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
Q24
jee_main_2026_21_jan_morning
Cayley-Hamilton Theorem Applications
For some
alpha, beta in mathbbR$\alpha, \beta \in \mathbb{R}$ , let
A = beginbmatrix alpha & 2 \\ 1 & 2 endbmatrix$A = \begin{bmatrix} \alpha & 2 \\ 1 & 2 \end{bmatrix}$ and
B = beginbmatrix 1 & 1 \\ 1 & beta endbmatrix$B = \begin{bmatrix} 1 & 1 \\ 1 & \beta \end{bmatrix}$ be such that
A^2 - 4A + 2I = B^2 - 3B + I = O$A^2 - 4A + 2I = B^2 - 3B + I = O$ . Then
(det (operatornameadj(A^3 - B^3)))^2$(\det (\operatorname{adj}(A^3 - B^3)))^2$ is equal to ....
Numerical Answer. Answer: 225 to 225
Solution
### Related Formula
For a 2 times 2$2 \times 2$ matrix M$M$, Cayley-Hamilton equation states:
M^2 - operatornameTr(M)M + operatornamedet(M)I = O$$M^2 - \operatorname{Tr}(M)M + \operatorname{det}(M)I = O$$
Also, |operatornameadj(M)| = |M|^n-1$|\operatorname{adj}(M)| = |M|^{n-1}$, and for a 2 times 2$2 \times 2$ matrix, |operatornameadj(M)| = |M|$|\operatorname{adj}(M)| = |M|$.
### Core Logic
Using the characteristic equation A^2 - operatornameTr(A)A + det(A)I = O$A^2 - \operatorname{Tr}(A)A + \det(A)I = O$:
Comparing with A^2 - 4A + 2I = O$A^2 - 4A + 2I = O$:
operatornameTr(A) = 4 Rightarrow alpha + 2 = 4 Rightarrow alpha = 2$\operatorname{Tr}(A) = 4 \Rightarrow \alpha + 2 = 4 \Rightarrow \alpha = 2$
Comparing with B^2 - 3B + I = O$B^2 - 3B + I = O$:
operatornameTr(B) = 3 Rightarrow 1 + beta = 3 Rightarrow beta = 2$\operatorname{Tr}(B) = 3 \Rightarrow 1 + \beta = 3 \Rightarrow \beta = 2$
### Step 1: Compute A^3 and B^3 via reduction
A = beginbmatrix 2 & 2 \\ 1 & 2 endbmatrix$$A = \begin{bmatrix} 2 & 2 \\ 1 & 2 \end{bmatrix}$$
A^2 = 4A - 2I$A^2 = 4A - 2I$
A^3 = A(4A - 2I) = 4A^2 - 2A = 4(4A - 2I) - 2A = 14A - 8I$$A^3 = A(4A - 2I) = 4A^2 - 2A = 4(4A - 2I) - 2A = 14A - 8I$$
A^3 = 14 beginbmatrix 2 & 2 \\ 1 & 2 endbmatrix - 8 beginbmatrix 1 & 0 \\ 0 & 1 endbmatrix = beginbmatrix 28 & 28 \\ 14 & 28 endbmatrix - beginbmatrix 8 & 0 \\ 0 & 8 endbmatrix = beginbmatrix 20 & 28 \\ 14 & 20 endbmatrix$$A^3 = 14 \begin{bmatrix} 2 & 2 \\ 1 & 2 \end{bmatrix} - 8 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 28 & 28 \\ 14 & 28 \end{bmatrix} - \begin{bmatrix} 8 & 0 \\ 0 & 8 \end{bmatrix} = \begin{bmatrix} 20 & 28 \\ 14 & 20 \end{bmatrix}$$
