Solution & Explanation
### Related Formula
Delta = beginvmatrix a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 endvmatrix$\Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$
If Delta neq 0$\Delta \neq 0$, unique solution.
If Delta = 0$\Delta = 0$ and Delta_x = Delta_y = Delta_z = 0$\Delta_x = \Delta_y = \Delta_z = 0$, infinitely many solutions.
If Delta = 0$\Delta = 0$ and at least one of Delta_x, Delta_y, Delta_z neq 0$\Delta_x, \Delta_y, \Delta_z \neq 0$, inconsistent (no solution).
### Core Logic
Given system:
x + y + z = 4mu$x + y + z = 4\mu$
x + 2y + 2lambda z = 10mu$x + 2y + 2\lambda z = 10\mu$
x + 3y + 4lambda^2 z = mu^2 + 15$x + 3y + 4\lambda^2 z = \mu^2 + 15$
Compute the main determinant Delta$\Delta$:
Delta = beginvmatrix 1 & 1 & 1 \\ 1 & 2 & 2lambda \\ 1 & 3 & 4lambda^2 endvmatrix$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 2\lambda \\ 1 & 3 & 4\lambda^2 \end{vmatrix}$
Apply operations: R_2 to R_2 - R_1$R_2 \to R_2 - R_1$, R_3 to R_3 - R_1$R_3 \to R_3 - R_1$
Delta = beginvmatrix 1 & 1 & 1 \\ 0 & 1 & 2lambda-1 \\ 0 & 2 & 4lambda^2-1 endvmatrix = 1 cdot (4lambda^2 - 1 - 2(2lambda - 1))$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 0 & 1 & 2\lambda-1 \\ 0 & 2 & 4\lambda^2-1 \end{vmatrix} = 1 \cdot (4\lambda^2 - 1 - 2(2\lambda - 1))$
= 4lambda^2 - 1 - 4lambda + 2 = 4lambda^2 - 4lambda + 1 = (2lambda - 1)^2$= 4\lambda^2 - 1 - 4\lambda + 2 = 4\lambda^2 - 4\lambda + 1 = (2\lambda - 1)^2$
### Step 1: Analyzing Unique Solution
For a unique solution, Delta neq 0 Rightarrow 2lambda - 1 neq 0 Rightarrow lambda neq frac12$\Delta \neq 0 \Rightarrow 2\lambda - 1 \neq 0 \Rightarrow \lambda \neq \frac{1}{2}$.
Note: For unique solution, mu$\mu$ can be anything. Option (4) states the system is consistent if lambda neq frac12$\lambda \neq \frac{1}{2}$, which is purely correct. Option (1) says "unique solution if lambda neq frac12$\lambda \neq \frac{1}{2}$ and mu neq 1, 15$\mu \neq 1, 15$". While true that it has a unique solution under those conditions, it also has a unique solution for mu = 1, 15$\mu = 1, 15$. Let's check consistency conditions.
### Step 2: Checking Delta components
Let Delta = 0$\Delta = 0$, so lambda = frac12$\lambda = \frac{1}{2}$.
Compute Delta_x$\Delta_x$ and Delta_z$\Delta_z$ (or Delta_y$\Delta_y$):
Wait, substituting lambda = 1/2$\lambda = 1/2$, the equations become:
x + y + z = 4mu$x + y + z = 4\mu$
x + 2y + z = 10mu$x + 2y + z = 10\mu$
x + 3y + z = mu^2 + 15$x + 3y + z = \mu^2 + 15$
From the first two, (x+2y+z) - (x+y+z) = 10mu - 4mu Rightarrow y = 6mu$(x+2y+z) - (x+y+z) = 10\mu - 4\mu \Rightarrow y = 6\mu$.
From the second and third, (x+3y+z) - (x+2y+z) = mu^2 + 15 - 10mu Rightarrow y = mu^2 - 10mu + 15$(x+3y+z) - (x+2y+z) = \mu^2 + 15 - 10\mu \Rightarrow y = \mu^2 - 10\mu + 15$.
