4$. Then $|A^{-1} \text{ adj } A|$ is equal to:

Options: A. $2\sqrt{3}$, B. $\sqrt{6}$, C. 12, D. 1

Correct Answer: A

Year: JEE Main 2023 (Online) 24th January Evening Shift

Solution: Given $|\text{adj}(\text{adj}(\text{adj } A))| = 12^4$. $\Rightarrow |A|^{(n-1)^3} = 12^4$. Asked $|A^{-1} \cdot \text{adj } A| = |A^{-1}| \cdot |\text{adj } A| = \frac{1}{|A|} \cdot |A|^{3-1} = |A| = 2\sqrt{3}$.

Step Solution:

1.  Apply Adjoint Power Property: Use $|\text{adj}^k A| = |A|^{(n-1)^k}$. Here $k=3$ (triple adjoint) and $n=3$ (order).

2.  Solve for $|A|$: $|A|^{(3-1)^3} = 12^4 \Rightarrow |A|^8 = 12^4$. Taking the square root of both sides twice gives $|A| = 12^{4/8} = 12^{1/2} = \sqrt{12}$.

3.  Simplify target expression: Use the multiplicative property: $|A^{-1} \text{ adj } A| = |A^{-1}| \cdot |\text{adj } A|$.

4.  Substitute known properties: $|A^{-1}| = \frac{1}{|A|}$ and $|\text{adj } A| = |A|^{n-1} = |A|^2$.

5.  Final Calculation: The expression simplifies to $\frac{1}{|A|} \cdot |A|^2 = |A|$. Thus, $|A| = \sqrt{12} = 2\sqrt{3}$.

Difficulty Level: Medium

Concept Name: Properties of Adjoints and Inverse Determinants

Short cut solution: In a $3 \times 3$ matrix, $A^{-1} \text{ adj } A$ is always equal to $|A|^{-1} \cdot |A| I = I$ only if you are looking at the matrix, but for determinants: $|A^{-1} \text{ adj } A| = \frac{|A|^2}{|A|} = |A|$. Directly find $|A|$ from the given equation to get the answer.

Here are the detailed solutions for Problems 47, 51, and 53 from the source material:

 Question 47

Full Question: Let $x, y, z > 1$ and $A = \begin{bmatrix} 1 & \log_x y & \log_x z \\ \log_y x & 2 & \log_y z \\ \log_z x & \log_z y & 3 \end{bmatrix}$. Then $|\text{adj}(\text{adj } A^2)|$ is equal to:

Options: A. 64, B. $2^8$, C. 48, D. 24

Correct Answer: B

Year: 25-Jan-2023 Shift 1

Solution: $|A| = \frac{1}{\log x \cdot \log y \cdot \log z} \begin{vmatrix} \log x & \log y & \log z \\ \log x & 2\log y & \log z \\ \log x & \log y & 3\log z \end{vmatrix} = 2$. $\Rightarrow |\text{adj}(\text{adj } A^2)| = |A^2|^4 = 2^8$.

Step Solution:

1.  Change of Base: Express all log terms using natural logarithms (base $e$): $A = \begin{bmatrix} 1 & \frac{\ln y}{\ln x} & \frac{\ln z}{\ln x} \\ \frac{\ln x}{\ln y} & 2 & \frac{\ln z}{\ln y} \\ \frac{\ln x}{\ln z} & \frac{\ln y}{\ln z} & 3 \end{bmatrix}$.

2.  Evaluate $|A|$: Factor out $\frac{1}{\ln x}, \frac{1}{\ln y}, \frac{1}{\ln z}$ from rows 1, 2, and 3 respectively, and multiply columns 1, 2, and 3 by $\ln x, \ln y, \ln z$. $|A| = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 3 \end{vmatrix}$.

3.  Simplify Determinant: Perform row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$: $|A| = \begin{vmatrix} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{vmatrix} = 2$.

4.  Apply Adjoint Property: Use the property $|\text{adj}(\text{adj } M)| = |M|^{(n-1)^2}$. For order $n=3$, $|\text{adj}(\text{adj } A^2)| = |A^2|^{(3-1)^2} = |A^2|^4$.

5.  Final Calculation: Since $|A^2| = |A|^2 = 2^2 = 4$, the result is $4^4 = (2^2)^4 = 2^8$.

Difficulty Level: Hard

Concept Name: Logarithmic Determinants and Successive Adjoint Properties

Short cut solution: Recognize that for any matrix where $a_{ij} = \frac{f(j)}{f(i)}$ except for diagonal entries $a_{ii} = k_i$, the determinant simplifies significantly by factoring out $f(i)$ from rows and multiplying into columns.

 Question 51

Full Question: The set of all values of $t \in \mathbb{R}$, for which the matrix $\begin{bmatrix} e^t & e^{-t}(\sin t - 2\cos t) & e^{-t}(-2\sin t - \cos t) \\ e^t & e^{-t}(2\sin t + \cos t) & e^{-t}(\sin t - 2\cos t) \\ e^t & e^{-t}\cos t & e^{-t}\sin t \end{bmatrix}$ is invertible, is:

Options: A. $\{(2k+1)\frac{\pi}{2}, k \in \mathbb{Z}\}$, B. $\{k\pi + \frac{\pi}{4}, k \in \mathbb{Z}\}$, C. $\{k\pi, k \in \mathbb{Z}\}$, D. $\mathbb{R}$

Correct Answer: D

Year: 29-Jan-2023 Shift 2

Solution: If it's invertible, then determinant value $\neq 0$. Factoring out $e^t, e^{-t}, e^{-t}$ gives $e^{-t} \begin{vmatrix} 1 & \sin t - 2\cos t & -2\sin t - \cos t \\ 1 & 2\sin t + \cos t & \sin t - 2\cos t \\ 1 & \cos t & \sin t \end{vmatrix} \neq 0$. Expanding shows $e^{-t} \times 6 \neq 0$, which is true for all $t \in \mathbb{R}$.

Step Solution:

1.  Factor Scalar Terms: Pull out $e^t$ from column 1, $e^{-t}$ from column 2, and $e^{-t}$ from column 3. $|A| = e^t \cdot e^{-t} \cdot e^{-t} \begin{vmatrix} 1 & \sin t - 2\cos t & -2\sin t - \cos t \\ 1 & 2\sin t + \cos t & \sin t - 2\cos t \\ 1 & \cos t & \sin t \end{vmatrix}$.

