### Related Formula For infinite solutions of a system of equations, the main determinant D$D$ and auxiliary determinants D_1, D_2, D_3$D_1, D_2, D_3$ must all equal 0. ### Core Logic Since a, b, c$a, b, c$ are in A.P., we have 2b = a + c implies a - 2b + c = 0$2b = a + c \implies a - 2b + c = 0$. Comparing this identity with the line equation ax + by + c = 0$ax + by + c = 0$, we immediately find that the lines always pass through the fixed point configuration (1, -2)$(1, -2)$. Thus, P = (1, -2)$P = (1, -2)$. ### Step 1: Evaluation of the Matrix for Infinite Solutions Let us set the system determinant D = 0$D = 0$: D = beginbmatrix 1 & 1 & 1 \\ 2 & 5 & alpha \\ 1 & 2 & 3 endbmatrix = 0$D = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 5 & \alpha \\ 1 & 2 & 3 \end{bmatrix} = 0$ 1(15 - 2alpha) - 1(6 - alpha) + 1(4 - 5) = 0$1(15 - 2\alpha) - 1(6 - \alpha) + 1(4 - 5) = 0$ 15 - 2alpha - 6 + alpha - 1 = 0 implies 8 - alpha = 0 implies alpha = 8$15 - 2\alpha - 6 + \alpha - 1 = 0 \implies 8 - \alpha = 0 \implies \alpha = 8$ Now set D_1 = 0$D_1 = 0$ by substituting columns: D_1 = beginbmatrix 6 & 1 & 1 \\ beta & 5 & 8 \\ 4 & 2 & 3 endbmatrix = 0$D_1 = \begin{bmatrix} 6 & 1 & 1 \\ \beta & 5 & 8 \\ 4 & 2 & 3 \end{bmatrix} = 0$ 6(15 - 16) - 1(3beta - 32) + 1(2beta - 20) = 0$6(15 - 16) - 1(3\beta - 32) + 1(2\beta - 20) = 0$ -6 - 3beta + 32 + 2beta - 20 = 0 implies 6 - beta = 0 implies beta = 6$-6 - 3\beta + 32 + 2\beta - 20 = 0 \implies 6 - \beta = 0 \implies \beta = 6$ Thus, Q = (8, 6)$Q = (8, 6)$. ### Step 2: Distance Metric Calculation Using the distance formula between P(1, -2)$P(1, -2)$ and Q(8, 6)$Q(8, 6)$: (PQ)^2 = (8 - 1)^2 + (6 - (-2))^2 = 7^2 + 8^2 = 49 + 64 = 113$(PQ)^2 = (8 - 1)^2 + (6 - (-2))^2 = 7^2 + 8^2 = 49 + 64 = 113$ ### Pattern Recognition A.P. coefficients inside a standard linear equation reveal a fixed point of concurrency by mapping matching coefficient components (1, -2, 1$1, -2, 1$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants Class 11 Mathematics: Straight Lines

### Related Formula cos(-x) = cos x$\cos(-x) = \cos x$ sin(-x) = -sin x$\sin(-x) = -\sin x$ sin(x+y) = sin x cos y + cos x sin y$\sin(x+y) = \sin x \cos y + \cos x \sin y$ cos(x+y) = cos x cos y - sin x sin y$\cos(x+y) = \cos x \cos y - \sin x \sin y$ ### Core Logic Evaluate f(-x)$f(-x)$: f(-x) = beginbmatrix cos(-x) & -sin(-x) & 0 \\ sin(-x) & cos(-x) & 0 \\ 0 & 0 & 1 endbmatrix = beginbmatrix cos x & sin x & 0 \\ -sin x & cos x & 0 \\ 0 & 0 & 1 endbmatrix$f(-x) = \begin{bmatrix} \cos(-x) & -\sin(-x) & 0 \\ \sin(-x) & \cos(-x) & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$ Checking if f(-x)$f(-x)$ is the inverse by evaluating f(x) cdot f(-x)$f(x) \cdot f(-x)$: f(x) f(-x) = beginbmatrix cos^2 x + sin^2 x & cos x sin x - sin x cos