Similarly, B = beginbmatrix 1 & 1 \\ 1 & 2 endbmatrix$B = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}$
B^2 = 3B - I$B^2 = 3B - I$
B^3 = 3B^2 - B = 3(3B - I) - B = 8B - 3I$$B^3 = 3B^2 - B = 3(3B - I) - B = 8B - 3I$$
B^3 = 8 beginbmatrix 1 & 1 \\ 1 & 2 endbmatrix - beginbmatrix 3 & 0 \\ 0 & 3 endbmatrix = beginbmatrix 8 & 8 \\ 8 & 16 endbmatrix - beginbmatrix 3 & 0 \\ 0 & 3 endbmatrix = beginbmatrix 5 & 8 \\ 8 & 13 endbmatrix$$B^3 = 8 \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 8 & 8 \\ 8 & 16 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix}$$
### Step 2: Difference and Determinant
A^3 - B^3 = beginbmatrix 20 & 28 \\ 14 & 20 endbmatrix - beginbmatrix 5 & 8 \\ 8 & 13 endbmatrix = beginbmatrix 15 & 20 \\ 6 & 7 endbmatrix$$A^3 - B^3 = \begin{bmatrix} 20 & 28 \\ 14 & 20 \end{bmatrix} - \begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix} = \begin{bmatrix} 15 & 20 \\ 6 & 7 \end{bmatrix}$$
det(A^3 - B^3) = (15 times 7) - (20 times 6) = 105 - 120 = -15$$\det(A^3 - B^3) = (15 \times 7) - (20 \times 6) = 105 - 120 = -15$$
### Step 3: Final Answer
For a 2 times 2$2 \times 2$ matrix, |operatornameadj(M)| = |M|$|\operatorname{adj}(M)| = |M|$.
Thus, det(operatornameadj(A^3 - B^3)) = -15$\det(\operatorname{adj}(A^3 - B^3)) = -15$.
(det(operatornameadj(A^3 - B^3)))^2 = (-15)^2 = 225$$(\det(\operatorname{adj}(A^3 - B^3)))^2 = (-15)^2 = 225$$
### Pattern Recognition
Do not manually multiply matrices to the 3rd power. Cayley-Hamilton strictly reduces M^3$M^3$ down to a linear combination cM + dI$cM + dI$. Expanding this requires only basic scalar arithmetic.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Matrices
Q58
jee_main_2025_02_april_evening
System of Linear Equations
If the system of equations
beginaligned 2x + lambda y + 3z &= 5 \\ 3x + 2y - z &= 7 \\ 4x + 5y + mu z &= 9 endaligned$$\begin{aligned} 2x + \lambda y + 3z &= 5 \\ 3x + 2y - z &= 7 \\ 4x + 5y + \mu z &= 9 \end{aligned}$$
has infinitely many solutions, then (lambda^2 + mu^2)$(\lambda^2 + \mu^2)$ is equal to:
Solution
### Related Formula
textFor infinitely many solutions: Delta = 0 quad textand quad Delta_i = 0$$\text{For infinitely many solutions: } \Delta = 0 \quad \text{and} \quad \Delta_i = 0$$
### Core Logic
For a system of 3 linear equations to have infinitely many solutions, the determinant of coefficients and all Cramer determinants must equal zero.