For the system to be consistent (infinite solutions since Delta = 0$\Delta = 0$), the two values of y$y$ must match:
6mu = mu^2 - 10mu + 15$6\mu = \mu^2 - 10\mu + 15$
mu^2 - 16mu + 15 = 0$\mu^2 - 16\mu + 15 = 0$
(mu - 1)(mu - 15) = 0$(\mu - 1)(\mu - 15) = 0$
So, if lambda = frac12$\lambda = \frac{1}{2}$, the system is consistent (infinite solutions) ONLY when mu = 1$\mu = 1$ or mu = 15$\mu = 15$.
If lambda = frac12$\lambda = \frac{1}{2}$ and mu neq 1, 15$\mu \neq 1, 15$, it is inconsistent.
### Step 3: Checking Options
Option (2) states: "The system is inconsistent if lambda = frac12$\lambda = \frac{1}{2}$ and mu neq 1$\mu \neq 1$".
If mu = 15$\mu = 15$ (which is neq 1$\neq 1$), the system is actually CONSISTENT (infinite solutions). Therefore, Option (2) is NOT strictly correct because mu=15$\mu=15$ makes it consistent.
Thus, statement (2) is the incorrect statement.
### Pattern Recognition
Cramer's rule dependencies can be quickly identified using algebraic elimination. When variables align symmetrically (like z$z$ mapping identically), subtracting equations exposes the consistency constraint directly without resolving full 3times3$3\times3$ determinants.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Determinants
More Determinants Previous-Year Questions — Page 5
Q65
jee_main_2025_24_jan_evening
Evaluation of Determinants using Limits
For some a, b, let
f(x)=beginvmatrix a+fracsin xx & 1 & b \\ a & 1+fracsin xx & b \\ a & 1 & b+fracsin xx endvmatrix, xne0$f(x)=\begin{vmatrix} a+\frac{\sin x}{x} & 1 & b \\ a & 1+\frac{\sin x}{x} & b \\ a & 1 & b+\frac{\sin x}{x} \end{vmatrix}, x\ne0$ [cite: 3370, 3371, 3372, 3374, 3375]
lim_xrightarrow0f(x)=lambda+mu a+vb$\lim_{x\rightarrow0}f(x)=\lambda+\mu a+vb$ [cite: 3376, 3377]
Then (lambda+mu+nu)^2$(\lambda+\mu+\nu)^{2}$ is equal to:
- A. 25$25$
- B. 9$9$
- C. 36$36$
- D. 16$16$
Solution
### Related Formula
The fundamental trigonometric limit is given by:
lim_x to 0 fracsin xx = 1$\lim_{x \to 0} \frac{\sin x}{x} = 1$
### Core Logic
Apply the limit inside each element of the matrix determinant :
lim_x to 0 f(x) = beginvmatrix a+1 & 1 & b \\ a & 2 & b \\ a & 1 & b+1 endvmatrix$\lim_{x \to 0} f(x) = \begin{\vmatrix} a+1 & 1 & b \\ a & 2 & b \\ a & 1 & b+1 \end{\vmatrix}$
### Step 1: Simplify the Determinant via Row Operations
Perform rows reductions R_2 to R_2 - R_1$R_2 \to R_2 - R_1$ and R_3 to R_3 - R_1$R_3 \to R_3 - R_1$ to create zeros:
lim_x to 0 f(x) = beginvmatrix a+1 & 1 & b \\ -1 & 1 & 0 \\ -1 & 0 & 1 endvmatrix$\lim_{x \to 0} f(x) = \begin{\vmatrix} a+1 & 1 & b \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{\vmatrix}$
Expand across the first row :
= (a+1)[1(1) - 0] - 1[-1(1) - 0] + b[0 - (-1)]$= (a+1)[1(1) - 0] - 1[-1(1) - 0] + b[0 - (-1)]$
