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 6
Q70
jee_main_2025_29_jan_morning
Cofactors and Determinant Value
Let mathrmA = left[mathbfa_mathrmijright] = beginbmatrix log_5128 & log_45 \\ log_58 & log_425 endbmatrix$\mathrm{A} = \left[\mathbf{a}_{\mathrm{ij}}\right] = \begin{\bmatrix} \log_5128 & \log_45 \\ \log_58 & \log_425 \end{\bmatrix}$ . If A_ij$A_{ij}$ is the cofactor of a_ij$a_{ij}$ , C_ij = sum_k=1^2 a_ik A_jk$C_{ij} = \sum_{k=1}^{2} a_{ik} A_{jk}$ , 1 leq i, j leq 2$1 \leq i, j \leq 2$ , and C = [C_ij]$C = [C_{ij}]$ , then 8|C|$8|C|$ is equal to:
- A. 262
- B. 288
- C. 242
- D. 222
Solution
### Related Formula
sum_k a_ik A_jk = delta_ij |A| implies C = beginbmatrix |A| & 0 \\ 0 & |A| endbmatrix implies |C| = |A|^2$\sum_{k} a_{ik} A_{jk} = \delta_{ij} |A| \implies C = \begin{\bmatrix} |A| & 0 \\ 0 & |A| \end{\bmatrix} \implies |C| = |A|^2$
### Core Logic
Evaluate the determinant of matrix A$A$:
|A| = (log_5 128)(log_4 25) - (log_4 5)(log_5 8)$|A| = (\log_5 128)(\log_4 25) - (\log_4 5)(\log_5 8)$
Using change of base rules:
|A| = left(7log_5 2right)left(2log_4 5right) - left(frac12log_2 5right)left(3log_5 2right)$|A| = \left(7\log_5 2\right)\left(2\log_4 5\right) - \left(\frac{1}{2}\log_2 5\right)\left(3\log_5 2\right)$
|A| = 14left(log_5 2 cdot frac12log_2 5right) - frac32 = 7 - 1.5 = 5.5 = frac112$|A| = 14\left(\log_5 2 \cdot \frac{1}{2}\log_2 5\right) - \frac{3}{2} = 7 - 1.5 = 5.5 = \frac{11}{2}$
### Step 1: Compute |C| and evaluate response target
Since matrix properties dictate |C| = |A|^2$|C| = |A|^2$:
|C| = left(frac112right)^2 = frac1214$|C| = \left(\frac{11}{2}\right)^2 = \frac{121}{4}$
Evaluate targeted multiplier:
8|C| = 8 times frac1214 = 2 times 121 = 242$8|C| = 8 \times \frac{121}{4} = 2 \times 121 = 242$
### Pattern Recognition
Recognize the core cofactor theorem identity instantly: multiplying rows by cofactors of other rows creates zero elements, yielding basic diagonal scalar structures matching matrix attributes.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Matrices and Determinants
Q72
jee_main_2025_29_jan_morning
Matrix Powers
Let S = left| m in Z : A^m^2 + A^m = 3I - A^-6 right|$S = \left| m \in Z : A^{m^2} + A^m = 3I - A^{-6} \right|$ , where A = beginbmatrix 2 & -1 \\ 1 & 0 endbmatrix$A = \begin{\bmatrix} 2 & -1 \\ 1 & 0 \end{\bmatrix}$ . Then n(S) is equal to
Numerical Answer. Answer: 2
Solution
### Related Formula
textInductive formulation for exponent powers of a pattern matrix$\text{Inductive formulation for exponent powers of a pattern matrix}$
### Core Logic
Evaluate lower power forms of A$A$ to establish inductive sequence patterns:
A = beginbmatrix 2 & -1 \\ 1 & 0 endbmatrix, quad A^2 = beginbmatrix 3 & -2 \\ 2 & -1 endbmatrix, quad A^3 = beginbmatrix 4 & -3 \\ 3 & -2 endbmatrix$A = \begin{\bmatrix} 2 & -1 \\ 1 & 0 \end{\bmatrix}, \quad A^2 = \begin{\bmatrix} 3 & -2 \\ 2 & -1 \end{\bmatrix}, \quad A^3 = \begin{\bmatrix} 4 & -3 \\ 3 & -2 \end{\bmatrix}$
