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 8
Q1
jee_main_2024_30_january_evening
System of Linear Equations
Consider the
system of linear equations
x + y + z = 5, x + 2 y + lambda^ 2 z = 9$x + y + z = 5, x + 2 y + \lambda^ {2} z = 9$,
x + 3y + lambda z = mu$x + 3y + \lambda z = \mu$ , where
lambda ,mu in mathbbR$\lambda ,\mu \in \mathbb{R}$ . Then, which of the following statement is NOT correct?
- A. textSystem has infinite number of solution if lambda = 1 text and mu = 13$\text{System has infinite number of solution if } \lambda = 1 \text{ and } \mu = 13$
- B. textSystem is inconsistent if lambda = 1 text and mu neq 13$\text{System is inconsistent if } \lambda = 1 \text{ and } \mu \neq 13$
- C. textSystem is consistent if lambda neq 1 text and mu = 13$\text{System is consistent if } \lambda \neq 1 \text{ and } \mu = 13$
- D. textSystem has unique solution if lambda neq 1 text and mu neq 13$\text{System has unique solution if } \lambda \neq 1 \text{ and } \mu \neq 13$
Solution
### Related Formula
Delta = beginvmatrix a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 endvmatrix$\Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$
### Core Logic
For the given system, we find the determinant of the coefficient matrix:
Delta = beginvmatrix 1 & 1 & 1 \\ 1 & 2 & lambda^2 \\ 1 & 3 & lambda endvmatrix$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & \lambda^2 \\ 1 & 3 & \lambda \end{vmatrix}$
Expanding along the first row:
Delta = 1(2lambda - 3lambda^2) - 1(lambda - lambda^2) + 1(3 - 2)$\Delta = 1(2\lambda - 3\lambda^2) - 1(\lambda - \lambda^2) + 1(3 - 2)$
Delta = 2lambda - 3lambda^2 - lambda + lambda^2 + 1$\Delta = 2\lambda - 3\lambda^2 - \lambda + \lambda^2 + 1$
Delta = -2lambda^2 + lambda + 1 = 0$\Delta = -2\lambda^2 + \lambda + 1 = 0$
Rightarrow 2lambda^2 - lambda - 1 = 0$\Rightarrow 2\lambda^2 - \lambda - 1 = 0$
Rightarrow (2lambda + 1)(lambda - 1) = 0 Rightarrow lambda = 1, -frac12$\Rightarrow (2\lambda + 1)(\lambda - 1) = 0 \Rightarrow \lambda = 1, -\frac{1}{2}$
### Step 1: Analyzing Cases
For infinite solutions or inconsistency, we check Delta_z$\Delta_z$ (or others):
Delta_z = beginvmatrix 1 & 1 & 5 \\ 1 & 2 & 9 \\ 1 & 3 & mu endvmatrix = 0$\Delta_z = \begin{vmatrix} 1 & 1 & 5 \\ 1 & 2 & 9 \\ 1 & 3 & \mu \end{vmatrix} = 0$
Evaluating the determinant:
1(2mu - 27) - 1(mu - 9) + 5(3 - 2) = 0$1(2\mu - 27) - 1(\mu - 9) + 5(3 - 2) = 0$
2mu - 27 - mu + 9 + 5 = 0 Rightarrow mu - 13 = 0 Rightarrow mu = 13$2\mu - 27 - \mu + 9 + 5 = 0 \Rightarrow \mu - 13 = 0 \Rightarrow \mu = 13$
Thus, if lambda = 1$\lambda = 1$ and mu = 13$\mu = 13$, the system has an infinite number of solutions.
If lambda = 1$\lambda = 1$ and mu neq 13$\mu \neq 13$, the system is inconsistent.
### Step 2: Checking Options
Option 1 is correct (True statement).
Option 2 is correct (True statement).
Option 3: If lambda neq 1$\lambda \neq 1$ and lambda neq -frac12$\lambda \neq -\frac{1}{2}$, the system has a unique solution (which is consistent) regardless of mu$\mu$. If lambda = -frac12$\lambda = -\frac{1}{2}$ and mu = 13$\mu = 13$, it may have infinite solutions or be inconsistent, but typically it is consistent for unique solutions if lambda neq 1$\lambda \neq 1$ and lambda neq -1/2$\lambda \neq -1/2$. Wait, the question option 4 says "System has unique solution if lambda neq 1$\lambda \neq 1$ and mu neq 13$\mu \neq 13$". This is NOT correct because if lambda = -frac12$\lambda = -\frac{1}{2}$ and mu neq 13$\mu \neq 13$, the determinant is zero and it does not have a unique solution.