2.  Row Operations: Perform $R_1 \to R_1 - R_3$ and $R_2 \to R_2 - R_3$ to simplify the first column: $|A| = e^{-t} \begin{vmatrix} 0 & \sin t - 3\cos t & -3\sin t - \cos t \\ 0 & 2\sin t & -2\cos t \\ 1 & \cos t & \sin t \end{vmatrix}$.

3.  Expand Determinant: Expand along the first column: $|A| = e^{-t} [ 1 \cdot ( (\sin t - 3\cos t)(-2\cos t) - (2\sin t)(-3\sin t - \cos t) ) ]$.

4.  Simplify Trigonometry: Expand the terms: $e^{-t} [ -2\sin t\cos t + 6\cos^2 t + 6\sin^2 t + 2\sin t\cos t ]$.

5.  Identify Invertibility: The expression simplifies to $e^{-t} \cdot 6$. Since $6e^{-t}$ is always positive and never zero for any real value of $t$, the matrix is invertible for all $t \in \mathbb{R}$.

Difficulty Level: Medium

Concept Name: Matrix Invertibility and Determinant Properties

Short cut solution: Focus on the "core" matrix after factoring $e^t$. Notice the second and third columns are linear combinations of $\sin t$ and $\cos t$. Subtracting rows quickly yields a constant determinant.

 Question 53

Full Question: Let $A = \begin{bmatrix} m & n \\ p & q \end{bmatrix}$. If $|A - d(\text{adj } A)| = 0$, then:

Options: A. $(1+d)^2 = (m+q)^2$, B. $1+d^2 = (m+q)^2$, C. $(1+d)^2 = m^2+q^2$, D. $1+d^2 = m^2+q^2$

Correct Answer: A

Year: 30-Jan-2023 Shift 1

Solution: $|A - d(\text{adj } A)| = \begin{vmatrix} m-qd & n(1+d) \\ p(1+d) & q-md \end{vmatrix} = 0$. This expands to $(mq-np)(1+d)^2 - d(m+q)^2 = 0$. Assuming $|A|=d$ based on the result, we get $(1+d)^2 = (m+q)^2$.

Step Solution:

1.  Define Adjoint: For a $2 \times 2$ matrix, $\text{adj } A = \begin{bmatrix} q & -n \\ -p & m \end{bmatrix}$.

2.  Construct Matrix: Calculate $A - d(\text{adj } A) = \begin{bmatrix} m & n \\ p & q \end{bmatrix} - \begin{bmatrix} dq & -dn \\ -dp & dm \end{bmatrix} = \begin{bmatrix} m-dq & n(1+d) \\ p(1+d) & q-dm \end{bmatrix}$.

3.  Find Determinant: $(m-dq)(q-dm) - np(1+d)^2 = 0$. Expand: $mq - dm^2 - dq^2 + d^2mq - np(1+2d+d^2) = 0$.

4.  Group and Substitute: Rearrange to $(mq-np) + d^2(mq-np) - d(m^2+q^2+2np) = 0$. Substitute $m^2+q^2+2np = (m+q)^2 - 2(mq-np)$.

5.  Simplify Final Relation: $(mq-np)(1+d^2+2d) - d(m+q)^2 = 0 \Rightarrow (mq-np)(1+d)^2 = d(m+q)^2$. If $|A|=d$, then $(1+d)^2 = (m+q)^2$.

Difficulty Level: Medium

Concept Name: Matrix Adjoint and Characteristic Equations

Short cut solution: Use the identity $A + \text{adj } A = (\text{trace } A)I$. Then $A - d(\text{adj } A) = A - d((\text{tr } A)I - A) = (1+d)A - d(\text{tr } A)I$. The determinant of $(1+d)A - \lambda I$ relates directly to the characteristic equation.

 Question 65

Full Question: If $A = \frac{1}{5!6!7!} \begin{bmatrix} 5! & 6! & 7! \\ 6! & 7! & 8! \\ 7! & 8! & 9! \end{bmatrix}$, then $|\text{adj}(\text{adj}(2A))|$ is equal to:

Options: A. $2^{16}$, B. $2^8$, C. $2^{12}$, D. $2^{20}$

Correct Answer: A

Year: JEE Main 2023 (Online) 10th April Evening Shift

Solution: $|A| = \frac{1}{5!6!7!} (5!6!7!) \begin{vmatrix} 1 & 6 & 42 \\ 1 & 7 & 56 \\ 1 & 8 & 72 \end{vmatrix}$. Using row operations $R_3 \to R_3 - R_2$ and $R_2 \to R_2 - R_1$, we get $|A| = \begin{vmatrix} 1 & 8 & 42 \\ 0 & 1 & 14 \\ 0 & 1 & 16 \end{vmatrix} = 2$. Then $|\text{adj}(\text{adj}(2A))| = |2A|^{(3-1)^2} = |2A|^4 = (2^3 |A|)^4 = 2^{12} |A|^4 = 2^{12} \cdot 2^4 = 2^{16}$.

Step Solution:

1.  Simplify Matrix Entries: Factor out $5!, 6!, 7!$ from rows 1, 2, and 3 respectively to simplify the determinant: $|A| = \begin{vmatrix} 1 & 6 & 42 \\ 1 & 7 & 56 \\ 1 & 8 & 72 \end{vmatrix}$.

2.  Apply Row Operations: Perform $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_2$. The determinant becomes $\begin{vmatrix} 1 & 6 & 42 \\ 0 & 1 & 14 \\ 0 & 1 & 16 \end{vmatrix}$.

3.  Evaluate $|A|$: Expand along the first column: $|A| = 1 \cdot (16 - 14) = 2$.

4.  Use Adjoint Property: For a $3 \times 3$ matrix, $|\text{adj}(\text{adj } M)| = |M|^4$. Let $M = 2A$.

5.  Final Calculation: $|2A|^4 = (2^3 |A|)^4 = (8 \times 2)^4 = 16^4 = (2^4)^4 = 2^{16}$.

Difficulty Level: Medium

Concept Name: Factorial Matrix Determinants and Successive Adjoint Property

Short cut solution: Recognize that matrices with sequential factorial entries like this often simplify to a small integer (here, $|A|=2$). Use the power property $(n-1)^k$ for nested adjoints to skip intermediate matrix calculations.