### Step 1: Set up determinant equations
The determinant of coefficients is:
Delta = beginvmatrix 2 & lambda & 3 \\ 3 & 2 & -1 \\ 4 & 5 & mu endvmatrix = 0$$\Delta = \begin{vmatrix} 2 & \lambda & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \mu \end{vmatrix} = 0$$
2(2mu + 5) - lambda(3mu + 4) + 3(15 - 8) = 0$$2(2\mu + 5) - \lambda(3\mu + 4) + 3(15 - 8) = 0$$
4mu + 10 - 3lambdamu - 4lambda + 21 = 0$$4\mu + 10 - 3\lambda\mu - 4\lambda + 21 = 0$$
4mu - 3lambdamu - 4lambda + 31 = 0 quad text--- (1)$$4\mu - 3\lambda\mu - 4\lambda + 31 = 0 \quad \text{--- (1)}$$
Now, set Delta_3 = 0$\Delta_3 = 0$:
Delta_3 = beginvmatrix 2 & lambda & 5 \\ 3 & 2 & 7 \\ 4 & 5 & 9 endvmatrix = 0$$\Delta_3 = \begin{vmatrix} 2 & \lambda & 5 \\ 3 & 2 & 7 \\ 4 & 5 & 9 \end{vmatrix} = 0$$
2(18 - 35) - lambda(27 - 28) + 5(15 - 8) = 0$$2(18 - 35) - \lambda(27 - 28) + 5(15 - 8) = 0$$
-34 + lambda + 35 = 0 implies lambda = -1$$-34 + \lambda + 35 = 0 \implies \lambda = -1$$
### Step 2: Solve for mu and compute the sum of squares
Substitute lambda = -1$\lambda = -1$ into equation (1):
4mu - 3(-1)mu - 4(-1) + 31 = 0$$4\mu - 3(-1)\mu - 4(-1) + 31 = 0$$
4mu + 3mu + 4 + 31 = 0$$4\mu + 3\mu + 4 + 31 = 0$$
7mu = -35 implies mu = -5$$7\mu = -35 \implies \mu = -5$$
Now calculate the sum of squares:
lambda^2 + mu^2 = (-1)^2 + (-5)^2 = 1 + 25 = 26$$\lambda^2 + \mu^2 = (-1)^2 + (-5)^2 = 1 + 25 = 26$$
### Pattern Recognition
Whenever you need to solve for two variables in Cramer's theorem, identifying which determinant lacks the complex variable (like Delta_3$\Delta_3$ which lacks mu$\mu$) is the fastest way to solve for one variable independently.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Matrices and Determinants
Q70
jee_main_2025_02_april_evening
Properties of Matrices
Let A be a 3 times 3$3 \times 3$ real matrix such that mathrmA^2 (mathrmA - 2mathrmI) - 4(mathrmA - mathrmI) = mathrmO$\mathrm{A}^2 (\mathrm{A} - 2\mathrm{I}) - 4(\mathrm{A} - \mathrm{I}) = \mathrm{O}$, where I and O are the identity and null matrices, respectively. If mathrmA^5 = alpha mathrmA^2 + beta mathrmA + gamma mathrmI$\mathrm{A}^5 = \alpha \mathrm{A}^2 + \beta \mathrm{A} + \gamma \mathrm{I}$, where alpha, beta$\alpha, \beta$ and gamma$\gamma$ are real constants, then alpha + beta + gamma$\alpha + \beta + \gamma$ is equal to:
Solution
### Related Formula
textCharacteristic equation reduction: mathrmA^3 = 2mathrmA^2 + 4mathrmA - 4mathrmI$$\text{Characteristic equation reduction: } \mathrm{A}^3 = 2\mathrm{A}^2 + 4\mathrm{A} - 4\mathrm{I}$$
### Core Logic
We use the given cubic matrix equation recursively to express the fifth power of matrix A$A$ solely in terms of quadratic and linear terms.