### Pattern Recognition
Look for the roots of Delta = 0$\Delta = 0$. Since there are two roots (lambda=1, -1/2$\lambda=1, -1/2$), a blanket statement about "unique solution if lambda neq 1$\lambda \neq 1$" ignores the other root and is inherently false.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Matrices and Determinants
Q20
jee_main_2024_30_january_evening
Properties of Matrices
Let R = beginpmatrix x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z endpmatrix$R = \begin{pmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{pmatrix}$ be a non-zero 3 times 3$3 \times 3$ matrix, where xsin theta = ysin left(theta +frac2pi3right) = zsin left(theta +frac4pi3right) neq 0$x\sin \theta = y\sin \left(\theta +\frac{2\pi}{3}\right) = z\sin \left(\theta +\frac{4\pi}{3}\right) \neq 0$, theta in (0,2pi)$\theta \in (0,2\pi)$. For a square matrix M$M$, let trace (M)$(M)$ denote the sum of all the diagonal entries of M$M$. Then, among the statements:
(I) Trace (R) = 0$(R) = 0$
(II) If trace (textadj(textadj(R))) = 0$(\text{adj}(\text{adj}(R))) = 0$ , then R$R$ has exactly one non-zero entry.
Choose the correct statement:
- A. textBoth (I) and (II) are true$\text{Both (I) and (II) are true}$
- B. textNeither (I) nor (II) is true.$\text{Neither (I) nor (II) is true.}$
- C. textOnly (II) is true.$\text{Only (II) is true.}$
- D. textOnly (I) is true.$\text{Only (I) is true.}$
Solution
### Related Formula
textadj(textadj(A)) = |A|^n-2 A$\text{adj}(\text{adj}(A)) = |A|^{n-2} A$
sin theta + sin left(theta + frac2pi3right) + sin left(theta + frac4pi3right) = 0$\sin \theta + \sin \left(\theta + \frac{2\pi}{3}\right) + \sin \left(\theta + \frac{4\pi}{3}\right) = 0$
### Core Logic
Let xsin theta = ysin left(theta +frac2pi3right) = zsin left(theta +frac4pi3right) = lambda$x\sin \theta = y\sin \left(\theta +\frac{2\pi}{3}\right) = z\sin \left(\theta +\frac{4\pi}{3}\right) = \lambda$.
Since lambda neq 0$\lambda \neq 0$, none of x, y, z$x, y, z$ can be zero. Therefore, x = fraclambdasintheta, y = fraclambdasin(theta + 2pi/3), z = fraclambdasin(theta + 4pi/3)$x = \frac{\lambda}{\sin\theta}, y = \frac{\lambda}{\sin(\theta + 2\pi/3)}, z = \frac{\lambda}{\sin(\theta + 4\pi/3)}$.
### Step 1: Evaluating Statement (I)
The fundamental identity states that sintheta + sin(theta + 120^circ) + sin(theta + 240^circ) = 0$\sin\theta + \sin(\theta + 120^\circ) + \sin(\theta + 240^\circ) = 0$.
Trace (R) = x + y + z$(R) = x + y + z$.
x + y + z = lambda left( frac1sintheta + frac1sin(theta + 2pi/3) + frac1sin(theta + 4pi/3) right)$x + y + z = \lambda \left( \frac{1}{\sin\theta} + \frac{1}{\sin(\theta + 2\pi/3)} + \frac{1}{\sin(\theta + 4\pi/3)} \right)$
Summing these fractions produces the sum of products of pairs of sines in the numerator:
textNumerator = sin(theta+120^circ)sin(theta+240^circ) + sinthetasin(theta+240^circ) + sinthetasin(theta+120^circ)$\text{Numerator } = \sin(\theta+120^\circ)\sin(\theta+240^\circ) + \sin\theta\sin(\theta+240^\circ) + \sin\theta\sin(\theta+120^\circ)$
Using sum to product formulas, this simplifies to -frac34 neq 0$-\frac{3}{4} \neq 0$ for general theta$\theta$. Thus, x+y+z neq 0$x+y+z \neq 0$.
Statement (I) is FALSE.
### Step 2: Evaluating Statement (II)
For a 3 times 3$3 \times 3$ matrix, textadj(textadj(R)) = |R|^3-2 R = |R|R$\text{adj}(\text{adj}(R)) = |R|^{3-2} R = |R|R$.
|R| = xyz$|R| = xyz$.
Trace (textadj(textadj(R))) = |R| cdot textTrace(R) = xyz(x + y + z)$(\text{adj}(\text{adj}(R))) = |R| \cdot \text{Trace}(R) = xyz(x + y + z)$.
From the given condition, x, y, z neq 0$x, y, z \neq 0$, meaning |R| = xyz neq 0$|R| = xyz \neq 0$. As shown above, x+y+z neq 0$x+y+z \neq 0$ as well.
Hence, Trace (textadj(textadj(R))) neq 0$(\text{adj}(\text{adj}(R))) \neq 0$.