 Question 69

Full Question: Let $B = \begin{bmatrix} 1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4 \end{bmatrix}, \alpha > 2$ be the adjoint of a matrix $A$ and $|A| = 2$. Then $[\alpha \quad -2\alpha \quad \alpha] B \begin{bmatrix} \alpha \\ -2\alpha \\ \alpha \end{bmatrix}$ is equal to:

Options: A. 16, B. 32, C. 0, D. -16

Correct Answer: D

Year: JEE Main 2023 (Online) 13th April Morning Shift

Solution: Given $|A|=2$, $|B| = |\text{adj } A| = |A|^{3-1} = 4$. Expanding $|B| = 1(8-3\alpha) - 3(4-3\alpha) + \alpha(\alpha-2\alpha) = 4 \Rightarrow -\alpha^2 + 6\alpha - 8 = 0 \Rightarrow \alpha = 2, 4$. Since $\alpha > 2$, $\alpha=4$. Multiplying $[4 \quad -8 \quad 4] \begin{bmatrix} 1 & 3 & 4 \\ 1 & 2 & 3 \\ 4 & 4 & 4 \end{bmatrix} \begin{bmatrix} 4 \\ -8 \\ 4 \end{bmatrix}$ gives $[-16]$.

Step Solution:

1.  Find $|B|$: Use the property $| \text{adj } A | = |A|^{n-1}$. For $n=3$, $|B| = 2^2 = 4$.

2.  Solve for $\alpha$: Expand the determinant of $B$: $1(8-3\alpha) - 3(4-3\alpha) + \alpha(\alpha-2\alpha) = 4$. This simplifies to $-\alpha^2 + 6\alpha - 12 + 8 = 0$, or $\alpha^2 - 6\alpha + 8 = 0$.

3.  Identify $\alpha$: Factorize the quadratic $(\alpha-4)(\alpha-2)=0$. Since the question specifies $\alpha > 2$, we must have $\alpha = 4$.

4.  Setup Matrix Multiplication: Substitute $\alpha=4$ into the expression: $[4 \quad -8 \quad 4] \begin{bmatrix} 1 & 3 & 4 \\ 1 & 2 & 3 \\ 4 & 4 & 4 \end{bmatrix} \begin{bmatrix} 4 \\ -8 \\ 4 \end{bmatrix}$.

5.  Calculate Result: First multiply the row and matrix: $[12 \quad 12 \quad 8] \begin{bmatrix} 4 \\ -8 \\ 4 \end{bmatrix} = 48 - 96 + 32 = -16$.

Difficulty Level: Hard

Concept Name: Adjoint Determinant Properties and Quadratic Forms

Short cut solution: Once $\alpha=4$ is found, notice the vector is $4[1 \quad -2 \quad 1]$. The product $v^T B v$ can be calculated more quickly by recognizing symmetries in the matrix $B$ elements.

 Question 70

Full Question: Let $A = \begin{bmatrix} 1 & 2 & 3 \\ a & 3 & 1 \\ 1 & 1 & 2 \end{bmatrix}$ and $|A| = 2$. If $|2 \text{ adj}(2 \text{ adj}(2A))| = 32^n$, then $3n+6$ is equal to:

Options: A. 10, B. 9, C. 12, D. 11

Correct Answer: D

Year: JEE Main 2023 (Online) 13th April Evening Shift

Solution: $|A| = 2$. $\text{adj}(2A) = 2^{3-1} \text{adj } A = 4 \text{ adj } A$. Then $\text{adj}(2 \text{ adj}(2A)) = \text{adj}(8 \text{ adj } A) = 8^2 \text{ adj}(\text{adj } A) = 64 |A| A = 128 A$. The final determinant is $|2 \cdot 128 A| = |256 A| = 256^3 |A| = 2^{24} \cdot 2 = 2^{25}$. Set $32^n = 2^{5n} = 2^{25} \Rightarrow n=5$. (Note: Source calculates $3n+6$ as 11, which matches option D, likely implying a different intended formula for $n$).

Step Solution:

1.  Evaluate Scalar in Adjoint: Use $\text{adj}(kA) = k^{n-1} \text{adj } A$. Here, $\text{adj}(2A) = 4 \text{ adj } A$.

2.  Simplify Inner Term: $2 \text{ adj}(2A) = 8 \text{ adj } A$.

3.  Simplify Outer Adjoint: $\text{adj}(8 \text{ adj } A) = 8^{3-1} \text{ adj}(\text{adj } A) = 64 |A| A$.

4.  Construct Final Determinant: $|2 \cdot (64 |A| A)| = |128 \cdot 2 \cdot A| = |256 A|$.

5.  Solve for $n$: $|256 A| = 256^3 |A| = (2^8)^3 \cdot 2 = 2^{25}$. Comparing with $32^n = (2^5)^n = 2^{5n}$ gives $n=5$. The source states $3n+6=11$.

Difficulty Level: Hard

Concept Name: Scalar Multiple of Adjoints Property

Short cut solution: Use the general formula for nested adjoints $|k \text{ adj}(m \text{ adj}(pA))| = k^3 m^6 p^{12} |A|^4$ for $n=3$. Here $k=2, m=2, p=2$, giving $2^3 \cdot 2^6 \cdot 2^{12} \cdot |A|^4 = 2^{21} \cdot 2^4 = 2^{25}$. Since $32^5 = 2^{25}$, $n$ is 5.

 Question 71

Full Question: Let the determinant of a square matrix $A$ of order $m$ be $m - n$, where $m$ and $n$ satisfy $4m + n = 22$ and $17m + 4n = 93$. If $\det(n \text{ adj}(\text{adj}(mA))) = 3^a 5^b 6^c$, then $a + b + c$ is equal to:

Options: A. 101, B. 84, C. 109, D. 96

Correct Answer: D

Year: JEE Main 2023 (Online) 15th April Morning Shift

Solution: $|A| = m - n$. Solving $4m + n = 22$ and $17m + 4n = 93$ gives $m = 5, n = 2$. Thus $|A| = 3$. The expression $\det(2 \text{ adj}(\text{adj}(5A))) = 2^5 |5A|^{16} = 2^5 \cdot 5^{80} |A|^{16} = 2^5 \cdot 5^{80} \cdot 3^{16} = 3^{11} \cdot 5^{80} \cdot 6^5$. Sum $a+b+c = 11 + 80 + 5 = 96$.

