Q52
jee_main_2025_08_april_evening
System of Linear Equations
Let alpha$\alpha$ be a solution of x^2 + x + 1 = 0$x^2 + x + 1 = 0$, and for some a$a$ and b$b$ in mathbbR$\mathbb{R}$, [4 quad a quad b]beginbmatrix1 & 16 & 13\\ -1 & -1 & 2\\ -2 & -14 & -8 endbmatrix = [0 quad 0 quad 0]$[4 \quad a \quad b]\begin{bmatrix}1 & 16 & 13\\ -1 & -1 & 2\\ -2 & -14 & -8 \end{bmatrix} = [0 \quad 0 \quad 0]$. If frac4alpha^4 + fracmalpha^a + fracnalpha^b = 3$\frac{4}{\alpha^4} + \frac{m}{\alpha^a} + \frac{n}{\alpha^b} = 3$, then m + n$m + n$ is equal to
- A. 3$3$
- B. 11$11$
- C. 7$7$
- D. 8$8$
Solution
### Related Formula
alpha^2 + alpha + 1 = 0 implies alpha = omega quad textwhere omega^3 = 1$$\alpha^2 + \alpha + 1 = 0 \implies \alpha = \omega \quad \text{where } \omega^3 = 1$$
### Core Logic
Perform row-matrix vector multiplication to generate a system of linear equations in a$a$ and b$b$. Solve for the powers and reduce the algebraic equation using complex roots of unity.
### Step 1: Solve Matrix Vector Multiplication
4 - a - 2b = 0$4 - a - 2b = 0$
64 - a - 14b = 0$$64 - a - 14b = 0$$
52 + 2a - 8b = 0$$52 + 2a - 8b = 0$$
From the first two equations, subtracting them gives:
60 - 12b = 0 implies b = 5$$60 - 12b = 0 \implies b = 5$$
Substituting b = 5$b = 5$ into the first equation:
4 - a - 10 = 0 implies a = -6$$4 - a - 10 = 0 \implies a = -6$$
### Step 2: Evaluate Exponential Equation with Roots of Unity
Substitute a = -6, b = 5$a = -6, b = 5$ into the given equation:
frac4alpha^4 + fracmalpha^-6 + fracnalpha^5 = 3 implies frac4omega + m + fracnomega^2 = 3$$\frac{4}{\alpha^4} + \frac{m}{\alpha^{-6}} + \frac{n}{\alpha^5} = 3 \implies \frac{4}{\omega} + m + \frac{n}{\omega^2} = 3$$
4omega^2 + m + nomega = 3$$4\omega^2 + m + n\omega = 3$$
### Step 3: Resolve Real and Imaginary Components
Substitute standard values omega = -frac12 + fracsqrt32i$\omega = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$ and omega^2 = -frac12 - fracsqrt32i$\omega^2 = -\frac{1}{2} - \frac{\sqrt{3}}{2}i$:
4left(-frac12 - fracsqrt32iright) + m + nleft(-frac12 + fracsqrt32iright) = 3$$4\left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) + m + n\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) = 3$$
Equating the imaginary components:
frac-4sqrt32 + fracnsqrt32 = 0 implies n = 4$$\frac{-4\sqrt{3}}{2} + \frac{n\sqrt{3}}{2} = 0 \implies n = 4$$
Equating the real components:
-2 + m - fracn2 = 3 implies -2 + m - 2 = 3 implies m = 7$$-2 + m - \frac{n}{2} = 3 \implies -2 + m - 2 = 3 \implies m = 7$$
m + n = 7 + 4 = 11$$m + n = 7 + 4 = 11$$
### Pattern Recognition
Whenever an expression satisfies Aomega^2 + Bomega + C = 0$A\omega^2 + B\omega + C = 0$, it directly maps to a comparison with the standard identity omega^2 + omega + 1 = 0$\omega^2 + \omega + 1 = 0$ up to a linear translation shift.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Matrices and Determinants
Class 11 Mathematics: Complex Numbers
Q69
jee_main_2025_08_april_evening
Determinant Properties of Adjoint
Let mathrmA = beginbmatrix 2 & 2 + mathrmp & 2 + mathrmp + mathrmq \\ 4 & 6 + 2mathrmp & 8 + 3mathrmp + 2mathrmq \\ 6 & 12 + 3mathrmp & 20 + 6mathrmp + 3mathrmq endbmatrix$\mathrm{A} = \begin{bmatrix} 2 & 2 + \mathrm{p} & 2 + \mathrm{p} + \mathrm{q} \\ 4 & 6 + 2\mathrm{p} & 8 + 3\mathrm{p} + 2\mathrm{q} \\ 6 & 12 + 3\mathrm{p} & 20 + 6\mathrm{p} + 3\mathrm{q} \end{bmatrix}$. If det left( adj(adj(3A)) right) = 2^m cdot 3^n$\det \left( adj(adj(3A)) \right) = 2^m \cdot 3^n$, m, n in mathbbN$m, n \in \mathbb{N}$, then m + n$m + n$ is equal to
- A. 22$22$
- B. 24$24$
- C. 26$26$
- D. 20$20$
Solution
### Related Formula
|operatornameadj(operatornameadj(M))| = |M|^(n-1)^2$$|\operatorname{adj}(\operatorname{adj}(M))| = |M|^{(n-1)^2}$$
|kM| = k^n|M|$|kM| = k^n|M|$
### Core Logic
Perform determinant row reduction transforms to decouple tracking metrics p$p$ and q$q$, leaving a baseline numerical determinant value behind.