### Step 1: Simplify the cubic matrix equation
The given equation is:
mathrmA^2 (mathrmA - 2mathrmI) - 4(mathrmA - mathrmI) = mathrmO$$\mathrm{A}^2 (\mathrm{A} - 2\mathrm{I}) - 4(\mathrm{A} - \mathrm{I}) = \mathrm{O}$$
mathrmA^3 - 2mathrmA^2 - 4mathrmA + 4mathrmI = mathrmO implies mathrmA^3 = 2mathrmA^2 + 4mathrmA - 4mathrmI$$\mathrm{A}^3 - 2\mathrm{A}^2 - 4\mathrm{A} + 4\mathrm{I} = \mathrm{O} \implies \mathrm{A}^3 = 2\mathrm{A}^2 + 4\mathrm{A} - 4\mathrm{I}$$
Multiply by matrix A$A$ to find the fourth power:
mathrmA^4 = 2mathrmA^3 + 4mathrmA^2 - 4mathrmA$$\mathrm{A}^4 = 2\mathrm{A}^3 + 4\mathrm{A}^2 - 4\mathrm{A}$$
### Step 2: Reduce the fourth power term
Substitute the expression for mathrmA^3$\mathrm{A}^3$ into our formula for mathrmA^4$\mathrm{A}^4$:
mathrmA^4 = 2left( 2mathrmA^2 + 4mathrmA - 4mathrmI right) + 4mathrmA^2 - 4mathrmA$$\mathrm{A}^4 = 2\left( 2\mathrm{A}^2 + 4\mathrm{A} - 4\mathrm{I} \right) + 4\mathrm{A}^2 - 4\mathrm{A}$$
mathrmA^4 = 4mathrmA^2 + 8mathrmA - 8mathrmI + 4mathrmA^2 - 4mathrmA = 8mathrmA^2 + 4mathrmA - 8mathrmI$$\mathrm{A}^4 = 4\mathrm{A}^2 + 8\mathrm{A} - 8\mathrm{I} + 4\mathrm{A}^2 - 4\mathrm{A} = 8\mathrm{A}^2 + 4\mathrm{A} - 8\mathrm{I}$$
Multiply by matrix A$A$ to find the fifth power:
mathrmA^5 = 8mathrmA^3 + 4mathrmA^2 - 8mathrmA$$\mathrm{A}^5 = 8\mathrm{A}^3 + 4\mathrm{A}^2 - 8\mathrm{A}$$
### Step 3: Reduce the fifth power term and solve
Substitute the expression for mathrmA^3$\mathrm{A}^3$ again:
mathrmA^5 = 8left( 2mathrmA^2 + 4mathrmA - 4mathrmI right) + 4mathrmA^2 - 8mathrmA$$\mathrm{A}^5 = 8\left( 2\mathrm{A}^2 + 4\mathrm{A} - 4\mathrm{I} \right) + 4\mathrm{A}^2 - 8\mathrm{A}$$
mathrmA^5 = 16mathrmA^2 + 32mathrmA - 32mathrmI + 4mathrmA^2 - 8mathrmA = 20mathrmA^2 + 24mathrmA - 32mathrmI$$\mathrm{A}^5 = 16\mathrm{A}^2 + 32\mathrm{A} - 32\mathrm{I} + 4\mathrm{A}^2 - 8\mathrm{A} = 20\mathrm{A}^2 + 24\mathrm{A} - 32\mathrm{I}$$
Comparing this with mathrmA^5 = alpha mathrmA^2 + beta mathrmA + gamma mathrmI$\mathrm{A}^5 = \alpha \mathrm{A}^2 + \beta \mathrm{A} + \gamma \mathrm{I}$, we find:
- alpha = 20$\alpha = 20$
- beta = 24$\beta = 24$
- gamma = -32$\gamma = -32$
Sum the coefficients:
alpha + beta + gamma = 20 + 24 - 32 = 12$$\alpha + \beta + \gamma = 20 + 24 - 32 = 12$$
### Pattern Recognition
Cayley-Hamilton reduction: For any polynomial equation of a matrix, higher powers A^k$A^k$ can always be reduced down to polynomials of order less than the degree of the characteristic equation by recursive substitution.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Matrices and Determinants
Q
jee_main_2025_02_april_morning
Idempotent Matrices
Let A = beginbmatrix alpha & -1 \\ 6 & beta endbmatrix$A = \begin{bmatrix} \alpha & -1 \\ 6 & \beta \end{bmatrix}$, alpha > 0$\alpha > 0$, such that det(A) = 0$\det(A) = 0$ and alpha + beta = 1$\alpha + \beta = 1$. If I$I$ denotes the 2 times 2$2 \times 2$ identity matrix, then the matrix (I + A)^8$(I + A)^8$ is:
- A. beginbmatrix 4 & -1 \\ 6 & -1 endbmatrix$\begin{bmatrix} 4 & -1 \\ 6 & -1 \end{bmatrix}$
- B. beginbmatrix 257 & -64 \\ 514 & -127 endbmatrix$\begin{bmatrix} 257 & -64 \\ 514 & -127 \end{bmatrix}$
- C. beginbmatrix 1025 & -511 \\ 2024 & -1024 endbmatrix$\begin{bmatrix} 1025 & -511 \\ 2024 & -1024 \end{bmatrix}$
- D. beginbmatrix 766 & -255 \\ 1530 & -509 endbmatrix$\begin{bmatrix} 766 & -255 \\ 1530 & -509 \end{bmatrix}$
Solution
### Related Formula
For a matrix satisfying A^2 = A$A^2 = A$ (Idempotent Matrix):
(I+A)^n = I + (2^n - 1)A$$(I+A)^n = I + (2^n - 1)A$$
### Core Logic
Given det(A) = alphabeta + 6 = 0 implies alphabeta = -6$\det(A) = \alpha\beta + 6 = 0 \implies \alpha\beta = -6$ and alpha + beta = 1$\alpha + \beta = 1$. Solving these gives alpha = 3, beta = -2$\alpha = 3, \beta = -2$ (since alpha > 0$\alpha > 0$).
### Step 1: Check Powers of A
Substitute values into A$A$:
A = beginbmatrix 3 & -1 \\ 6 & -2 endbmatrix$$A = \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix}$$
Compute A^2$A^2$:
A^2 = beginbmatrix 3 & -1 \\ 6 & -2 endbmatrix beginbmatrix 3 & -1 \\ 6 & -2 endbmatrix = beginbmatrix 9-6 & -3+2 \\ 18-12 & -6+4 endbmatrix = beginbmatrix 3 & -1 \\ 6 & -2 endbmatrix = A$$A^2 = \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix} \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 9-6 & -3+2 \\ 18-12 & -6+4 \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix} = A$$
### Step 2: Expand Matrix Expression
Since A^2 = A$A^2 = A$, it follows that A^n = A$A^n = A$ for all integers n ge 1$n \ge 1$.
(I + A)^8 = I + sum_k=1^8 binom8k A^k = I + A sum_k=1^8 binom8k = I + (2^8 - 1)A = I + 255A$$(I + A)^8 = I + \sum_{k=1}^{8} \binom{8}{k} A^k = I + A \sum_{k=1}^{8} \binom{8}{k} = I + (2^8 - 1)A = I + 255A$$
### Step 3: Construct the Final Matrix
(I + A)^8 = beginbmatrix 1 & 0 \\ 0 & 1 endbmatrix + 255 beginbmatrix 3 & -1 \\ 6 & -2 endbmatrix = beginbmatrix 1 + 765 & -255 \\ 1530 & 1 - 510 endbmatrix = beginbmatrix 766 & -255 \\ 1530 & -509 endbmatrix$$(I + A)^8 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + 255 \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 1 + 765 & -255 \\ 1530 & 1 - 510 \end{bmatrix} = \begin{bmatrix} 766 & -255 \\ 1530 & -509 \end{bmatrix}$$
### Pattern Recognition
Whenever texttr(A) = 1$\text{tr}(A) = 1$ and det(A) = 0$\det(A) = 0$ for a 2 times 2$2 \times 2$ matrix, Cayley-Hamilton theorem gives A^2 - texttr(A)A + det(A)I = 0 implies A^2 = A$A^2 - \text{tr}(A)A + \det(A)I = 0 \implies A^2 = A$. Thus A$A$ is idempotent, simplifying polynomial expansions exponentially.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Matrices and Determinants
Q
jee_main_2025_02_april_morning
System of Linear Equations
If the system of linear equations
3x + y + beta z = 3$$3x + y + \beta z = 3$$
2x + alpha y - z = -3$$2x + \alpha y - z = -3$$
x + 2y + z = 4$x + 2y + z = 4$
has infinitely many solutions, then the value of 22beta - 9alpha$22\beta - 9\alpha$ is:
- A. 49$49$
- B. 31$31$
- C. 43$43$
- D. 37$37$
Solution
### Related Formula
Cramer's Rule for infinite solutions specifies that the main determinant and all component determinants must vanish:
Delta = 0 quad textand quad Delta_1 = Delta_2 = Delta_3 = 0$$\Delta = 0 \quad \text{and} \quad \Delta_1 = \Delta_2 = \Delta_3 = 0$$
### Core Logic
Set the key system determinants to zero to form equations linking alpha$\alpha$ and beta$\beta$, then isolate the constants.