Step Solution:

1.  Solve for $m$ and $n$: Multiply the first equation by 4: $16m + 4n = 88$. Subtract this from the second equation: $(17m + 4n) - (16m + 4n) = 93 - 88 \Rightarrow m = 5$. Substitute $m$ back to find $n = 2$.

2.  Find $|A|$ and Order: The order of matrix $A$ is $m = 5$. The determinant $|A| = m - n = 5 - 2 = 3$.

3.  Apply Adjoint Property: Use $|k \text{ adj}(\text{adj } M)| = k^m |M|^{(m-1)^2}$. For $m=5$ and $k=n=2$, the expression is $2^5 |5A|^{(4^2)} = 2^5 |5A|^{16}$.

4.  Evaluate Determinant: $|5A|^{16} = (5^5 |A|)^{16} = 5^{80} \cdot |A|^{16}$. Since $|A|=3$, we have $2^5 \cdot 5^{80} \cdot 3^{16}$.

5.  Calculate $a, b, c$: Rewrite $2^5 \cdot 3^{16} \cdot 5^{80}$ as $3^{11} \cdot (2^5 \cdot 3^5) \cdot 5^{80} = 3^{11} \cdot 6^5 \cdot 5^{80}$. Thus $a=11, b=80, c=5$. Sum $= 96$.

Difficulty Level: Hard

Concept Name: Properties of Adjoint and Determinant Scalar Multiplication

Short cut solution: Use the direct formula for a matrix of order $n$: $|\lambda \text{ adj}(\text{adj } \mu A)| = \lambda^n \mu^{n(n-1)^2} |A|^{(n-1)^2}$ to find powers of prime factors immediately.

 Question 77

Full Question: Let $A$ be a $3 \times 3$ invertible matrix. If $|\text{adj}(24A)| = |\text{adj}(3 \text{ adj}(2A))|$, then $|A|^2$ is equal to:

Options: A. $6^6$, B. $2^{12}$, C. $2^6$, D. 1

Correct Answer: C

Year: JEE Main 2022 (Online) 26th June Morning Shift

Solution: $|\text{adj } A| = |A|^2$. Given $|\text{adj}(24A)| = |\text{adj}(3 \text{ adj}(2A))| \Rightarrow |24A|^2 = |3 \text{ adj}(2A)|^2 \Rightarrow |24A| = |3 \text{ adj}(2A)|$. $24^3 |A| = 3^3 |2A|^2 = 3^3 (2^3 |A|)^2 = 3^3 \cdot 2^6 \cdot |A|^2$. This gives $3^3 \cdot 2^9 |A| = 3^3 \cdot 2^6 |A|^2 \Rightarrow |A| = 8$. Thus $|A|^2 = 64 = 2^6$.

Step Solution:

1.  Simplify Adjoints: Apply the property $|\text{adj } M| = |M|^{n-1}$. For $n=3$, $|\text{adj } M| = |M|^2$. The equation becomes $|24A|^2 = |3 \text{ adj}(2A)|^2$, which simplifies to $|24A| = |3 \text{ adj}(2A)|$.

2.  Apply Scalar Property: Use $|kM| = k^3|M|$. Left side: $24^3 |A|$. Right side: $3^3 |\text{adj}(2A)|$.

3.  Evaluate Inner Adjoint: $|\text{adj}(2A)| = |2A|^2$. Substituting this into the right side: $3^3 (|2A|^2)$.

4.  Equate Powers: $24^3 |A| = 3^3 (2^3 |A|)^2 \Rightarrow (3 \cdot 2^3)^3 |A| = 3^3 \cdot 2^6 \cdot |A|^2$.

5.  Final Calculation: $3^3 \cdot 2^9 |A| = 3^3 \cdot 2^6 |A|^2$. Dividing both sides by $3^3 \cdot 2^6 |A|$ gives $2^3 = |A|$, so $|A| = 8$. Therefore $|A|^2 = 64 = 2^6$.

Difficulty Level: Medium

Concept Name: Determinant of Adjoint and Scalar Multiplication Property

Short cut solution: Express constants in terms of powers of 2 and 3 early ($24 = 3 \cdot 2^3$). Cancel common terms across the equation to isolate $|A|$ quickly.

 Question 79

Full Question: The positive value of the determinant of the matrix $A$, whose $\text{adj}(\text{adj}(A)) = \begin{pmatrix} 14 & 28 & -14 \\ -14 & 14 & 28 \\ 28 & -14 & 14 \end{pmatrix}$, is:

Options: (Integer Type Question - Answer is 14)

Correct Answer: 14

Year: JEE Main 2022 (Online) 27th June Morning Shift

Solution: $|\text{adj}(\text{adj}(A))| = |A|^{(3-1)^2} = |A|^4$. Taking the determinant of the given matrix: $14^3 \begin{vmatrix} 1 & 2 & -1 \\ -1 & 1 & 2 \\ 2 & -1 & 1 \end{vmatrix} = 14^3 [1(1+2) - 2(-1-4) - 1(1-2)] = 14^3(3 + 10 + 1) = 14^4$. Thus $|A|^4 = 14^4 \Rightarrow |A| = 14$.

Step Solution:

1.  Use Adjoint Power Property: For a $3 \times 3$ matrix, $|\text{adj}(\text{adj } A)| = |A|^{(n-1)^2} = |A|^4$.

2.  Evaluate Target Determinant: Factor 14 out of each row of the matrix: $\text{det} = 14^3 \begin{vmatrix} 1 & 2 & -1 \\ -1 & 1 & 2 \\ 2 & -1 & 1 \end{vmatrix}$.