Because the hypothesis "If trace (textadj(textadj(R))) = 0$(\text{adj}(\text{adj}(R))) = 0$" is strictly false, the implication itself is vacuously true in propositional logic. (False implies$\implies$ Anything is True).
Statement (II) is TRUE.
### Pattern Recognition
Propositional logic strictly dictates that an 'If P then Q' statement evaluates to True if the premise P is inherently impossible (vacuous truth). Calculating the adj-adj trace purely checks for premise invalidity.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Matrices and Determinants
Q17
jee_main_2024_30_jan_morning
Properties of Determinants
If
f(x) = beginvmatrix 2 cos^4 x & 2 sin^4 x & 3 + sin^2 2x \\ 3 + 2 cos^4 x & 2 sin^4 x & sin^2 2x \\ 2 cos^4 x & 3 + 2 sin^4 x & sin^2 2x endvmatrix$f(x) = \begin{vmatrix} 2 \cos^4 x & 2 \sin^4 x & 3 + \sin^2 2x \\ 3 + 2 \cos^4 x & 2 \sin^4 x & \sin^2 2x \\ 2 \cos^4 x & 3 + 2 \sin^4 x & \sin^2 2x \end{vmatrix}$
then frac15 f'(0)$\frac{1}{5} f'(0)$ is equal to
- A. 0$0$
- B. 1$1$
- C. 2$2$
- D. 6$6$
Solution
### Related Formula
Row Operations in Determinants:
R_i to R_i + kR_j text does not change the determinant.$R_i \to R_i + kR_j \text{ does not change the determinant.}$
### Core Logic
Apply row operations to simplify the determinant:
R_2 to R_2 - R_1$R_2 \to R_2 - R_1$ and R_3 to R_3 - R_1$R_3 \to R_3 - R_1$
f(x) = beginvmatrix 2 cos^4 x & 2 sin^4 x & 3 + sin^2 2x \\ (3 + 2cos^4 x) - 2cos^4 x & 2sin^4 x - 2sin^4 x & sin^2 2x - (3 + sin^2 2x) \\ 2cos^4 x - 2cos^4 x & (3 + 2sin^4 x) - 2sin^4 x & sin^2 2x - (3 + sin^2 2x) endvmatrix$f(x) = \begin{vmatrix} 2 \cos^4 x & 2 \sin^4 x & 3 + \sin^2 2x \\ (3 + 2\cos^4 x) - 2\cos^4 x & 2\sin^4 x - 2\sin^4 x & \sin^2 2x - (3 + \sin^2 2x) \\ 2\cos^4 x - 2\cos^4 x & (3 + 2\sin^4 x) - 2\sin^4 x & \sin^2 2x - (3 + \sin^2 2x) \end{vmatrix}$
### Step 1: Expanding the simplified determinant
f(x) = beginvmatrix 2 cos^4 x & 2 sin^4 x & 3 + sin^2 2x \\ 3 & 0 & -3 \\ 0 & 3 & -3 endvmatrix$f(x) = \begin{vmatrix} 2 \cos^4 x & 2 \sin^4 x & 3 + \sin^2 2x \\ 3 & 0 & -3 \\ 0 & 3 & -3 \end{vmatrix}$
Expand along the 2nd row:
f(x) = -3 [ 2sin^4 x(-3) - 3(3 + sin^2 2x) ] + (-3) [ 2cos^4 x(3) - 0 ]$f(x) = -3 [ 2\sin^4 x(-3) - 3(3 + \sin^2 2x) ] + (-3) [ 2\cos^4 x(3) - 0 ]$
Wait, let's expand properly:
f(x) = -3 beginvmatrix 2sin^4 x & 3 + sin^2 2x \\ 3 & -3 endvmatrix - (-3) beginvmatrix 2cos^4 x & 2sin^4 x \\ 0 & 3 endvmatrix$f(x) = -3 \begin{vmatrix} 2\sin^4 x & 3 + \sin^2 2x \\ 3 & -3 \end{vmatrix} - (-3) \begin{vmatrix} 2\cos^4 x & 2\sin^4 x \\ 0 & 3 \end{vmatrix}$
f(x) = -3 [ -6sin^4 x - 9 - 3sin^2 2x ] + 3 [ 6cos^4 x ]$f(x) = -3 [ -6\sin^4 x - 9 - 3\sin^2 2x ] + 3 [ 6\cos^4 x ]$
f(x) = 18sin^4 x + 27 + 9sin^2 2x + 18cos^4 x$f(x) = 18\sin^4 x + 27 + 9\sin^2 2x + 18\cos^4 x$
f(x) = 18(sin^4 x + cos^4 x) + 9(4sin^2 x cos^2 x) + 27$f(x) = 18(\sin^4 x + \cos^4 x) + 9(4\sin^2 x \cos^2 x) + 27$
f(x) = 18(sin^2 x + cos^2 x)^2 - 36sin^2 x cos^2 x + 36sin^2 x cos^2 x + 27$f(x) = 18(\sin^2 x + \cos^2 x)^2 - 36\sin^2 x \cos^2 x + 36\sin^2 x \cos^2 x + 27$
f(x) = 18(1) + 27 = 45$f(x) = 18(1) + 27 = 45$
### Step 2: Differentiating constant
Since f(x) = 45$f(x) = 45$ is a constant function,
f'(x) = 0$f'(x) = 0$
Therefore, frac15 f'(0) = 0$\frac{1}{5} f'(0) = 0$.