This cleanly establishes general power state rule configuration expression:
A^m = beginbmatrix m+1 & -m \\ m & -m+1 endbmatrix$A^m = \begin{\bmatrix} m+1 & -m \\ m & -m+1 \end{\bmatrix}$
### Step 1: Setup Matrix Power Equation
Using the pattern, find expressions for terms:
A^6 = beginbmatrix 7 & -6 \\ 6 & -5 endbmatrix, quad A^-6 = (A^6)^-1 = beginbmatrix -5 & 6 \\ -6 & 7 endbmatrix$A^6 = \begin{\bmatrix} 7 & -6 \\ 6 & -5 \end{\bmatrix}, \quad A^{-6} = (A^6)^{-1} = \begin{\bmatrix} -5 & 6 \\ -6 & 7 \end{\bmatrix}$
Substitute into the targeted equation block:
A^m^2 + A^m = 3beginbmatrix 1 & 0 \\ 0 & 1 endbmatrix - beginbmatrix -5 & 6 \\ -6 & 7 endbmatrix = beginbmatrix 8 & -6 \\ 6 & -4 endbmatrix$A^{m^2} + A^m = 3\begin{\bmatrix} 1 & 0 \\ 0 & 1 \end{\bmatrix} - \begin{\bmatrix} -5 & 6 \\ -6 & 7 \end{\bmatrix} = \begin{\bmatrix} 8 & -6 \\ 6 & -4 \end{\bmatrix}$
### Step 2: Equate Corresponding Elements
From the bottom-\left entry [2,1]$[2,1]$:
m^2 + m = 6 implies m^2 + m - 6 = 0$m^2 + m = 6 \implies m^2 + m - 6 = 0$
(m+3)(m-2) = 0 implies m = -3, 2$(m+3)(m-2) = 0 \implies m = -3, 2$
Both integer states satisfy all matrix components seamlessly. Therefore, the number of elements n(S) = 2$n(S) = 2$.
### Pattern Recognition
Always calculate first 2–3 matrix powers to spot linear arithmetic trends across specific cell values (m+1, -m$m+1, -m$), avoiding tedious full Cayley-Hamilton character expansions.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Matrices
Q3
jee_main_2024_01_february_morning
Properties of Determinants
If A=beginbmatrixsqrt2 & 1 \\ -1 & sqrt2endbmatrix$A=\begin{bmatrix}\sqrt{2} & 1 \\ -1 & \sqrt{2}\end{bmatrix}$, B=beginbmatrix1 & 0 \\ 1 & 1endbmatrix$B=\begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}$, C=ABA^T$C=ABA^{T}$ and X=A^TC^2A$X=A^{T}C^{2}A$, then det X$\det X$ is equal to:
- A. 243$243$
- B. 729$729$
- C. 27$27$
- D. 891$891$
Solution
### Related Formula
Properties of Determinants:
1. det(AB) = det(A) cdot det(B)$\det(AB) = \det(A) \cdot \det(B)$
2. det(A^T) = det(A)$\det(A^T) = \det(A)$
3. det(A^n) = (det(A))^n$\det(A^n) = (\det(A))^n$
### Core Logic
First, find the determinant of matrix A$A$ and matrix B$B$:
det(A) = |A| = beginvmatrixsqrt2 & 1 \\ -1 & sqrt2endvmatrix = (sqrt2)(sqrt2) - (1)(-1) = 2 + 1 = 3 $\det(A) = |A| = \begin{vmatrix}\sqrt{2} & 1 \\ -1 & \sqrt{2}\end{vmatrix} = (\sqrt{2})(\sqrt{2}) - (1)(-1) = 2 + 1 = 3 $
det(B) = |B| = beginvmatrix1 & 0 \\ 1 & 1endvmatrix = (1)(1) - (0)(1) = 1 $\det(B) = |B| = \begin{vmatrix}1 & 0 \\ 1 & 1\end{vmatrix} = (1)(1) - (0)(1) = 1 $
### Step 1: Calculate determinant of C
Given C = ABA^T$C = ABA^{T}$, we calculate its determinant:
det(C) = det(ABA^T) = det(A) cdot det(B) cdot det(A^T) $\det(C) = \det(ABA^{T}) = \det(A) \cdot \det(B) \cdot \det(A^{T}) $
Using det(A^T) = det(A) = 3$\det(A^T) = \det(A) = 3$:
det(C) = 3 cdot 1 cdot 3 = 9 $\det(C) = 3 \cdot 1 \cdot 3 = 9 $
### Step 2: Calculate determinant of X
Given X = A^TC^2A$X = A^{T}C^{2}A$, we determine |X|$|X|$:
det(X) = |A^TC^2A| = |A^T| cdot |C^2| cdot |A| $\det(X) = |A^{T}C^{2}A| = |A^{T}| \cdot |C^{2}| \cdot |A| $
det(X) = |A| cdot |C|^2 cdot |A| = |A|^2 cdot |C|^2 $\det(X) = |A| \cdot |C|^{2} \cdot |A| = |A|^{2} \cdot |C|^{2} $
Substitute the values |A| = 3$|A| = 3$ and |C| = 9$|C| = 9$:
det(X) = (3)^2 cdot (9)^2 = 9 cdot 81 = 729 $\det(X) = (3)^{2} \cdot (9)^{2} = 9 \cdot 81 = 729 $
### Pattern Recognition
Sees: Transpose and multiplication operations nested inside a determinant statement.
Shortcut: Never explicitly compute product matrices like ABA^T$ABA^T$ or A^TC^2A$A^TC^2A$. Apply the distributive identity of determinants |XYZ| = |X||Y||Z|$|XYZ| = |X||Y||Z|$ entirely to work strictly with scalar multiplication.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Matrices and Determinants
Q12
jee_main_2024_01_february_morning
System of Linear Equations
If the system of equations
2x+3y-z=5$2x+3y-z=5$
x+alpha y+3z=-4$x+\alpha y+3z=-4$
3x-y+beta z=7$3x-y+\beta z=7$
has infinitely many solutions, then 13 alpha beta$13 \alpha \beta$ is equal to:
- A. 1110$1110$
- B. 1120$1120$
- C. 1210$1210$
- D. 1220$1220$
Solution
### Related Formula
For a system of linear equations representing planes to have infinitely many solutions, the planes must belong to a single family sharing a common line of intersection:
P_1 = k_1 P_2 + k_2 P_3$P_1 = k_1 P_2 + k_2 P_3$
### Core Logic
Let the planes be defined as:
- P_1: 2x + 3y - z - 5 = 0$P_1: 2x + 3y - z - 5 = 0$
- P_2: x + alpha y + 3z + 4 = 0$P_2: x + \alpha y + 3z + 4 = 0$
- P_3: 3x - y + beta z - 7 = 0$P_3: 3x - y + \beta z - 7 = 0$
Expressing P_1$P_1$ as a linear combination of P_2$P_2$ and P_3$P_3$:
2x + 3y - z - 5 = k_1(x + alpha y + 3z + 4) + k_2(3x - y + beta z - 7)$2x + 3y - z - 5 = k_1(x + \alpha y + 3z + 4) + k_2(3x - y + \beta z - 7)$
### Step 1: Evaluate parameters k1 and k2
Comparing the coefficients of x$x$ and the constant terms on both sides:
- For x$x$:
k_1 + 3k_2 = 2 quad implies (1)$k_1 + 3k_2 = 2 \quad \implies (1)$
- For the constants:
4k_1 - 7k_2 = -5 quad implies (2)$4k_1 - 7k_2 = -5 \quad \implies (2)$
Multiplying equation (1) by 4 gives 4k_1 + 12k_2 = 8$4k_1 + 12k_2 = 8$. Subtracting equation (2) from this result:
(4k_1 + 12k_2) - (4k_1 - 7k_2) = 8 - (-5)$(4k_1 + 12k_2) - (4k_1 - 7k_2) = 8 - (-5)$
19k_2 = 13 implies k_2 = frac1319$19k_2 = 13 \implies k_2 = \frac{13}{19}$
Substituting k_2$k_2$ back into equation (1):
k_1 + 3left(frac1319right) = 2 implies k_1 = 2 - frac3919 = -frac119$k_1 + 3\left(\frac{13}{19}\right) = 2 \implies k_1 = 2 - \frac{39}{19} = -\frac{1}{19}$
### Step 2: Calculate alpha, beta and final product
Comparing coefficients for y$y$ and z$z$:
- For y$y$:
k_1alpha - k_2 = 3 implies -frac119alpha - frac1319 = 3$k_1\alpha - k_2 = 3 \implies -\frac{1}{19}\alpha - \frac{13}{19} = 3$
-alpha - 13 = 57 implies alpha = -70$-\alpha - 13 = 57 \implies \alpha = -70$
- For z$z$:
3k_1 + k_2beta = -1 implies 3left(-frac119right) + frac1319beta = -1$3k_1 + k_2\beta = -1 \implies 3\left(-\frac{1}{19}\right) + \frac{13}{19}\beta = -1$
-3 + 13beta = -19 implies 13beta = -16 implies beta = -frac1613$-3 + 13\beta = -19 \implies 13\beta = -16 \implies \beta = -\frac{16}{13}$
Now, compute 13 alpha beta$13 \alpha \beta$:
13 alpha beta = 13 times (-70) times left(-frac1613right) = 70 times 16 = 1120$13 \alpha \beta = 13 \times (-70) \times \left(-\frac{16}{13}\right) = 70 \times 16 = 1120$
### Pattern Recognition
Sees: Infinite solution framework for three linear planes.
Shortcut: Using the family of planes equation is significantly less prone to fractional algebraic mistakes compared to establishing Cramer's rule determinants (D = D_x = D_y = D_z = 0$D = D_x = D_y = D_z = 0$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Matrices and Determinants
Q1
jee_main_2024_29_january_evening
Properties of Determinants
Let A = beginbmatrix 2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2 endbmatrix$A = \begin{bmatrix} 2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2 \end{bmatrix}$ and P = beginbmatrix 1 & 2 & 0 \\ 5 & 0 & 2 \\ 7 & 1 & 5 endbmatrix$P = \begin{bmatrix} 1 & 2 & 0 \\ 5 & 0 & 2 \\ 7 & 1 & 5 \end{bmatrix}$. The sum of the prime factors of |P^-1AP - 2I|$|P^{-1}AP - 2I|$ is equal to
Solution
### Related Formula
|P^-1AP - 2I| = |P^-1(A - 2I)P| = |P^-1| cdot |A - 2I| cdot |P| = |A - 2I|$|P^{-1}AP - 2I| = |P^{-1}(A - 2I)P| = |P^{-1}| \cdot |A - 2I| \cdot |P| = |A - 2I|$
### Core Logic
Since |P^-1| cdot |P| = 1$|P^{-1}| \cdot |P| = 1$, the expression simplifies completely to the determinant of A - 2I$A - 2I$.
First, let us construct the matrix A - 2I$A - 2I$:
A - 2I = beginbmatrix 2-2 & 1 & 2 \\ 6 & 2-2 & 11 \\ 3 & 3 & 2-2 endbmatrix = beginbmatrix 0 & 1 & 2 \\ 6 & 0 & 11 \\ 3 & 3 & 0 endbmatrix$A - 2I = \begin{bmatrix} 2-2 & 1 & 2 \\ 6 & 2-2 & 11 \\ 3 & 3 & 2-2 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 2 \\ 6 & 0 & 11 \\ 3 & 3 & 0 \end{bmatrix}$
Now, evaluating the determinant:
|A - 2I| = 0(0 - 33) - 1(0 - 33) + 2(18 - 0) = 33 + 36 = 69$|A - 2I| = 0(0 - 33) - 1(0 - 33) + 2(18 - 0) = 33 + 36 = 69$
### Step 1: Finding Prime Factors
The number obtained is 69$69$. Let us find its prime factorization:
69 = 3 times 23$69 = 3 \times 23$
Both 3$3$ and 23$23$ are prime numbers. Their sum is:
textSum = 3 + 23 = 26$\text{Sum} = 3 + 23 = 26$
### Pattern Recognition
Whenever you encounter a matrix expression of the form P^-1AP - kI$P^{-1}AP - kI$, always factor out P^-1$P^{-1}$ and P$P$ to simplify it to |A - kI|$|A - kI|$. This saves tremendous time over computing matrix multiplications.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Matrices and Determinants