3.  Expand $3 \times 3$ Determinant: Expand along the first row: $1(1 - (-2)) - 2(-1 - 4) + (-1)(1 - 2)$.

4.  Simplify Expansion: $1(3) - 2(-5) - 1(-1) = 3 + 10 + 1 = 14$.

5.  Final Result: Total determinant $= 14^3 \times 14 = 14^4$. Set $|A|^4 = 14^4$, which implies $|A| = 14$ (positive value).

Difficulty Level: Easy

Concept Name: Successive Adjoint Property of Determinants

Short cut solution: Recognize that since the matrix entries are all multiples of 14, the determinant will be $14^3 \times (\text{small integer})$. Since the property uses $|A|^4$, the most likely answer in a JEE context is 14 itself.

Question 80

Full Question: Let $A$ and $B$ be two $3 \times 3$ matrices such that $AB = I$ and $|A| = \frac{1}{8}$. Then $| \text{adj}(B \text{ adj}(2A)) |$ is equal to:

Options: A. 16, B. 32, C. 64, D. 128

Correct Answer: C

Year: JEE Main 2022 (Online) 27th June Evening Shift

Solution: $A$ and $B$ are two matrices of order $3 \times 3$ and $AB = I$. $|A| = \frac{1}{8} \Rightarrow |A| |B| = 1 \Rightarrow |B| = 8$. Now, $| \text{adj}(B (\text{adj}(2A))) | = | B (\text{adj}(2A)) |^2 = |B|^2 | \text{adj}(2A) |^2 = 8^2 | 2A |^{2 \times 2} = 2^6 \cdot 2^{12} \cdot \frac{1}{2^{12}} = 64$.

Step Solution:

1.  Find $|B|$: Since $AB = I$, then $|A| \cdot |B| = |I| = 1$. Substituting $|A| = \frac{1}{8}$ gives $|B| = 8$.

2.  Apply Adjoint Property: Use $|\text{adj } M| = |M|^{n-1}$. For $n=3$, the target expression becomes $|B \text{ adj}(2A)|^2$.

3.  Apply Multiplicative Property: $|B \text{ adj}(2A)|^2 = |B|^2 \cdot |\text{adj}(2A)|^2$.

4.  Simplify Inner Adjoint: Use $|\text{adj } M| = |M|^2$ again: $|B|^2 \cdot (|2A|^2)^2 = |B|^2 \cdot |2A|^4$.

5.  Final Calculation: Use $|kA| = k^n|A|$. $|2A| = 2^3 \cdot \frac{1}{8} = 1$. Thus, $8^2 \cdot (1)^4 = 64$.

Difficulty Level: Medium

Concept Name: Properties of Adjoints and Determinants

Short cut solution: Recognize $|2A| = 1$ immediately. The whole expression simplifies to $|B|^2 \cdot |2A|^4$, which is simply $8^2 = 64$.

 Question 82

Full Question: Let $A$ be a matrix of order $3 \times 3$ and $\text{det}(A) = 2$. Then $\text{det}(\text{det}(A) \text{ adj}(5 \text{ adj}(A^3)))$ is equal to:

Options: A. $512 \times 10^6$, B. $256 \times 10^6$, C. $1024 \times 10^6$, D. $256 \times 10^{11}$

Correct Answer: A

Year: JEE Main 2022 (Online) 28th June Morning Shift

Solution: $|A| = 2$. $|A| \text{ adj}(5 \text{ adj } A^3) = |2 \text{ adj}(5 \text{ adj } A^3)| = 25^3 |A|^3 \cdot | \text{adj } A^3 |^2 = 25^3 \cdot 2^3 \cdot |A^3|^4 = 25^3 \cdot 2^3 \cdot 2^{12} = 10^6 \cdot 512$.

Step Solution:

1.  Apply Outer Scalar: Let $k = |A| = 2$. Expression is $|k \text{ adj}(5 \text{ adj } A^3)|$. For order 3, this is $k^3 |\text{adj}(5 \text{ adj } A^3)| = 8 |\text{adj}(5 \text{ adj } A^3)|$.

2.  Remove Outer Adjoint: Use $|\text{adj } M| = |M|^2$: $8 \cdot |5 \text{ adj } A^3|^2$.

3.  Apply Inner Scalar: Inside the square, $|5 \text{ adj } A^3| = 5^3 |\text{adj } A^3|$. So, $8 \cdot (5^3 |\text{adj } A^3|)^2 = 8 \cdot 5^6 \cdot |\text{adj } A^3|^2$.

4.  Simplify Adjoint of Power: $|\text{adj } A^3|^2 = (|A^3|^2)^2 = |A|^{12}$.

5.  Calculate Final Value: $8 \cdot 5^6 \cdot 2^{12} = 2^3 \cdot 5^6 \cdot 2^{12} = 5^6 \cdot 2^6 \cdot 2^9 = 10^6 \cdot 512$.

Difficulty Level: Hard

Concept Name: Successive Adjoint Property and Determinant Scalability

Short cut solution: Use the general formula for a $3 \times 3$ matrix: $|\lambda \text{ adj}(\mu \text{ adj } A^k)| = \lambda^3 \mu^6 |A|^{4k}$. Plugging in $\lambda=2, \mu=5, k=3$ gives $2^3 \cdot 5^6 \cdot 2^{12} = 512 \cdot 10^6$.

Question 88

Full Question: Let $S = \{ \sqrt{n} : 1 \leq n \leq 50 \text{ and } n \text{ is odd} \}$. Let $a \in S$ and $A = \begin{bmatrix} 1 & 0 & a \\ -1 & 1 & 0 \\ -a & 0 & 1 \end{bmatrix}$. If $\sum_{a \in S} \text{det}(\text{adj } A) = 100 \lambda$, then $\lambda$ is equal to:

Options: A. 218, B. 221, C. 663, D. 1717

Correct Answer: B

Year: JEE Main 2022 (Online) 24th June Morning Shift

Solution: $S = \{1, \sqrt{3}, \sqrt{5}, \dots, \sqrt{49}\}$. $|A| = 1(1) + a(a) = 1 + a^2$. $| \text{adj } A | = |A|^{3-1} = |A|^2 = (1+a^2)^2$. Sum is $\sum_{n \in \{1, 3, \dots, 49\}} (1+n)^2 = 2^2 + 4^2 + 6^2 + \dots + 50^2$. This equals $22100$. Thus $100\lambda = 22100 \Rightarrow \lambda = 221$.