= (a+1)(1) + 1 + b = a + b + 2$= (a+1)(1) + 1 + b = a + b + 2$ [cite: 3999, 4000]
### Step 2: Match Coefficients
Equate this outcome with the target parameter template lambda + mu a + nu b$\lambda + \mu a + \nu b$ [cite: 3377, 4000]:
lambda = 2, quad mu = 1, quad nu = 1$\lambda = 2, \quad \mu = 1, \quad \nu = 1$
Calculate (lambda + mu + nu)^2$(\lambda + \mu + \nu)^2$ [cite: 3378, 4001]:
(2 + 1 + 1)^2 = 4^2 = 16$(2 + 1 + 1)^2 = 4^2 = 16$
### Pattern Recognition
Standard row manipulations on identity-shifted arrays quickly eliminate complex parameter symbols, reducing determinant calculations into straightforward polynomial expansions.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Matrices and Determinants
Class 11 Mathematics: Limits and Derivatives
Q70
jee_main_2025_24_jan_morning
System of Linear Equations
If the system of equations
2x - y + z = 4$2x - y + z = 4$
5x + lambda y + 3z = 12$5x + \lambda y + 3z = 12$
100x - 47y + mu z = 212$100x - 47y + \mu z = 212$
has infinitely many solutions, then mu - 2lambda$\mu - 2\lambda$ is equal to :
- A. 56$56$
- B. 59$59$
- C. 55$55$
- D. 57$57$
Solution
### Related Formula
According to Cramer's Rule, for a system of linear equations to have infinitely many solutions, the main determinant Delta$\Delta$ and all directional determinants Delta_1, Delta_2, Delta_3$\Delta_1, \Delta_2, \Delta_3$ must equal zero simultaneously.
### Core Logic
Set up the directional determinant equation Delta_3 = 0$\Delta_3 = 0$ by replacing the third column with the constant vector:
Delta_3 = left| beginmatrix 2 & -1 & 4 \\ 5 & lambda & 12 \\ 100 & -47 & 212 endmatrix right| = 0$\Delta_3 = \left| \begin{matrix} 2 & -1 & 4 \\ 5 & \lambda & 12 \\ 100 & -47 & 212 \end{matrix} \right| = 0$
Expand the determinant along the first row:
2[212lambda - 12(-47)] - (-1)[5(212) - 12(100)] + 4[5(-47) - 100lambda] = 0$2[212\lambda - 12(-47)] - (-1)[5(212) - 12(100)] + 4[5(-47) - 100\lambda] = 0$
2[212lambda + 564] + 1[1060 - 1200] + 4[-235 - 100lambda] = 0$2[212\lambda + 564] + 1[1060 - 1200] + 4[-235 - 100\lambda] = 0$
424lambda + 1128 - 140 - 940 - 400lambda = 0$424\lambda + 1128 - 140 - 940 - 400\lambda = 0$
24lambda + 48 = 0 implies lambda = -2$24\lambda + 48 = 0 \implies \lambda = -2$
### Step 1: Solve for Mu using the main determinant
Set the primary coefficient matrix determinant Delta = 0$\Delta = 0$ and substitute lambda = -2$\lambda = -2$:
Delta = left| beginmatrix 2 & -1 & 1 \\ 5 & -2 & 3 \\ 100 & -47 & mu endmatrix right| = 0$\Delta = \left| \begin{matrix} 2 & -1 & 1 \\ 5 & -2 & 3 \\ 100 & -47 & \mu \end{matrix} \right| = 0$
Expand the determinant along the first row:
2[-2mu - 3(-47)] - (-1)[5mu - 3(100)] + 1[5(-47) - (-2)(100)] = 0$2[-2\mu - 3(-47)] - (-1)[5\mu - 3(100)] + 1[5(-47) - (-2)(100)] = 0$