### Step 1: Set Main Determinant to Zero
Delta = beginvmatrix 3 & 1 & beta \\ 2 & alpha & -1 \\ 1 & 2 & 1 endvmatrix = 0$$\Delta = \begin{vmatrix} 3 & 1 & \beta \\ 2 & \alpha & -1 \\ 1 & 2 & 1 \end{vmatrix} = 0$$
Expand along the first row:
3(alpha + 2) - 1(2 + 1) + beta(4 - alpha) = 0$$3(\alpha + 2) - 1(2 + 1) + \beta(4 - \alpha) = 0$$
3alpha + 6 - 3 + 4beta - alphabeta = 0 implies 3alpha + 4beta - alphabeta + 3 = 0 quad dots (1)$$3\alpha + 6 - 3 + 4\beta - \alpha\beta = 0 \implies 3\alpha + 4\beta - \alpha\beta + 3 = 0 \quad \dots (1)$$
### Step 2: Set Subsidiary Determinant to Zero
Using Delta_3 = 0$\Delta_3 = 0$ by substituting the constants vector into the third column:
Delta_3 = beginvmatrix 3 & 1 & 3 \\ 2 & alpha & -3 \\ 1 & 2 & 4 endvmatrix = 0$$\Delta_3 = \begin{vmatrix} 3 & 1 & 3 \\ 2 & \alpha & -3 \\ 1 & 2 & 4 \end{vmatrix} = 0$$
Expand along the first row:
3(4alpha + 6) - 1(8 + 3) + 3(4 - alpha) = 0$$3(4\alpha + 6) - 1(8 + 3) + 3(4 - \alpha) = 0$$
12alpha + 18 - 11 + 12 - 3alpha = 0 implies 9alpha + 19 = 0 implies alpha = -frac199$$12\alpha + 18 - 11 + 12 - 3\alpha = 0 \implies 9\alpha + 19 = 0 \implies \alpha = -\frac{19}{9}$$
### Step 3: Solve for Beta and Final Expression
Substitute alpha = -frac199$\alpha = -\frac{19}{9}$ into equation (1):
3left(-frac199right) + 4beta - left(-frac199right)beta + 3 = 0$$3\left(-\frac{19}{9}\right) + 4\beta - \left(-\frac{19}{9}\right)\beta + 3 = 0$$
-frac193 + 3 + betaleft(4 + frac199right) = 0 implies -frac103 + betaleft(frac559right) = 0$$-\frac{19}{3} + 3 + \beta\left(4 + \frac{19}{9}\right) = 0 \implies -\frac{10}{3} + \beta\left(\frac{55}{9}\right) = 0$$
frac559beta = frac103 implies beta = frac103 cdot frac955 = frac611$$\frac{55}{9}\beta = \frac{10}{3} \implies \beta = \frac{10}{3} \cdot \frac{9}{55} = \frac{6}{11}$$
Now compute 22beta - 9alpha$22\beta - 9\alpha$:
22left(frac611right) - 9left(-frac199
ight) = 12 + 19 = 31$$22\left(\frac{6}{11}\right) - 9\left(-\frac{19}{9}
ight) = 12 + 19 = 31$$
### Pattern Recognition
Choosing Delta_3$\Delta_3$ over Delta_1$\Delta_1$ or Delta_2$\Delta_2$ eliminates beta$\beta$ entirely because the variable parameters are localized in specific positions. This yields alpha$\alpha$ directly without requiring a coupled system solution.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Matrices and Determinants