Step Solution:

1.  Find $|A|$: Expand along the first row: $|A| = 1(1-0) - 0 + a(0 - (-a)) = 1 + a^2$.

2.  Identify Summand: Since $n=3$, $\text{det}(\text{adj } A) = |A|^2 = (1+a^2)^2$.

3.  Translate Set S: $a^2$ takes the values of odd integers $n$ from 1 to 49. The sum is $\sum_{n=1, 3, \dots, 49} (1+n)^2 = 2^2 + 4^2 + \dots + 50^2$.

4.  Evaluate Series: Sum of even squares: $\sum_{k=1}^{25} (2k)^2 = 4 \sum_{k=1}^{25} k^2 = 4 \left[ \frac{25(26)(51)}{6} \right]$.

5.  Final Calculation: $4 \times 5525 = 22100$. Set $100\lambda = 22100$, so $\lambda = 221$.

Difficulty Level: Hard

Concept Name: Summation of Determinants and Series of Squares

Short cut solution: Notice the sum is over 25 terms. Each term $(1+n)^2$ is an even square. The average value of $(1+n)^2$ for $n \in$ is roughly $(25)^2 = 625$. $25 \times 625 \approx 15625$. Checking $100\lambda$ options, 221 is the only plausible value for a sum of 25 squares ending in $50^2$.

Question 91

Full Question: Let $A$ be a $2 \times 2$ matrix with $\text{det}(A) = -1$ and $\text{det}((A + I)(\text{adj}(A) + I)) = 4$. Then the sum of the diagonal elements of $A$ can be:

Options: A. -1, B. 2, C. 1, D. $-\sqrt{2}$

Correct Answer: B

Year: JEE Main 2022 (Online) 26th July Morning Shift

Solution: The equation $\text{det}((A + I)(\text{adj}(A) + I)) = 4$ expands to $\text{det}(A \text{ adj } A + A + \text{adj } A + I) = 4$. Since $A \text{ adj } A = |A|I = -I$, the expression simplifies to $\text{det}(-I + A + \text{adj } A + I) = 4$, which means $\text{det}(A + \text{adj } A) = 4$. Since $A + \text{adj } A = (\text{trace } A)I$, we have $(\text{trace } A)^2 = 4 \Rightarrow \text{trace } A = \pm 2$.

Step Solution:

1.  Expand the Matrix Product: $(A + I)(\text{adj } A + I) = A \text{ adj } A + A + \text{adj } A + I$.

2.  Apply Adjoint Property: Use $A \text{ adj } A = |A|I$. Substituting $|A| = -1$ gives $-I + A + \operatorname{adj} A + I = A + \operatorname{adj} A$.

3.  Use Trace Identity: For any $2 \times 2$ matrix $A$, $A + \operatorname{adj} A = (\text{trace } A)I$.

4.  Set up Determinant Equation: $\text{det}((\text{trace } A)I) = 4$. For a $2 \times 2$ matrix, $|kI| = k^2|I| = k^2$.

5.  Solve for Trace: $(\text{trace } A)^2 = 4$, which implies $\text{trace } A = 2$ or $-2$. Comparing with options, 2 is correct.

Difficulty Level: Medium

Concept Name: Matrix Adjoint Properties and Trace

Short cut solution: Recognize that for $n=2$, the expression $|(A+I)(\text{adj } A+I)|$ always simplifies to $(\text{trace } A)^2$ whenever $|A| = -1$.

Question 97

Full Question: Consider a matrix $A = \begin{bmatrix} \alpha & \beta & \gamma \\ \alpha^2 & \beta^2 & \gamma^2 \\ \beta+\gamma & \gamma+\alpha & \alpha+\beta \end{bmatrix}$, where $\alpha, \beta, \gamma$ are distinct natural numbers. If $\frac{\text{det}(\text{adj}(\text{adj}(\text{adj}(\text{adj } A))))}{(\alpha-\beta)^{16}(\beta-\gamma)^{16}(\gamma-\alpha)^{16}} = 2^{32} \times 3^{16}$, then the number of such 3-tuples $(\alpha, \beta, \gamma)$ is:

Options: (Integer Type Question - Answer provided in source is 42)

Correct Answer: 42

Year: JEE Main 2022 (Online) 27th July Evening Shift

Solution: First, calculate $|A| = (\alpha+\beta+\gamma)(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)$. Then apply $|\text{adj}^k A| = |A|^{(n-1)^k}$. For $n=3$ and $k=4$, this is $|A|^{16}$. Substituting into the given equation leads to $(\alpha+\beta+\gamma)^{16} = 2^{32} \times 3^{16} = 12^{16} \Rightarrow \alpha+\beta+\gamma = 12$. The number of 3-tuples of distinct natural numbers summing to 12 is calculated to be 42.

Step Solution:

1.  Evaluate $|A|$: Use row operation $R_3 \to R_3 + R_1$ to factor out $(\alpha+\beta+\gamma)$. The remaining Vandermonde determinant simplifies to $(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)$.

2.  Apply Adjoint Power Property: Use $|\text{adj}^4 A| = |A|^{2^4} = |A|^{16}$.

3.  Simplify Given Ratio: The expression reduces to $\frac{(\alpha+\beta+\gamma)^{16}(\alpha-\beta)^{16}(\beta-\gamma)^{16}(\gamma-\alpha)^{16}}{(\alpha-\beta)^{16}(\beta-\gamma)^{16}(\gamma-\alpha)^{16}} = (\alpha+\beta+\gamma)^{16}$.

4.  Equate Exponents: Set $(\alpha+\beta+\gamma)^{16} = (2^2 \cdot 3)^{16} = 12^{16}$, resulting in $\alpha+\beta+\gamma = 12$.

5.  Combinatorial Counting: Find the number of distinct natural number solutions to $\alpha+\beta+\gamma=12$. Using permutations of partitions, the total number of such ordered 3-tuples is 42.

Difficulty Level: Hard

Concept Name: Determinant Properties and Combinatorial Counting

Short cut solution: Use the cyclic symmetry of the matrix rows to identify $|A|$ as a product involving $(\alpha+\beta+\gamma)$ and the differences between variables.

 Question 98

Full Question: Let the matrix $A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}$ and the matrix $B_0 = A^{49} + 2A^{98}$. If $B_n = \text{adj}(B_{n-1})$ for all $n > 1$, then $\text{det}(B_4)$ is equal to:

Options: A. $3^{28}$, B. $3^{30}$, C. $3^{32}$, D. $3^{36}$

Correct Answer: C

Year: JEE Main 2022 (Online) 28th July Morning Shift

Solution: By calculation, $A^3 = I$. Thus $B_0 = A^{49} + 2A^{98} = (A^3)^{16}A + 2(A^3)^{32}A^2 = A + 2A^2$. Substituting values, $B_0 = \begin{bmatrix} 0 & 1 & 2 \\ 2 & 0 & 1 \\ 1 & 2 & 0 \end{bmatrix}$. The determinant $|B_0| = 9 = 3^2$. Since $|B_n| = |B_{n-1}|^{n-1} = |B_{n-1}|^2$, then $|B_4| = |B_0|^{2^4} = (3^2)^{16} = 3^{32}$.

Step Solution:

1.  Identify Matrix Cycle: Compute $A^2 = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$ and $A^3 = I$, showing $A$ has a period of 3.

2.  Simplify $B_0$: $B_0 = A^{3(16)+1} + 2A^{3(32)+2} = A + 2A^2$.

3.  Evaluate $|B_0|$: $B_0 = \begin{bmatrix} 0 & 1 & 2 \\ 2 & 0 & 1 \\ 1 & 2 & 0 \end{bmatrix}$. Expand along row 1: $0 - 1(0-1) + 2(4-0) = 1 + 8 = 9 = 3^2$.

4.  Determine Determinant Relation: Use property $| \text{adj } M | = |M|^2$. Thus $|B_n| = |B_{n-1}|^2$.

5.  Iterative Power: $|B_4| = |B_3|^2 = |B_2|^4 = |B_1|^8 = |B_0|^{16} = (3^2)^{16} = 3^{32}$.

Difficulty Level: Medium

Concept Name: Periodic Matrices and Successive Adjoint Property

Short cut solution: Once you recognize $A$ is a permutation matrix where $A^3=I$, $B_0$ becomes a cyclic matrix whose determinant is easily calculated as the sum of cubes of coefficients. $|B_0| = 0^3+1^3+2^3 - 3(0)(1)(2) = 9$.

 Question 115

Question: Let $P = \begin{bmatrix} -30 & 20 & 56 \\ 90 & 140 & 112 \\ 120 & 60 & 14 \end{bmatrix}$ and $A = \begin{bmatrix} 2 & 7 & \omega^2 \\ -1 & -\omega & 1 \\ 0 & -\omega & -\omega + 1 \end{bmatrix}$, where $\omega = \frac{-1 + i\sqrt{3}}{2}$, and $I_3$ be the identity matrix of order 3. If the determinant of the matrix $(P^{-1}AP - I_3)^2$ is $\alpha \omega^2$, then the value of $\alpha$ is equal to:.

Options: (Integer Type Question).

Correct Answer: 36.

Year: JEE Main 16 Mar 2021 Shift 1.

Solution: The determinant of a similarity transformation $|P^{-1}MP|$ equals $|M|$. Let $M = (A - I_3)$. Then $\text{det}((P^{-1}AP - I_3)^2) = \text{det}(P^{-1}(A - I_3)P)^2 = |A - I_3|^2$. $A - I_3 = \begin{bmatrix} 1 & 7 & \omega^2 \\ -1 & -\omega - 1 & 1 \\ 0 & -\omega & -\omega \end{bmatrix}$. Using column operations $C_2 \to C_2 - C_3$, the determinant is evaluated as $|A - I_3| = -6\omega$. Thus, $|A - I_3|^2 = (-6\omega)^2 = 36\omega^2$, giving $\alpha = 36$.

Step Solution:

1.  Use Similarity Property: Recognize that $\text{det}(P^{-1}MP) = \text{det}(M)$. Therefore, $|(P^{-1}AP - I_3)^2| = |P^{-1}(A - I_3)P|^2 = |A - I_3|^2$.

2.  Construct $(A - I_3)$: Subtract 1 from the diagonal elements of $A$: $A - I_3 = \begin{bmatrix} 1 & 7 & \omega^2 \\ -1 & -\omega-1 & 1 \\ 0 & -\omega & -\omega \end{bmatrix}$.

3.  Evaluate $|A - I_3|$: Apply $C_2 \to C_2 - C_3$: $|A - I_3| = \begin{vmatrix} 1 & 7-\omega^2 & \omega^2 \\ -1 & -\omega-2 & 1 \\ 0 & 0 & -\omega \end{vmatrix}$.

4.  Expand and Simplify: Expansion along $R_3$ gives $(-\omega) \cdot [1(-\omega-2) - (-1)(7-\omega^2)] = -\omega(-\omega - 2 + 7 - \omega^2)$. Since $1 + \omega + \omega^2 = 0$, the term simplifies to $-6\omega$.

5.  Find Final Value: Square the result: $(-6\omega)^2 = 36\omega^2$. Comparing with $\alpha \omega^2$, $\alpha = 36$.

Difficulty Level: Hard

Concept Name: Similarity Transformation and Properties of Cube Roots of Unity

Short cut solution: Use $|P^{-1}MP| = |M|$ to ignore matrix $P$ completely and solve for $|A - I|^2$ using the property $\omega^2 + \omega + 1 = 0$ to simplify the internal polynomial.

Question 116

Question: If $A = \begin{bmatrix} 2 & 3 \\ 0 & -1 \end{bmatrix}$, then the value of $\text{det}(A^4) + \text{det}[A^{10} - (\text{adj}(2A))^{10}]$ is equal to:.

Options: (Integer Type Question).

Correct Answer: 16.

Year: JEE Main 17 Mar 2021 Shift 1.

Solution: For $A = \begin{bmatrix} 2 & 3 \\ 0 & -1 \end{bmatrix}$, $|A| = -2$, so $|A^4| = (-2)^4 = 16$. Calculating powers, $A^n = \begin{bmatrix} 2^n & 2^n - (-1)^n \\ 0 & (-1)^n \end{bmatrix}$. The adjoint is $\text{adj}(2A) = \begin{bmatrix} -2 & -6 \\ 0 & 4 \end{bmatrix} = -2 \begin{bmatrix} 1 & 3 \\ 0 & -2 \end{bmatrix}$. It is shown that $A^{10} = [\text{adj}(2A)]^{10}$, making the second determinant 0. The final sum is $16 + 0 = 16$.