### Step 1: Simplify the Determinant of Matrix A
Execute columns adjustments: C_3 rightarrow C_3 - C_2 - C_1 times fracq2$C_3 \rightarrow C_3 - C_2 - C_1 \times \frac{q}{2}$, then C_2 rightarrow C_2 - C_1 times left(1 + fracp2right)$C_2 \rightarrow C_2 - C_1 \times \left(1 + \frac{p}{2}\right)$:
|A| = beginvmatrix 2 & 0 & 0 \\ 4 & 2 & 2+p \\ 6 & 6 & 8+3p endvmatrix = 2left(16 + 6p - 12 - 6pright) = 8 = 2^3$$|A| = \begin{vmatrix} 2 & 0 & 0 \\ 4 & 2 & 2+p \\ 6 & 6 & 8+3p \end{vmatrix} = 2\left(16 + 6p - 12 - 6p\right) = 8 = 2^3$$
### Step 2: Apply Adjoint Exponent Transforms
For a matrix of dimensional profile size 3:
|operatornameadj(operatornameadj(3A))| = |3A|^(3-1)^2 = |3A|^4$$|\operatorname{adj}(\operatorname{adj}(3A))| = |3A|^{(3-1)^2} = |3A|^4$$
### Step 3: Resolve Exponential System Size
|3A| = 3^3 |A| = 3^3 times 2^3$$|3A| = 3^3 |A| = 3^3 \times 2^3$$
|3A|^4 = (3^3 times 2^3)^4 = 2^12 times 3^12$$|3A|^4 = (3^3 \times 2^3)^4 = 2^{12} \times 3^{12}$$
Matching base parameter targets: m = 12, n = 12 implies m + n = 24$m = 12, n = 12 \implies m + n = 24$
### Pattern Recognition
Linear parameter shifts down secondary column profiles usually dissolve cleanly during forward element elimination column steps.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Matrices and Determinants
Q61
jee_main_2025_29_jan_evening
System of Linear Equations
Let alpha, beta$\alpha, \beta$ ( alpha neq beta$\alpha \neq \beta$ ) be the values of m$m$, for which the equations x + y + z = 1$x + y + z = 1$ ; x + 2y + 4z = m$x + 2y + 4z = m$ and x + 4y + 10z = m^2$x + 4y + 10z = m^2$ have infinitely many solutions. Then the value of sum_n=1^10 (n^alpha + n^beta)$\sum_{n=1}^{10} (n^\alpha + n^\beta)$ is equal to:
- A. 440$440$
- B. 3080$3080$
- C. 3410$3410$
- D. 560$560$
Solution
### Related Formula
Cramer's rule for infinite solutions in a 3 variable system requires:
Delta = Delta_x = Delta_y = Delta_z = 0$$\Delta = \Delta_x = \Delta_y = \Delta_z = 0$$
### Core Logic
Set up the primary matrix determinant Delta$\Delta$:
Delta = beginvmatrix 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 10 endvmatrix$$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 10 \end{vmatrix}$$
= 1(20 - 16) - 1(10 - 4) + 1(4 - 2) = 4 - 6 + 2 = 0$$= 1(20 - 16) - 1(10 - 4) + 1(4 - 2) = 4 - 6 + 2 = 0$$
Since Delta = 0$\Delta = 0$ is true independent of m$m$, analyze secondary delta constraints to maintain consistency for infinite paths.
