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:

### 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

### 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