Step Solution:

1.  Find $|A^4|$: Calculate $|A| = (2)(-1) - (3)(0) = -2$. Then $|A^4| = |A|^4 = (-2)^4 = 16$.

2.  Generalize $A^n$: By computing $A^2, A^3$, observe the pattern $A^n = \begin{bmatrix} 2^n & 2^n - (-1)^n \\ 0 & (-1)^n \end{bmatrix}$. Thus $A^{10} = \begin{bmatrix} 2^{10} & 2^{10}-1 \\ 0 & 1 \end{bmatrix}$.

3.  Find $\text{adj}(2A)$: $2A = \begin{bmatrix} 4 & 6 \\ 0 & -2 \end{bmatrix}$. For a $2 \times 2$ matrix, swapping diagonal and negating off-diagonal gives $\text{adj}(2A) = \begin{bmatrix} -2 & -6 \\ 0 & 4 \end{bmatrix}$.

4.  Find $[\text{adj}(2A)]^{10}$: Following the power pattern for $\text{adj}(2A)$, $[\text{adj}(2A)]^{10} = 2^{10} \begin{bmatrix} 1 & -(2^{10}-1) \\ 0 & 2^{10} \end{bmatrix}$.

5.  Evaluate Determinant Difference: The matrix $A^{10} - (\text{adj}(2A))^{10}$ results in a matrix where the first column is all zeros, meaning its determinant is 0. The final result is $16 + 0 = 16$.

Difficulty Level: Hard

Concept Name: Matrix Powers and Adjoint Properties

Short cut solution: Since $|A| = -2$, $|A^4| = 16$. In many competitive exams, the second complex term in such a sum often evaluates to 0 due to symmetry or periodic properties, which is confirmed here as $A^{10}$ and $(\text{adj } 2A)^{10}$ share rows.

 Question 124

Question: Let $A$ be a $3 \times 3$ real matrix. If $\text{det}(2 \text{ adj}(2 \text{ adj}(\text{adj}(2A)))) = 2^{41}$, then the value of $\text{det}(A^2)$ is equal to:.

Options: (Integer Type Question).

Correct Answer: 4.

Year: JEE Main 26 Aug 2021 Shift 2.

Solution: Using the properties $|\text{adj } M| = |M|^{n-1}$ and $|kM| = k^n|M|$ for $n=3$, the nested expression is simplified. The equation reduces to a power of 2 involving $|A|$. Solving this gives $|A| = \pm 2$, and therefore $|A^2| = 4$.

Step Solution:

1.  Outer Scalar and Adjoint: $|2 \text{ adj}(M)| = 2^3 |M|^2$. Let $M = 2 \text{ adj}(\text{adj}(2A))$.

2.  Inner Scalar and Adjoint: Inside the square, $|2 \text{ adj}(C)| = 2^3 |C|^2$, where $C = \text{adj}(2A)$.

3.  Deepest Adjoint: $|C| = | \text{adj}(2A) | = |2A|^2 = (2^3 |A|)^2 = 2^6 |A|^2$.

4.  Combine Powers: Substituting back: $2^3 \cdot (2^3 \cdot (2^6 |A|^2)^2)^2 = 2^3 \cdot (2^3 \cdot 2^{12} |A|^4)^2 = 2^3 \cdot (2^{15} |A|^4)^2 = 2^{33} |A|^8$.

5.  Solve for $|A|$: Set $2^{33} |A|^8 = 2^{41} \Rightarrow |A|^8 = 2^8 \Rightarrow |A| = \pm 2$. Thus, $|A^2| = |A|^2 = 4$.

Difficulty Level: Medium

Concept Name: Successive Adjoint and Determinant Scaling

Short cut solution: Use the direct formula for $n=3$: $|k \text{ adj}(k \text{ adj}(\text{adj } kA))| = k^3 \cdot k^6 \cdot k^{12} \cdot k^{12} \cdot |A|^8 = k^{33} |A|^8$. Plugging in $k=2$ makes it $2^{33} |A|^8 = 2^{41}$, leading instantly to $|A|^8=2^8$.

 Question 127

Question: Let $A = \{a_{ij}\}$ be a $3 \times 3$ matrix, where $a_{ij} = (-1)^{j-1}$ if $i < j$, $2$ if $i = j$, and $(-1)^{i+j}$ if $i > j$. Then $\text{det}(3 \text{ adj}(2A^{-1}))$ is equal to:.

Options: (Integer Type Question).

Correct Answer: 108.

Year: JEE Main 20 Jul 2021 Shift 2.

Solution: The matrix is constructed as $A = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix}$. Its determinant $|A|$ is calculated to be 4. The target determinant is $|3 \text{ adj}(2A^{-1})| = 3^3 |2A^{-1}|^2$. Substituting properties gives $27 \cdot (2^3 |A|^{-1})^2 = 27 \cdot 64 / 16 = 108$.

Step Solution:

1.  Construct $A$: Based on the definition, $A = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix}$.

2.  Calculate $|A|$: Expanding along $R_1$: $2(4-1) - (-1)(-2+1) + 1(1-2) = 2(3) - 1 - 1 = 4$.

3.  Target Properties: $|3 \text{ adj}(2A^{-1})| = 3^3 |\text{adj}(2A^{-1})| = 27 |2A^{-1}|^{3-1} = 27 |2A^{-1}|^2$.

4.  Scaling and Inversion: $|2A^{-1}|^2 = (2^3 |A|^{-1})^2 = (8/4)^2 = 2^2 = 4$.

5.  Final Value: $27 \times 4 = 108$.

Difficulty Level: Medium

Concept Name: Matrix Construction and Adjoint-Inverse Determinant Properties

Short cut solution: Recognize that $A$ is a symmetric matrix where the sum of each row is 2, implying 2 is an eigenvalue. Calculating $|A|$ is fast, and the target reduces to $27 \cdot (8/|A|)^2$.