x & 0 \\ sin x cos x - cos x sin x & sin^2 x + cos^2 x & 0 \\ 0 & 0 & 1 endbmatrix$f(x) f(-x) = \begin{bmatrix} \cos^2 x + \sin^2 x & \cos x \sin x - \sin x \cos x & 0 \\ \sin x \cos x - \cos x \sin x & \sin^2 x + \cos^2 x & 0 \\ 0 & 0 & 1 \end{bmatrix}$ f(x) f(-x) = beginbmatrix 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 endbmatrix = I$f(x) f(-x) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$ Thus, Statement I is true. ### Step 1: Checking Statement II Evaluate the matrix multiplication f(x) cdot f(y)$f(x) \cdot f(y)$: f(x) f(y) = beginbmatrix cos x & -sin x & 0 \\ sin x & cos x & 0 \\ 0 & 0 & 1 endbmatrix beginbmatrix cos y & -sin y & 0 \\ sin y & cos y & 0 \\ 0 & 0 & 1 endbmatrix$f(x) f(y) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix}$ = beginbmatrix cos x cos y - sin x sin y & -cos x sin y - sin x cos y & 0 \\ sin x cos y + cos x sin y & -sin x sin y + cos x cos y & 0 \\ 0 & 0 & 1 endbmatrix$= \begin{bmatrix} \cos x \cos y - \sin x \sin y & -\cos x \sin y - \sin x \cos y & 0 \\ \sin x \cos y + \cos x \sin y & -\sin x \sin y + \cos x \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix}$ Apply standard trigonometric compound angle formulas: = beginbmatrix cos(x+y) & -sin(x+y) & 0 \\ sin(x+y) & cos(x+y) & 0 \\ 0 & 0 & 1 endbmatrix = f(x+y)$= \begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix} = f(x+y)$ Thus, Statement II is also true. ### Step 2: Final Conclusion Both Statement I and Statement II are true. ### Pattern Recognition This specific matrix represents a standard 2D rotation matrix embedded in 3D space. Rotation matrices naturally follow R(x)R(y) = R(x+y)$R(x)R(y) = R(x+y)$ (additive property of angles) and their inverse is always obtained by negating the angle R(-x) = R(x)^-1$R(-x) = R(x)^{-1}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Matrices Class 11 Maths: Trigonometric Functions

### Related Formula |AB| = |A||B|$|AB| = |A||B|$ Trace(B$B$) = sum b_ii$\sum b_{ii}$ ### Core Logic Since B = [B_1, B_2, B_3]$B = [B_1, B_2, B_3]$, the matrix multiplication AB$AB$ effectively applies A$A$ to each column of B$B$: AB = [AB_1, AB_2, AB_3] = beginbmatrix 1 & 2 & 3 \\ 0 & 3 & 2 \\ 0 & 0 & 1 endbmatrix$AB = [AB_1, AB_2, AB_3] = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 3 & 2 \\ 0 & 0 & 1 \end{bmatrix}$ Let this target matrix be C$C$. We know AB = C$AB = C$. Thus, taking determinants on both sides: |A| |B| = |C|$|A| |B| = |C|$ ### Step 1: Finding Determinants Calculate determinant of A$A$: |A| = 2(1 - 0) - 0 + 1(0 - 1) = 2 - 1 = 1$|A| = 2(1 - 0) - 0 + 1(0 - 1) = 2 - 1 = 1$ Calculate determinant of C$C$. Since C$C$ is an upper triangular matrix, its determinant is simply the product of its main diagonal: |C| = 1 times 3 times 1 = 3$|C| = 1 \times 3 \times 1 = 3$ Therefore, 1 times |B| = 3 Rightarrow |B| = 3$1 \times |B| = 3 \Rightarrow |B| = 3$. So alpha = 3$\alpha = 3$. ### Step 2: Finding Matrix B To find beta$\beta$ (the trace of B$B$), we must explicitly find B = A^-1C$B = A^{-1}C$. Rather than finding the full inverse, manually solve A B_i = C_i$A B_i = C_i$: For B_1 = [x, y, z]^T$B_1 = [x, y, z]^T$: 2x+z=1$2x+z=1$, x+y=0$x+y=0$, x+z=0 Rightarrow x=1, z=-1, y=-1$x+z=0 \Rightarrow x=1, z=-1, y=-1$. Thus B_1 = [1, -1, -1]^T$B_1 = [1, -1, -1]^T$. For B_2$B_2$: 2x+z=2$2x+z=2$, x+y=3$x+y=3$, x+z=0 Rightarrow x=2, z=-2, y=1$x+z=0 \Rightarrow x=2, z=-2, y=1$. Thus B_2 = [2, 1, -2]^T$B_2 = [2, 1, -2]^T$. For B_3$B_3$: 2x+z=3$2x+z=3$, x+y=2$x+y=2$, x+z=1 Rightarrow x=2, z=-1, y=0$x+z=1 \Rightarrow x=2, z=-1, y=0$. Thus B_3 = [3, 0, -1]^T$B_3 = [3, 0, -1]^T$. ### Step 3: Calculating Trace and Final Output Constructing Matrix B: B = beginbmatrix 1 & 2 & 3 \\ -1 & 1 & 0 \\ -1 & -2 & -1 endbmatrix$B = \begin{bmatrix} 1 & 2 & 3 \\ -1 & 1 & 0 \\ -1 & -2 & -1 \end{bmatrix}$ The diagonal elements are 1, 1, -1$1, 1, -1$. Trace beta = 1 + 1 - 1 = 1$\beta = 1 + 1 - 1 = 1$. Finally compute alpha^3 + beta^3$\alpha^3 + \beta^3$: 3^3 + 1^3 = 27 + 1 = 28$3^3 + 1^3 = 27 + 1 = 28$ ### Pattern Recognition When given AX_i = Y_i$AX_i = Y_i$ for multiple columns, they collectively form A X = Y$A X = Y$. Using |A||X| = |Y|$|A||X| = |Y|$ bypasses full matrix inversion if you strictly need determinants. To grab the trace, solving equations systematically column-by-column is generally less error-prone than forming the full adjoint inverse matrix. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Matrices

### Related Formula |kA| = k^n |A|$|kA| = k^n |A|$ Where A$A$ is an n times n$n \times n$ matrix, and k$k$ is a scalar. ### Core Logic Find the determinant of the 3 times 3$3 \times 3$ matrix A$A$: |A| = 1(alpha cdot alpha - beta cdot beta) - 0 + 0$|A| = 1(\alpha \cdot \alpha - \beta \cdot \beta) - 0 + 0$ |A| = alpha^2 - beta^2$|A| = \alpha^2 - \beta^2$ Given the condition |2A|^3 = 2^21$|2A|^3 = 2^{21}$. Since A$A$ is a 3 times 3$3 \times 3$ matrix, applying the scalar property |kA| = k^3|A|$|kA| = k^3|A|$: |2A| = 2^3|A| = 8|A|$|2A| = 2^3|A| = 8|A|$ Substitute this back into the original condition: (2^3|A|)^3 = 2^21$(2^3|A|)^3 = 2^{21}$ 2^9 |A|^3 = 2^21$2^9 |A|^3 = 2^{21}$ |A|^3 = frac2^212^9 = 2^12$|A|^3 = \frac{2^{21}}{2^9} = 2^{12}$ Taking the cube root of both sides: |A| = 2^4 = 16$|A| = 2^4 = 16$ ### Step 1: Solve the Diophantine Equation We have: alpha^2 - beta^2 = 16$\alpha^2 - \beta^2 = 16$ (alpha - beta)(alpha + beta) = 16$(\alpha - \beta)(\alpha + \beta) = 16$ Since alpha$\alpha$ and beta$\beta$ are integers, their sum and difference must also be integers. Also, (alpha + beta)$(\alpha + \beta)$ and (alpha - beta)$(\alpha - \beta)$ must share the same parity (both even or both odd) because their sum is 2alpha$2\alpha$ (an even number). Since their product is 16, the only valid integer factor pairs of 16 that share the same parity are (8, 2)$(8, 2)$ and (-8, -2)$(-8, -2)$ and (4, 4)$(4, 4)$ and (-4, -4)$(-4, -4)$. Case 1: (alpha + beta) = 8$(\alpha + \beta) = 8$ and (alpha - beta) = 2$(\alpha - \beta) = 2$ Adding them gives 2alpha = 10 Rightarrow alpha = 5$2\alpha = 10 \Rightarrow \alpha = 5$. Thus beta = 3$\beta = 3$. Case 2: (alpha + beta) = 4$(\alpha + \beta) = 4$ and (alpha - beta) = 4$(\alpha - \beta) = 4$ Adding them gives 2alpha = 8 Rightarrow alpha = 4$2\alpha = 8 \Rightarrow \alpha = 4$. Thus beta = 0$\beta = 0$. Looking at the options provided (3, 5, 17, 9), the value alpha = 5$\alpha = 5$ is listed. ### Pattern Recognition Extracting scalar multipliers from determinants always depends on the dimension n$n$ of the matrix. For Diophantine equations like x^2 - y^2 = k$x^2 - y^2 = k$, factoring into (x-y)(x+y)$(x-y)(x+y)$ and analyzing parity constraints restricts the solution space instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Determinants

### Related Formula (X+Y)^2 = X^2 + XY + YX + Y^2$(X+Y)^2 = X^2 + XY + YX + Y^2$ Because matrix multiplication is generally non-commutative, you cannot assume XY = YX$XY = YX$ unless explicitly proven. ### Core Logic Given A A^T = I$A A^T = I$. Since A is a square matrix, this implies A$A$ is orthogonal, so A^T A = I$A^T A = I$ as well. Expand the inner binomials strictly respecting matrix order: (A+A^T)^2 = A^2 + AA^T + A^TA + (A^T)^2$(A+A^T)^2 = A^2 + AA^T + A^TA + (A^T)^2$ (A-A^T)^2 = A^2 - AA^T - A^TA + (A^T)^2$(A-A^T)^2 = A^2 - AA^T - A^TA + (A^T)^2$ Add both expanded expressions together. Notice that the cross terms perfectly cancel each other out: (A+A^T)^2 + (A-A^T)^2 = 2A^2 + 2(A^T)^2 = 2[A^2 + (A^T)^2]$(A+A^T)^2 + (A-A^T)^2 = 2A^2 + 2(A^T)^2 = 2[A^2 + (A^T)^2]$ ### Step 1: Simplify Final Expression Substitute this simplified sum back into the original expression: frac12A left[ 2(A^2 + (A^T)^2) right]$\frac{1}{2}A \left[ 2(A^2 + (A^T)^2) \right]$ = A[A^2 + (A^T)^2]$= A[A^2 + (A^T)^2]$ Distribute A$A$ from the left side: = A cdot A^2 + A cdot (A^T)^2$= A \cdot A^2 + A \cdot (A^T)^2$ = A^3 + A A^T A^T$= A^3 + A A^T A^T$ Since we are given that A A^T = I$A A^T = I$, substitute I$I$ into the second term: = A^3 + I A^T$= A^3 + I A^T$ = A^3 + A^T$= A^3 + A^T$ ### Pattern Recognition Identities like (X+Y)^2 + (X-Y)^2 = 2(X^2+Y^2)$(X+Y)^2 + (X-Y)^2 = 2(X^2+Y^2)$ natively hold for matrices as well because the restrictive cross terms +XY+YX$+XY+YX$ and -XY-YX$-XY-YX$ cancel strictly linearly regardless of commutativity. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices