### Related Formula Using standard matrix properties: - M \, textadj(M) = |M|I$M \, \text{adj}(M) = |M|I$ - textadj(kM) = k^n-1textadj(M)$\text{adj}(kM) = k^{n-1}\text{adj}(M)$ - |M^k| = |M|^k$|M^k| = |M|^k$ - [textadj(M^-1)] = [textadj(M)]^-1$[\text{adj}(M^{-1})] = [\text{adj}(M)]^{-1}$ ### Core Logic First, find lambda$\lambda$ by evaluating |A| = -1$|A| = -1$: |A| = lambda(10 - (-6)) - 2(8 - 42) + 3(-4 - 35) = -1$|A| = \lambda(10 - (-6)) - 2(8 - 42) + 3(-4 - 35) = -1$ 16lambda - 2(-34) + 3(-39) = -1$16\lambda - 2(-34) + 3(-39) = -1$ 16lambda + 68 - 117 = -1 implies 16lambda - 49 = -1 implies 16lambda = 48 implies lambda = 3$16\lambda + 68 - 117 = -1 \implies 16\lambda - 49 = -1 \implies 16\lambda = 48 \implies \lambda = 3$ ### Step 1: Simplifying matrix expression B$B$ Let C = A \, textadj(A^2)$C = A \, \text{adj}(A^2)$. We know: A^2 \, textadj(A^2) = |A^2|I = |A|^2 I$A^2 \, \text{adj}(A^2) = |A^2|I = |A|^2 I$ Multiply C$C$ by A$A$ on the left: AC = A^2 \, textadj(A^2) = |A|^2 I$AC = A^2 \, \text{adj}(A^2) = |A|^2 I$ Since |A| = -1 implies |A|^2 = 1$|A| = -1 \implies |A|^2 = 1$: AC = I implies C = A^-1$AC = I \implies C = A^{-1}$ Now, we are given B^-1 = textadj(C) = textadj(A^-1)$B^{-1} = \text{adj}(C) = \text{adj}(A^{-1})$: B = [textadj(A^-1)]^-1 = textadj(A)$B = [\text{adj}(A^{-1})]^{-1} = \text{adj}(A)$ ### Step 2: Determinant calculation of lambda B + I$\lambda B + I$ We need to find |lambda B + I| = |3textadj(A) + I|$|\lambda B + I| = |3\text{adj}(A) + I|$: Let P = 3textadj(A) + I$P = 3\text{adj}(A) + I$. AP = 3Atextadj(A) + A = 3|A|I + A = -3I + A = A - 3I$AP = 3A\text{adj}(A) + A = 3|A|I + A = -3I + A = A - 3I$ Taking determinants on both sides: |A| cdot |P| = |A - 3I| implies -|P| = |A - 3I| implies |P| = -|A - 3I|$|A| \cdot |P| = |A - 3I| \implies -|P| = |A - 3I| \implies |P| = -|A - 3I|$ Let's calculate matrix A - 3I$A - 3I$: A - 3I = beginbmatrix 0 & 2 & 3 \\ 4 & 2 & 6 \\ 7 & -1 & -1 endbmatrix$A - 3I = \begin{bmatrix} 0 & 2 & 3 \\ 4 & 2 & 6 \\ 7 & -1 & -1 \end{bmatrix}$ textDet(A-3I) = 0 - 2(4(-1) - 42) + 3(-4 - 14) = -2(-46) + 3(-18) = 92 - 54 = 38$\text{Det}(A-3I) = 0 - 2(4(-1) - 42) + 3(-4 - 14) = -2(-46) + 3(-18) = 92 - 54 = 38$ Thus: |P| = -38$|P| = -38$ Taking absolute value of determinant / magnitude: ||lambda B + I|| = 38$||\lambda B + I|| = 38$ ### Pattern Recognition For products of adjoint matrices, always relate back to basic identity M \, textadj(M) = |M|I$M \, \text{adj}(M) = |M|I$. Pre-multiplying by A$A$ or A^2$A^2$ collapses long chains of adjoints instantly into standard scalar multiples. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants

### Related Formula For any n times n$n \times n$ matrix A$A$, the determinant properties of adjoints scale iteratively as follows: |textadj A| = |A|^n-1$|\text{adj } A| = |A|^{n-1}$ |textadj(adj A)| = |A|^(n-1)^2$|\text{adj(adj } A)| = |A|^{(n-1)^2}$ |textadj(adj(adj A))| = |A|^(n-1)^3$|\text{adj(adj(adj } A))| = |A|^{(n-1)^3}$ ### Core Logic Since A$A$ is a 3 times 3$3 \times 3$ matrix (n=3$n=3$): |textadj(adj(adj A))| = |A|^(3-1)^3 = |A|^8 = 81$|\text{adj(adj(adj } A))| = |A|^{(3-1)^3} = |A|^8 = 81$ |A|^8 = 3^4 implies |A|^2 = 3 implies |A| = 3^1/2 = sqrt3$|A|^8 = 3^4 \implies |A|^2 = 3 \implies |A| = 3^{1/2} = \sqrt{3}$ Now look at the power base for the equation: |textadj(adj A)| = |A|^(3-1)^2 = |A|^4$|\text{adj(adj } A)| = |A|^{(3-1)^2} = |A|^4$. Substitute this into the matching requirement equation set: left(|A|^4right)^frac(n-1)^22 = |A|^3n^2 - 5n - 4$\left(|A|^4\right)^{\frac{(n-1)^2}{2}} = |A|^{3n^2 - 5n - 4}$ |A|^2(n-1)^2 = |A|^3n^2 - 5n - 4$|A|^{2(n-1)^2} = |A|^{3n^2 - 5n - 4}$ Equating exponents since bases are identical: 2(n - 1)^2 = 3n^2 - 5n - 4$2(n - 1)^2 = 3n^2 - 5n - 4$ 2(n^2 - 2n + 1) = 3n^2 - 5n - 4$2(n^2 - 2n + 1) = 3n^2 - 5n - 4$ 2n^2 - 4n + 2 = 3n^2 - 5n - 4$2n^2 - 4n + 2 = 3n^2 - 5n - 4$ n^2 - n - 6 = 0$n^2 - n - 6 = 0$ ### Step 1: Solve for Exponent Parameter Factoring the quadratic parameter relation: (n - 3)(n + 2) = 0 implies n = 3 quad textor quad n = -2$(n - 3)(n + 2) = 0 \implies n = 3 \quad \text{or} \quad n = -2$ Both choices are valid integers, so the set S = \-2, 3\$S = \{-2, 3\}$. ### Step 2: Calculate the Target Summation We need to evaluate sum_nin S |A^n^2 + n| = |A^(-2)^2 + (-2)| + |A^(3)^2 + 3|$\sum_{n\in S} |A^{n^2 + n}| = |A^{(-2)^2 + (-2)}| + |A^{(3)^2 + 3}|$: - For n = -2$n = -2$, n^2 + n = 4 - 2 = 2 implies |A^2| = |A|^2 = 3$n^2 + n = 4 - 2 = 2 \implies |A^2| = |A|^2 = 3$ - For n = 3$n = 3$, n^2 + n = 9 + 3 = 12 implies |A^12| = |A|^12 = (sqrt3)^12 = 3^6 = 729$n^2 + n = 9 + 3 = 12 \implies |A^{12}| = |A|^{12} = (\sqrt{3})^{12} = 3^6 = 729$ Summing these evaluated values: textTotal = 3 + 729 = 732$\text{Total} = 3 + 729 = 732$ ### Pattern Recognition Always remember that |A^k| = |A|^k$|A^k| = |A|^k$. Calculating determinant transformations directly as scalar power factors first prevents rendering high order numerical values prematurely. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants

### Related Formula For a system of three linear equations to possess infinitely many solutions, the main determinant Delta$\Delta$ must equal 0$0$, along with all auxiliary determinants: Delta_x = Delta_y = Delta_z = 0$\Delta_x = \Delta_y = \Delta_z = 0$. The perpendicular radius distance from a point (x_0, y_0)$(x_0, y_0)$ to a line Ax + By + C = 0$Ax + By + C = 0$ is: d = frac|Ax_0 + By_0 + C|sqrtA^2 + B^2$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$ ### Core Logic Set the coefficient matrix determinant Delta = 0$\Delta = 0$: left| beginmatrix 2 & 3 & 5 \\ 7 & 3 & -2 \\ 12 & 3 & -(lambda + 4) endmatrix right| = 0$\left| \begin{matrix} 2 & 3 & 5 \\ 7 & 3 & -2 \\ 12 & 3 & -(\lambda + 4) \end{matrix} \right| = 0$ Expanding row layout variations or applying column actions yields: 2[-3(lambda + 4) + 6] - 3[-7(lambda + 4) + 24] + 5[21 - 36] = 0$2[-3(\lambda + 4) + 6] - 3[-7(\lambda + 4) + 24] + 5[21 - 36] = 0$ -6lambda - 24 + 12 + 21lambda + 84 - 72 - 75 = 0$-6\lambda - 24 + 12 + 21\lambda + 84 - 72 - 75 = 0$ 15lambda - 75 = 0 implies lambda = 5$15\lambda - 75 = 0 \implies \lambda = 5$ ### Step 1: Solve for the Constant Parameter Using Cramer's condition for the infinite solution constraint, evaluate the structural augmented row columns component matrix Delta_y = 0$\Delta_y = 0$: left| beginmatrix 2 & 9 & 5 \\ 7 & 8 & -2 \\ 12 & 16 - mu & -9 endmatrix right| = 0$\left| \begin{matrix} 2 & 9 & 5 \\ 7 & 8 & -2 \\ 12 & 16 - \mu & -9 \end{matrix} \right| = 0$ Solving this configuration yields: mu = 9$\mu = 9$ Hence, the circle center is (lambda, mu) = (5, 9)$(\lambda, \mu) = (5, 9)$. ### Step 2: Calculate Radius via Distance The line is given as 4x - 3y = 0$4x - 3y = 0$. Calculate the perpendicular distance from (5, 9)$(5, 9)$ to this line: textRadius = frac|4(5) - 3(9)|sqrt4^2 + (-3)^2 = frac|20 - 27|5 = frac|-7|5 = frac75$\text{Radius} = \frac{|4(5) - 3(9)|}{\sqrt{4^2 + (-3)^2}} = \frac{|20 - 27|}{5} = \frac{|-7|}{5} = \frac{7}{5}$ ### Pattern Recognition Notice that since the y-coefficients are identical (3, 3, 3$3, 3, 3$) across all lines, performing raw row subtractions (R_2 - R_1$R_2 - R_1$ and R_3 - R_2$R_3 - R_2$) strips away the y$y$-variable instantly during structural matrix processing. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants Class 11 Mathematics: Straight Lines

### Related Formula A 2 times 2$2 \times 2$ matrix left[ beginmatrix a & d \\ b & c endmatrix right]$\left[ \begin{matrix} a & d \\ b & c \end{matrix} \right]$ is singular if its determinant equals 0$0$: ad - bc = 0 implies ad = bc$ad - bc = 0 \implies ad = bc$ ### Core Logic Elements must be chosen from the set S = \2, 3, 6, 9\$S = \{2, 3, 6, 9\}$. We need to count all valid quadruplets (a, b, c, d)$(a, b, c, d)$ such that the product of the main diagonal equals the product of the off-diagonal. Let's analyze systematic product matching cases: - **Case 1**: Exactly 1 distinct number is used across all four slots. Example: 2 times 2 = 2 times 2$2 \times 2 = 2 \times 2$. Since there are 4 distinct choices in S$S$, this provides: textWays = ^4C_1 = 4$\text{Ways} = ^4C_1 = 4$ ### Step 1: Evaluate Multi Number Configurations - **Case 2**: Exactly 2 distinct numbers are used. The numbers must pair up to provide identical products (e.g., a=d$a=d$ and b=c$b=c$). Choosing 2 numbers from 4: ^4C_2 = 6$^4C_2 = 6$ pairs. Each pair can be arranged in 4 unique layouts (2 times 2$2 \times 2$ arrangements). textWays = 6 times 4 = 24$\text{Ways} = 6 \times 4 = 24$ ### Step 2: Evaluate Four Distinct Element Sets - **Case 3**: Exactly 3 distinct numbers are used. None will satisfy the strict ad = bc$ad = bc$ zero determinant equality constraints without repeating products. textWays = 0$\text{Ways} = 0$ - **Case 4**: Exactly 4 distinct numbers are used. We need ad = bc$ad = bc$ using the entire set \2, 3, 6, 9\$\{2, 3, 6, 9\}$. Notice that 2 times 9 = 18$2 \times 9 = 18$ and 3 times 6 = 18$3 \times 6 = 18$. This