2[-2mu + 141] + [5mu - 300] + [-235 + 200] = 0$2[-2\mu + 141] + [5\mu - 300] + [-235 + 200] = 0$
-4mu + 282 + 5mu - 300 - 35 = 0$-4\mu + 282 + 5\mu - 300 - 35 = 0$
mu - 53 = 0 implies mu = 53$\mu - 53 = 0 \implies \mu = 53$
### Step 2: Calculate the Target Value
Substitute the values of mu$\mu$ and lambda$\lambda$ into the expression:
mu - 2lambda = 53 - 2(-2) = 53 + 4 = 57$\mu - 2\lambda = 53 - 2(-2) = 53 + 4 = 57$
### Pattern Recognition
When solving systems of equations for infinite solution parameters, choosing a directional determinant that excludes one of the variables simplifies the problem into two separate single-variable equations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Matrices and Determinants
Q74
jee_main_2025_24_jan_morning
Properties of Matrices and Adjoints
Let A$A$ be a 3 times 3$3 \times 3$ matrix such that mathbfX^mathrmTmathbfAmathbfX = mathbf0$\mathbf{X}^{\mathrm{T}}\mathbf{A}\mathbf{X} = \mathbf{0}$ for all nonzero 3 times 1$3 \times 1$ matrices mathbfX = beginbmatrix x \\ y \\ z endbmatrix$\mathbf{X} = \begin{\bmatrix} x \\ y \\ z \end{\bmatrix}$ . If A beginbmatrix 1\\ 1\\ 1 endbmatrix = beginbmatrix 1\\ 4\\ -5 endbmatrix$A \begin{\bmatrix} 1\\ 1\\ 1 \end{\bmatrix} = \begin{\bmatrix} 1\\ 4\\ -5 \end{\bmatrix}$, Abeginbmatrix 1\\ 2\\ 1 endbmatrix = beginbmatrix 0\\ 4\\ -8 endbmatrix$A\begin{\bmatrix} 1\\ 2\\ 1 \end{\bmatrix} = \begin{\bmatrix} 0\\ 4\\ -8 \end{\bmatrix}$, and det(mathrmadj(2(A + I))) = 2^alpha3^beta5^gamma$\det(\mathrm{adj}(2(A + I))) = 2^{\alpha}3^{\beta}5^{\gamma}$ for alpha, beta, gamma in mathbbN$\alpha, \beta, \gamma \in \mathbb{N}$, then alpha^2 + beta^2 + gamma^2$\alpha^2 + \beta^2 + \gamma^2$ is ________.
Numerical Answer. Answer: 44
Solution
### Related Formula
The quadratic form condition mathbfX^mathrmTmathbfAmathbfX = mathbf0$\mathbf{X}^{\mathrm{T}}\mathbf{A}\mathbf{X} = \mathbf{0}$ holds for all non-zero vectors mathbfX$\mathbf{X}$ if and only if A$A$ is a skew-symmetric matrix. For a 3 times 3$3 \times 3$ skew-symmetric matrix, the components satisfy:
A = beginbmatrix 0 & x_1 & x_2 \\ -x_1 & 0 & x_3 \\ -x_2 & -x_3 & 0 endbmatrix$A = \begin{\bmatrix} 0 & x_1 & x_2 \\ -x_1 & 0 & x_3 \\ -x_2 & -x_3 & 0 \end{\bmatrix}$
### Core Logic
Let's define the matrix A$A$ using the parameters of a standard skew-symmetric form:
A = beginbmatrix 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 endbmatrix$A = \begin{\bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{\bmatrix}$
Apply the first given matrix multiplication vector condition:
A beginbmatrix 1 \\ 1 \\ 1 endbmatrix = beginbmatrix a+b \\ -a+c \\ -b-c endbmatrix = beginbmatrix 1 \\ 4 \\ -5 endbmatrix$A \begin{\bmatrix} 1 \\ 1 \\ 1 \end{\bmatrix} = \begin{\bmatrix} a+b \\ -a+c \\ -b-c \end{\bmatrix} = \begin{\bmatrix} 1 \\ 4 \\ -5 \end{\bmatrix}$
This gives the system of linear equations:
a + b = 1 quad dots (1)$a + b = 1 \quad \dots (1)$
-a + c = 4 quad dots (2)$-a + c = 4 \quad \dots (2)$
-b - c = -5 implies b + c = 5 quad dots (3)$-b - c = -5 \implies b + c = 5 \quad \dots (3)$
Apply the second given matrix multiplication vector condition:
A beginbmatrix 1 \\ 2 \\ 1 endbmatrix = beginbmatrix 2a+b \\ -a+c \\ -b-2c endbmatrix = beginbmatrix 0 \\ 4 \\ -8 endbmatrix$A \begin{\bmatrix} 1 \\ 2 \\ 1 \end{\bmatrix} = \begin{\bmatrix} 2a+b \\ -a+c \\ -b-2c \end{\bmatrix} = \begin{\bmatrix} 0 \\ 4 \\ -8 \end{\bmatrix}$
This gives the equation:
2a + b = 0 quad dots (4)$2a + b = 0 \quad \dots (4)$
### Step 1: Solve for Matrix Elements
Subtract equation (1) from equation (4):
(2a + b) - (a + b) = 0 - 1 implies a = -1$(2a + b) - (a + b) = 0 - 1 \implies a = -1$
Substitute a = -1$a = -1$ back into equation (1):
-1 + b = 1 implies b = 2$-1 + b = 1 \implies b = 2$
Substitute a = -1$a = -1$ into equation (2):
-(-1) + c = 4 implies 1 + c = 4 implies c = 3$-(-1) + c = 4 \implies 1 + c = 4 \implies c = 3$
Thus, the explicit matrix A$A$ is:
A = beginbmatrix 0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0 endbmatrix$A = \begin{\bmatrix} 0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0 \end{\bmatrix}$
### Step 2: Compute Target Matrix Determinant
Construct the modified target matrix 2(A+I)$2(A+I)$:
A + I = beginbmatrix 1 & -1 & 2 \\ 1 & 1 & 3 \\ -2 & -3 & 1 endbmatrix$A + I = \begin{\bmatrix} 1 & -1 & 2 \\ 1 & 1 & 3 \\ -2 & -3 & 1 \end{\bmatrix}$
2(A + I) = beginbmatrix 2 & -2 & 4 \\ 2 & 2 & 6 \\ -4 & -6 & 2 endbmatrix$2(A + I) = \begin{\bmatrix} 2 & -2 & 4 \\ 2 & 2 & 6 \\ -4 & -6 & 2 \end{\bmatrix}$
Calculate its determinant value:
det(2(A+I)) = 2[4 - (-36)] - (-2)[4 - (-24)] + 4[-12 - (-8)]$\det(2(A+I)) = 2[4 - (-36)] - (-2)[4 - (-24)] + 4[-12 - (-8)]$
det(2(A+I)) = 2[40] + 2[28] + 4[-4] = 80 + 56 - 16 = 120$\det(2(A+I)) = 2[40] + 2[28] + 4[-4] = 80 + 56 - 16 = 120$
### Step 3: Analyze Adjoint Power and Prime Factors
Using the standard determinant identity for adjoints, det(mathrmadj(M)) = (det M)^n-1$\det(\mathrm{adj}(M)) = (\det M)^{n-1}$ where n=3$n=3$:
det(mathrmadj(2(A+I))) = (120)^3-1 = 120^2$\det(\mathrm{adj}(2(A+I))) = (120)^{3-1} = 120^2$
Find the prime factorization of the result:
120 = 2^3 cdot 3^1 cdot 5^1 implies 120^2 = (2^3 cdot 3^1 cdot 5^1)^2 = 2^6 cdot 3^2 cdot 5^2$120 = 2^3 \cdot 3^1 \cdot 5^1 \implies 120^2 = (2^3 \cdot 3^1 \cdot 5^1)^2 = 2^6 \cdot 3^2 \cdot 5^2$
This maps the exponents directly to our target variables:
alpha = 6, quad beta = 2, quad gamma = 2$\alpha = 6, \quad \beta = 2, \quad \gamma = 2$
Finally, calculate the \sum of their squares:
alpha^2 + beta^2 + gamma^2 = 6^2 + 2^2 + 2^2 = 36 + 4 + 4 = 44$\alpha^2 + \beta^2 + \gamma^2 = 6^2 + 2^2 + 2^2 = 36 + 4 + 4 = 44$
### Pattern Recognition
The condition mathbfX^T A mathbfX = 0$\mathbf{X}^T A \mathbf{X} = 0$ always implies that A$A$ is a skew-symmetric matrix, which instantly forces the diagonal entries to be zero, reducing the number of unknown parameters from 9 down to 3.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Matrices and Determinants
Q60
jee_main_2025_28_jan_evening
Properties of Matrices and Powers
Let A=left[beginmatrixfrac1sqrt2&-2\\ 0&1endmatrixright]$A=\left[\begin{matrix}\frac{1}{\sqrt{2}}&-2\\ 0&1\end{matrix}\right]$ and P=left[beginmatrixcostheta&-sintheta\\ sintheta&costhetaendmatrixright]$P=\left[\begin{matrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{matrix}\right]$, theta>0$\theta>0$. If B=PAP^T$B=PAP^{T}$, C=P^TB^10P$C=P^{T}B^{10}P$ and the sum of the diagonal elements of C is fracmn$\frac{m}{n}$ where textgcd(m,n)=1$\text{\gcd}(m,n)=1$, then m+n$m+n$ is:
- A. 65$65$
- B. 127$127$
- C. 258$258$
- D. 2049$2049$
Solution
### Related Formula
For orthogonal matrix P$P$, P^T P = P P^T = I$P^T P = P P^T = I$.
If B = PAP^T$B = PAP^T$, then:
B^k = (PAP^T)(PAP^T)dots(PAP^T) = PA^kP^T$B^k = (PAP^T)(PAP^T)\dots(PAP^T) = PA^kP^T$
### Core Logic
Given C = P^T B^10 P$C = P^T B^{10} P$. Substitute B^10 = P A^10 P^T$B^{10} = P A^{10} P^T$ into the expression:
C = P^T (P A^10 P^T) P$C = P^T (P A^{10} P^T) P$
C = (P^T P) A^10 (P^T P)$C = (P^T P) A^{10} (P^T P)$
Since P$P$ is an orthogonal rotation matrix, P^T P = I$P^T P = I$, meaning:
C = I cdot A^10 cdot I = A^10$C = I \cdot A^{10} \cdot I = A^{10}$
Therefore, the sum of diagonal elements of C$C$ is simply the trace of A^10$A^{10}$.
### Step 1: Analyze Powers of Upper Triangular Matrix A
Matrix A$A$ is upper triangular:
A = left[beginmatrixfrac1sqrt2&-2\\ 0&1endmatrixright]$A = \left[\begin{matrix}\frac{1}{\sqrt{2}}&-2\\ 0&1\end{matrix}\right]$
For any upper triangular matrix, any integer power k$k$ preserves the main diagonal entries as simply the powers of the individual diagonal elements:
A^10 = left[beginmatrixleft(frac1sqrt2right)^10&*\\ 0&1^10endmatrixright] = left[beginmatrixfrac132&*\\ 0&1endmatrixright]$A^{10} = \left[\begin{matrix}\left(\frac{1}{\sqrt{2}}\right)^{10}&*\\ 0&1^{10}\end{matrix}\right] = \left[\begin{matrix}\frac{1}{32}&*\\ 0&1\end{matrix}\right]$
### Step 2: Calculate the Trace and sum m+n
textSum of diagonal elements = textTrace(C) = textTrace(A^10) = frac132 + 1 = frac3332$\text{Sum of diagonal elements } = \text{Trace}(C) = \text{Trace}(A^{10}) = \frac{1}{32} + 1 = \frac{33}{32}$
Given fracmn = frac3332$\frac{m}{n} = \frac{33}{32}$ with textgcd(33, 32) = 1$\text{\gcd}(33, 32) = 1$:
m = 33, quad n = 32$m = 33, \quad n = 32$
m + n = 33 + 32 = 65$m + n = 33 + 32 = 65$
### Pattern Recognition
Traces of matrices are invariant under cyclic permutations, so textTr(P^T B^10 P) = textTr(P P^T B^10) = textTr(B^10)$\text{Tr}(P^T B^{10} P) = \text{Tr}(P P^T B^{10}) = \text{Tr}(B^{10})$. Furthermore, textTr(P A^10 P^T) = textTr(A^10)$\text{Tr}(P A^{10} P^T) = \text{Tr}(A^{10})$. This identity bypasses the need to evaluate any outer matrix multiplication.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Matrices and Determinants
Q
jee_main_2025_29_jan_morning
Properties of Determinants
Let M and m respectively be the maximum and the minimum values of
f (x) = left| beginarrayc c c 1 + sin^ 2 x & cos^ 2 x & 4 sin 4 x \\ sin^ 2 x & 1 + cos^ 2 x & 4 sin 4 x \\ sin^ 2 x & cos^ 2 x & 1 + 4 sin 4 x endarray right|, x in R$f (x) = \left| \begin{array}{c c c} 1 + \sin^ {2} x & \cos^ {2} x & 4 \sin 4 x \\ \sin^ {2} x & 1 + \cos^ {2} x & 4 \sin 4 x \\ \sin^ {2} x & \cos^ {2} x & 1 + 4 \sin 4 x \end{array} \right|, x \in R$
Then mathbfM^4 -mathbfm^4$\mathbf{M}^4 -\mathbf{m}^4$ is equal to :
- A. 1280
- B. 1295
- C. 1040
- D. 1215
Solution
### Related Formula
sin^2 x + cos^2 x = 1$\sin^2 x + \cos^2 x = 1$
-1 le sin 4x le 1$-1 \le \sin 4x \le 1$
### Core Logic
Apply the row operations R_2 to R_2 - R_1$R_2 \to R_2 - R_1$ and R_3 to R_3 - R_1$R_3 \to R_3 - R_1$ to simplify the determinant:
f(x) = left| beginarrayc c c 1 + sin^ 2 x & cos^ 2 x & 4 sin 4 x \\ -1 & 1 & 0 \\ -1 & 0 & 1 endarray right|$f(x) = \left| \begin{array}{c c c} 1 + \sin^ {2} x & \cos^ {2} x & 4 \sin 4 x \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array} \right|$
### Step 1: Expand the Determinant
Expanding along the first row:
f(x) = (1 + sin^2 x)(1 - 0) - cos^2 x(-1 - 0) + 4sin 4x(0 - (-1))$f(x) = (1 + \sin^2 x)(1 - 0) - \cos^2 x(-1 - 0) + 4\sin 4x(0 - (-1))$
f(x) = 1 + sin^2 x + cos^2 x + 4sin 4x$f(x) = 1 + \sin^2 x + \cos^2 x + 4\sin 4x$
Since sin^2 x + cos^2 x = 1$\sin^2 x + \cos^2 x = 1$, we get:
f(x) = 2 + 4sin 4x$f(x) = 2 + 4\sin 4x$
### Step 2: Find Maximum and Minimum Values
The range of sin 4x$\sin 4x$ is [-1, 1]$[-1, 1]$.
M = 2 + 4(1) = 6$M = 2 + 4(1) = 6$
m = 2 + 4(-1) = -2$m = 2 + 4(-1) = -2$
### Step 3: Calculate M^4 - m^4$M^4 - m^4$
M^4 - m^4 = 6^4 - (-2)^4 = 1296 - 16 = 1280$M^4 - m^4 = 6^4 - (-2)^4 = 1296 - 16 = 1280$
### Pattern Recognition
Look for repeated structures or cyclic additions in rows. Subtracting rows quickly creates zeros, reducing complex trigonometric matrices into elementary algebraic expressions.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Matrices and Determinants
Class 11 Mathematics: Trigonometric Functions