### Pattern Recognition
If row subtractions reduce variable rows to pure constants, the entire determinant expansion often resolves into a pure trigonometric identity producing a global constant, immediately rendering its derivative 0.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Determinants
Class 12 Maths: Continuity and Differentiability
Q19
jee_main_2024_31_jan_evening
System of Linear Equations
Let A$A$ be a 3times3$3\times3$ real matrix such that Abeginbmatrix 1 \\ 0 \\ 1 endbmatrix=2beginbmatrix 1 \\ 0 \\ 1 endbmatrix, Abeginbmatrix -1 \\ 0 \\ 1 endbmatrix=4beginbmatrix -1 \\ 0 \\ 1 endbmatrix, Abeginbmatrix 0 \\ 1 \\ 0 endbmatrix=2beginbmatrix 0 \\ 1 \\ 0 endbmatrix$A\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}=2\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, A\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}=4\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}, A\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}=2\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$. Then, the system (A-3I)beginbmatrix x \\ y \\ z endbmatrix=beginbmatrix 1 \\ 2 \\ 3 endbmatrix$(A-3I)\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ has
- A. textunique solution$\text{unique solution}$
- B. textexactly two solutions$\text{exactly two solutions}$
- C. textno solution$\text{no solution}$
- D. textinfinitely many solutions$\text{infinitely many solutions}$
Solution
### Core Logic
Let A = beginbmatrix x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 endbmatrix$A = \begin{bmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{bmatrix}$. Using the given column transformations:
1) Abeginbmatrix 1 \\ 0 \\ 1 endbmatrix = beginbmatrix 2 \\ 0 \\ 2 endbmatrix implies x_1+z_1=2,\; x_2+z_2=0,\; x_3+z_3=2$A\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \\ 2 \end{bmatrix} \implies x_1+z_1=2,\; x_2+z_2=0,\; x_3+z_3=2$.
2) Abeginbmatrix -1 \\ 0 \\ 1 endbmatrix = beginbmatrix -4 \\ 0 \\ 4 endbmatrix implies -x_1+z_1=-4,\; -x_2+z_2=0,\; -x_3+z_3=4$A\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} -4 \\ 0 \\ 4 \end{bmatrix} \implies -x_1+z_1=-4,\; -x_2+z_2=0,\; -x_3+z_3=4$.
3) Abeginbmatrix 0 \\ 1 \\ 0 endbmatrix = beginbmatrix 0 \\ 2 \\ 0 endbmatrix implies y_1=0,\; y_2=2,\; y_3=0$A\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 2 \\ 0 \end{bmatrix} \implies y_1=0,\; y_2=2,\; y_3=0$.
Solving pairs from 1) and 2):
x_1+z_1=2$x_1+z_1=2$ and -x_1+z_1=-4 implies 2z_1=-2 implies z_1=-1, x_1=3$-x_1+z_1=-4 \implies 2z_1=-2 \implies z_1=-1, x_1=3$.
x_2+z_2=0$x_2+z_2=0$ and -x_2+z_2=0 implies x_2=0, z_2=0$-x_2+z_2=0 \implies x_2=0, z_2=0$.
x_3+z_3=2$x_3+z_3=2$ and -x_3+z_3=4 implies 2z_3=6 implies z_3=3, x_3=-1$-x_3+z_3=4 \implies 2z_3=6 \implies z_3=3, x_3=-1$.
So, A = beginbmatrix 3 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 3 endbmatrix$A = \begin{bmatrix} 3 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 3 \end{bmatrix}$.
Now, compute A-3I$A-3I$:
A-3I = beginbmatrix 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 endbmatrix$A-3I = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}$.
Determinant |A-3I| = -1(-1)(-1) = 1 neq 0$|A-3I| = -1(-1)(-1) = 1 \neq 0$.
Since the determinant is non-zero, the system (A-3I)X = B$(A-3I)X = B$ has a unique solution.
### Pattern Recognition
Whenever independent eigenvectors are given, the matrix can be uniquely constructed. If det != 0 for (A-3I)$(A-3I)$, uniqueness is guaranteed unconditionally.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Matrices