### Step 1: Compute Dependent Variable Constraints
Evaluate Delta_x = 0$\Delta_x = 0$:
Delta_x = beginvmatrix 1 & 1 & 1 \\ m & 2 & 4 \\ m^2 & 4 & 10 endvmatrix = 0$$\Delta_x = \begin{vmatrix} 1 & 1 & 1 \\ m & 2 & 4 \\ m^2 & 4 & 10 \end{vmatrix} = 0$$
1(20 - 16) - 1(10m - 4m^2) + 1(4m - 2m^2) = 0$$1(20 - 16) - 1(10m - 4m^2) + 1(4m - 2m^2) = 0$$
4 - 10m + 4m^2 + 4m - 2m^2 = 0$$4 - 10m + 4m^2 + 4m - 2m^2 = 0$$
2m^2 - 6m + 4 = 0 implies m^2 - 3m + 2 = 0$$2m^2 - 6m + 4 = 0 \implies m^2 - 3m + 2 = 0$$
Thus, m = 1, 2$m = 1, 2$, which gives alpha = 1, beta = 2$\alpha = 1, \beta = 2$.
### Step 2: Calculate Sigma Expression
sum_n=1^10 (n^1 + n^2) = sum_n=1^10 n + sum_n=1^10 n^2$$\sum_{n=1}^{10} (n^1 + n^2) = \sum_{n=1}^{10} n + \sum_{n=1}^{10} n^2$$
= frac10(11)2 + frac10(11)(21)6 = 55 + 385 = 440$$= \frac{10(11)}{2} + \frac{10(11)(21)}{6} = 55 + 385 = 440$$
### Pattern Recognition
When infinitely many solutions are required, solving the determinant created by replacing one column with the constant vector provides parameter roots directly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Matrices and Determinants
Q62
jee_main_2025_29_jan_evening
Matrix Multiplication and Powers
Let mathbfA = [a_ij]$\mathbf{A} = [a_{ij}]$ be a matrix of order 3 times 3$3 \times 3$, with a_ij = left(sqrt2right)^i+j$a_{ij} = \left(\sqrt{2}\right)^{i+j}$. If the sum of all the elements in the third row of A^2$A^2$ is alpha + beta sqrt2$\alpha + \beta \sqrt{2}$, alpha, beta in mathbfZ$\alpha, \beta \in \mathbf{Z}$, then alpha + beta$\alpha + \beta$ is equal to
- A. 280$280$
- B. 168$168$
- C. 210$210$
- D. 224$224$
Solution
### Related Formula
Element formula entry rule:
a_ij = (sqrt2)^i+j$$a_{ij} = (\sqrt{2})^{i+j}$$
### Core Logic
Constructing the initial matrix structure from the formula entries:
A = beginbmatrix 2 & 2sqrt2 & 4 \\ 2sqrt2 & 4 & 4sqrt2 \\ 4 & 4sqrt2 & 8 endbmatrix$$A = \begin{bmatrix} 2 & 2\sqrt{2} & 4 \\ 2\sqrt{2} & 4 & 4\sqrt{2} \\ 4 & 4\sqrt{2} & 8 \end{bmatrix}$$
Factoring scalar factor 4 out to ease squaring multiplication lines:
A = 2beginbmatrix 1 & sqrt2 & 2 \\ sqrt2 & 2 & 2sqrt2 \\ 2 & 2sqrt2 & 4 endbmatrix$$A = 2\begin{bmatrix} 1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2\sqrt{2} \\ 2 & 2\sqrt{2} & 4 \end{bmatrix}$$
### Step 1: Calculate Rows of Power Matrix
Squaring matrix A^2$A^2$ matches scalar multipliers:
A^2 = 4 beginbmatrix 1 & sqrt2 & 2 \\ sqrt2 & 2 & 2sqrt2 \\ 2 & 2sqrt2 & 4 endbmatrix beginbmatrix 1 & sqrt2 & 2 \\ sqrt2 & 2 & 2sqrt2 \\ 2 & 2sqrt2 & 4 endbmatrix$$A^2 = 4 \begin{bmatrix} 1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2\sqrt{2} \\ 2 & 2\sqrt{2} & 4 \end{bmatrix} \begin{bmatrix} 1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2\sqrt{2} \\ 2 & 2\sqrt{2} & 4 \end{bmatrix}$$
Extract third row entries explicitly:
textRow 3 = 4 beginbmatrix (2+4+8) & (2sqrt2+4sqrt2+8sqrt2) & (4+8+16) endbmatrix$$\text{Row 3} = 4 \begin{bmatrix} (2+4+8) & (2\sqrt{2}+4\sqrt{2}+8\sqrt{2}) & (4+8+16) \end{bmatrix}$$
= 4 beginbmatrix 14 & 14sqrt2 & 28 endbmatrix$$= 4 \begin{bmatrix} 14 & 14\sqrt{2} & 28 \end{bmatrix}$$
### Step 2: Aggregate Entries
textSum of row elements = 4(14 + 14sqrt2 + 28)$$\text{Sum of row elements} = 4(14 + 14\sqrt{2} + 28)$$