forms a matching product combination pair. The number of unique matrices that can be built by assigning these elements to the diagonal slots is: textWays = 2 times 4 = 8$\text{Ways} = 2 \times 4 = 8$ ### Step 3: Total Summation Sum the total count of valid singular configurations: textTotal Matrices = 4 + 24 + 0 + 8 = 36$\text{Total Matrices} = 4 + 24 + 0 + 8 = 36$ ### Pattern Recognition Always break down element counting problems involving determinants into distinct number-repetition subsets (1 number repeated, 2 numbers repeated, etc.) to ensure no configuration is missed. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants Class 11 Mathematics: Permutations and Combinations

### Related Formula alpha^2 + alpha + 1 = 0 implies alpha = omega quad textwhere omega^3 = 1$\alpha^2 + \alpha + 1 = 0 \implies \alpha = \omega \quad \text{where } \omega^3 = 1$ ### Core Logic Perform row-matrix vector multiplication to generate a system of linear equations in a$a$ and b$b$. Solve for the powers and reduce the algebraic equation using complex roots of unity. ### Step 1: Solve Matrix Vector Multiplication 4 - a - 2b = 0$4 - a - 2b = 0$ 64 - a - 14b = 0$64 - a - 14b = 0$ 52 + 2a - 8b = 0$52 + 2a - 8b = 0$ From the first two equations, subtracting them gives: 60 - 12b = 0 implies b = 5$60 - 12b = 0 \implies b = 5$ Substituting b = 5$b = 5$ into the first equation: 4 - a - 10 = 0 implies a = -6$4 - a - 10 = 0 \implies a = -6$ ### Step 2: Evaluate Exponential Equation with Roots of Unity Substitute a = -6, b = 5$a = -6, b = 5$ into the given equation: frac4alpha^4 + fracmalpha^-6 + fracnalpha^5 = 3 implies frac4omega + m + fracnomega^2 = 3$\frac{4}{\alpha^4} + \frac{m}{\alpha^{-6}} + \frac{n}{\alpha^5} = 3 \implies \frac{4}{\omega} + m + \frac{n}{\omega^2} = 3$ 4omega^2 + m + nomega = 3$4\omega^2 + m + n\omega = 3$ ### Step 3: Resolve Real and Imaginary Components Substitute standard values omega = -frac12 + fracsqrt32i$\omega = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$ and omega^2 = -frac12 - fracsqrt32i$\omega^2 = -\frac{1}{2} - \frac{\sqrt{3}}{2}i$: 4left(-frac12 - fracsqrt32iright) + m + nleft(-frac12 + fracsqrt32iright) = 3$4\left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) + m + n\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) = 3$ Equating the imaginary components: frac-4sqrt32 + fracnsqrt32 = 0 implies n = 4$\frac{-4\sqrt{3}}{2} + \frac{n\sqrt{3}}{2} = 0 \implies n = 4$ Equating the real components: -2 + m - fracn2 = 3 implies -2 + m - 2 = 3 implies m = 7$-2 + m - \frac{n}{2} = 3 \implies -2 + m - 2 = 3 \implies m = 7$ m + n = 7 + 4 = 11$m + n = 7 + 4 = 11$ ### Pattern Recognition Whenever an expression satisfies Aomega^2 + Bomega + C = 0$A\omega^2 + B\omega + C = 0$, it directly maps to a comparison with the standard identity omega^2 + omega + 1 = 0$\omega^2 + \omega + 1 = 0$ up to a linear translation shift. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants Class 11 Mathematics: Complex Numbers