= 4(42 + 14sqrt2) = 168 + 56sqrt2$$= 4(42 + 14\sqrt{2}) = 168 + 56\sqrt{2}$$
Matching structural parameters:
alpha = 168, quad beta = 56$$\alpha = 168, \quad \beta = 56$$
alpha + beta = 168 + 56 = 224$$\alpha + \beta = 168 + 56 = 224$$
### Pattern Recognition
Pull common scaling scalar integers out of matrices before running large multiplications. It eliminates algebraic tracking errors across geometric indices.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Matrices and Determinants
Q71
jee_main_2025_28_jan_morning
Symmetric and Skew Symmetric Matrices
Let M$M$ denote the set of all real matrices of order 3 times 3$3 \times 3$ and let S = \-3, -2, -1, 1, 2\$S = \{-3, -2, -1, 1, 2\}$. Let
S_1 = left\A = left[ a_ij right] in M: A = A^T text and a_ij in S, forall i, j right\$$S_1 = \left\{A = \left[ a_{ij} \right] \in M: A = A^T \text{ and } a_{ij} \in S, \forall i, j \right\}$$
S_2 = left\A = left[ a_ij right] in M: A = -A^T text and a_ij in S, forall i, j right\$$S_2 = \left\{A = \left[ a_{ij} \right] \in M: A = -A^T \text{ and } a_{ij} \in S, \forall i, j \right\}$$
S_3 = left\A = left[ a_ij right] in M: a_11 + a_22 + a_33 = 0 text and a_ij in S, forall i, j right\$$S_3 = \left\{A = \left[ a_{ij} \right] \in M: a_{11} + a_{22} + a_{33} = 0 \text{ and } a_{ij} \in S, \forall i, j \right\}$$
If n(S_1 cup S_2 cup S_3) = 125alpha$n(S_1 \cup S_2 \cup S_3) = 125\alpha$, then alpha$\alpha$ equals.
Numerical Answer. Answer: 1613 to 1613
Solution
### Related Formula
Set Principle of Inclusion-Exclusion:
n(S_1 cup S_2 cup S_3) = n(S_1) + n(S_2) + n(S_3) - n(S_1 cap S_2) - n(S_2 cap S_3) - n(S_1 cap S_3) + n(S_1 cap S_2 cap S_3)$$n(S_1 \cup S_2 \cup S_3) = n(S_1) + n(S_2) + n(S_3) - n(S_1 \cap S_2) - n(S_2 \cap S_3) - n(S_1 \cap S_3) + n(S_1 \cap S_2 \cap S_3)$$
### Core Logic
Let's count each subset based on the 5 elements available in S$S$:
1. For S_1$S_1$ (Symmetric matrices): 6 independent element choices implies n(S_1) = 5^6$\implies n(S_1) = 5^6$.
2. For S_2$S_2$ (Skew-symmetric matrices): Diagonal elements must be 0, but 0 notin S$0 \notin S$, so n(S_2) = 0$n(S_2) = 0$.
Since n(S_2) = 0$n(S_2) = 0$, any intersection term involving S_2$S_2$ also becomes 0.
### Step 1: Calculating Trace Matrix Variations
For S_3$S_3$ (Trace equal to zero conditions):
The condition a_11 + a_22 + a_33 = 0$a_{11} + a_{22} + a_{33} = 0$ over S = \-3, -2, -1, 1, 2\$S = \{-3, -2, -1, 1, 2\}$ has exactly 12 valid tuple combinations. The remaining 6 elements can be chosen freely.
n(S_3) = 12 times 5^6$$n(S_3) = 12 \times 5^6$$
For the intersection n(S_1 cap S_3)$n(S_1 \cap S_3)$:
n(S_1 cap S_3) = 12 times 5^3$$n(S_1 \cap S_3) = 12 \times 5^3$$
### Step 2: Final Inclusion-Exclusion Assembly
n(S_1 cup S_2 cup S_3) = 5^6 + 12 times 5^6 - 12 times 5^3$$n(S_1 \cup S_2 \cup S_3) = 5^6 + 12 \times 5^6 - 12 \times 5^3$$
= 5^3 times [13 times 5^3 - 12] = 125 times 1613$$= 5^3 \times [13 \times 5^3 - 12] = 125 \times 1613$$
Thus, alpha = 1613$\alpha = 1613$.
### Pattern Recognition
Always check if the set contains 0$0$. Missing zero elements in skew-symmetric matrix setups instantly zeros out large blocks of permutations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Matrices and Determinants
