Let for any three distinct consecutive terms a, b, c$a, b, c$ of an A.P, the lines ax + by + c = 0$ax + by + c = 0$ be concurrent at the point P$P$ and Q (alpha, beta)$Q (\alpha, \beta)$ be a point such that the system of equations x + y + z = 6$x + y + z = 6$, 2 x + 5 y + alpha z = beta$2 x + 5 y + \alpha z = \beta$ and x + 2y + 3z = 4$x + 2y + 3z = 4$ has infinitely many solutions. Then (PQ)^2$(PQ)^2$ is equal to

### Related Formula For a 2 times 2$2 \times 2$ matrix M$M$, Cayley-Hamilton equation states: M^2 - operatornameTr(M)M + operatornamedet(M)I = O$$M^2 - \operatorname{Tr}(M)M + \operatorname{det}(M)I = O$$ Also, |operatornameadj(M)| = |M|^n-1$|\operatorname{adj}(M)| = |M|^{n-1}$, and for a 2 times 2$2 \times 2$ matrix, |operatornameadj(M)| = |M|$|\operatorname{adj}(M)| = |M|$. ### Core Logic Using the characteristic equation A^2 - operatornameTr(A)A + det(A)I = O$A^2 - \operatorname{Tr}(A)A + \det(A)I = O$: Comparing with A^2 - 4A + 2I = O$A^2 - 4A + 2I = O$: operatornameTr(A) = 4 Rightarrow alpha + 2 = 4 Rightarrow alpha = 2$\operatorname{Tr}(A) = 4 \Rightarrow \alpha + 2 = 4 \Rightarrow \alpha = 2$ Comparing with B^2 - 3B + I = O$B^2 - 3B + I = O$: operatornameTr(B) = 3 Rightarrow 1 + beta = 3 Rightarrow beta = 2$\operatorname{Tr}(B) = 3 \Rightarrow 1 + \beta = 3 \Rightarrow \beta = 2$ ### Step 1: Compute A^3 and B^3 via reduction A = beginbmatrix 2 & 2 \\ 1 & 2 endbmatrix$$A = \begin{bmatrix} 2 & 2 \\ 1 & 2 \end{bmatrix}$$ A^2 = 4A - 2I$A^2 = 4A - 2I$ A^3 = A(4A - 2I) = 4A^2 - 2A = 4(4A - 2I) - 2A = 14A - 8I$$A^3 = A(4A - 2I) = 4A^2 - 2A = 4(4A - 2I) - 2A = 14A - 8I$$ A^3 = 14 beginbmatrix 2 & 2 \\ 1 & 2 endbmatrix - 8 beginbmatrix 1 & 0 \\ 0 & 1 endbmatrix = beginbmatrix 28 & 28 \\ 14 & 28 endbmatrix - beginbmatrix 8 & 0 \\ 0 & 8 endbmatrix = beginbmatrix 20 & 28 \\ 14 & 20 endbmatrix$$A^3 = 14 \begin{bmatrix} 2 & 2 \\ 1 & 2 \end{bmatrix} - 8 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 28 & 28 \\ 14 & 28 \end{bmatrix} - \begin{bmatrix} 8 & 0 \\ 0 & 8 \end{bmatrix} = \begin{bmatrix} 20 & 28 \\ 14 & 20 \end{bmatrix}$$ Similarly, B = beginbmatrix 1 & 1 \\ 1 & 2 endbmatrix$B = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}$ B^2 = 3B - I$B^2 = 3B - I$ B^3 = 3B^2 - B = 3(3B - I) - B = 8B - 3I$$B^3 = 3B^2 - B = 3(3B - I) - B = 8B - 3I$$ B^3 = 8 beginbmatrix 1 & 1 \\ 1 & 2 endbmatrix - beginbmatrix 3 & 0 \\ 0 & 3 endbmatrix = beginbmatrix 8 & 8 \\ 8 & 16 endbmatrix - beginbmatrix 3 & 0 \\ 0 & 3 endbmatrix = beginbmatrix 5 & 8 \\ 8 & 13 endbmatrix$$B^3 = 8 \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 8 & 8 \\ 8 & 16 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix}$$ ### Step 2: Difference and Determinant A^3 - B^3 = beginbmatrix 20 & 28 \\ 14 & 20 endbmatrix - beginbmatrix 5 & 8 \\ 8 & 13 endbmatrix = beginbmatrix 15 & 20 \\ 6 & 7 endbmatrix$$A^3 - B^3 = \begin{bmatrix} 20 & 28 \\ 14 & 20 \end{bmatrix} - \begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix} = \begin{bmatrix} 15 & 20 \\ 6 & 7 \end{bmatrix}$$ det(A^3 - B^3) = (15 times 7) - (20 times 6) = 105 - 120 = -15$$\det(A^3 - B^3) = (15 \times 7) - (20 \times 6) = 105 - 120 = -15$$ ### Step 3: Final Answer For a 2 times 2$2 \times 2$ matrix, |operatornameadj(M)| = |M|$|\operatorname{adj}(M)| = |M|$. Thus, det(operatornameadj(A^3 - B^3)) = -15$\det(\operatorname{adj}(A^3 - B^3)) = -15$. (det(operatornameadj(A^3 - B^3)))^2 = (-15)^2 = 225$$(\det(\operatorname{adj}(A^3 - B^3)))^2 = (-15)^2 = 225$$ ### Pattern Recognition Do not manually multiply matrices to the 3rd power. Cayley-Hamilton strictly reduces M^3$M^3$ down to a linear combination cM + dI$cM + dI$. Expanding this requires only basic scalar arithmetic. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Matrices

### Related Formula textFor infinitely many solutions: Delta = 0 quad textand quad Delta_i = 0$$\text{For infinitely many solutions: } \Delta = 0 \quad \text{and} \quad \Delta_i = 0$$ ### Core Logic For a system of 3 linear equations to have infinitely many solutions, the determinant of coefficients and all Cramer determinants must equal zero. ### Step 1: Set up determinant equations The determinant of coefficients is: Delta = beginvmatrix 2 & lambda & 3 \\ 3 & 2 & -1 \\ 4 & 5 & mu endvmatrix = 0$$\Delta = \begin{vmatrix} 2 & \lambda & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \mu \end{vmatrix} = 0$$ 2(2mu + 5) - lambda(3mu + 4) + 3(15 - 8) = 0$$2(2\mu + 5) - \lambda(3\mu + 4) + 3(15 - 8) = 0$$ 4mu + 10 - 3lambdamu - 4lambda + 21 = 0$$4\mu + 10 - 3\lambda\mu - 4\lambda + 21 = 0$$ 4mu - 3lambdamu - 4lambda + 31 = 0 quad text--- (1)$$4\mu - 3\lambda\mu - 4\lambda + 31 = 0 \quad \text{--- (1)}$$ Now, set Delta_3 = 0$\Delta_3 = 0$: Delta_3 = beginvmatrix 2 & lambda & 5 \\ 3 & 2 & 7 \\ 4 & 5 & 9 endvmatrix = 0$$\Delta_3 = \begin{vmatrix} 2 & \lambda & 5 \\ 3 & 2 & 7 \\ 4 & 5 & 9 \end{vmatrix} = 0$$ 2(18 - 35) - lambda(27 - 28) + 5(15 - 8) = 0$$2(18 - 35) - \lambda(27 - 28) + 5(15 - 8) = 0$$ -34 + lambda + 35 = 0 implies lambda = -1$$-34 + \lambda + 35 = 0 \implies \lambda = -1$$ ### Step 2: Solve for mu and compute the sum of squares Substitute lambda = -1$\lambda = -1$ into equation (1): 4mu - 3(-1)mu - 4(-1) + 31 = 0$$4\mu - 3(-1)\mu - 4(-1) + 31 = 0$$ 4mu + 3mu + 4 + 31 = 0$$4\mu + 3\mu + 4 + 31 = 0$$ 7mu = -35 implies mu = -5$$7\mu = -35 \implies \mu = -5$$ Now calculate the sum of squares: lambda^2 + mu^2 = (-1)^2 + (-5)^2 = 1 + 25 = 26$$\lambda^2 + \mu^2 = (-1)^2 + (-5)^2 = 1 + 25 = 26$$ ### Pattern Recognition Whenever you need to solve for two variables in Cramer's theorem, identifying which determinant lacks the complex variable (like Delta_3$\Delta_3$ which lacks mu$\mu$) is the fastest way to solve for one variable independently. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants

### Related Formula textCharacteristic equation reduction: mathrmA^3 = 2mathrmA^2 + 4mathrmA - 4mathrmI$$\text{Characteristic equation reduction: } \mathrm{A}^3 = 2\mathrm{A}^2 + 4\mathrm{A} - 4\mathrm{I}$$ ### Core Logic We use the given cubic matrix equation recursively to express the fifth power of matrix A$A$ solely in terms of quadratic and linear terms. ### Step 1: Simplify the cubic matrix equation The given equation is: mathrmA^2 (mathrmA - 2mathrmI) - 4(mathrmA - mathrmI) = mathrmO$$\mathrm{A}^2 (\mathrm{A} - 2\mathrm{I}) - 4(\mathrm{A} - \mathrm{I}) = \mathrm{O}$$ mathrmA^3 - 2mathrmA^2 - 4mathrmA + 4mathrmI = mathrmO implies mathrmA^3 = 2mathrmA^2 + 4mathrmA - 4mathrmI$$\mathrm{A}^3 - 2\mathrm{A}^2 - 4\mathrm{A} + 4\mathrm{I} = \mathrm{O} \implies \mathrm{A}^3 = 2\mathrm{A}^2 + 4\mathrm{A} - 4\mathrm{I}$$ Multiply by matrix A$A$ to find the fourth power: mathrmA^4 = 2mathrmA^3 + 4mathrmA^2 - 4mathrmA$$\mathrm{A}^4 = 2\mathrm{A}^3 + 4\mathrm{A}^2 - 4\mathrm{A}$$ ### Step 2: Reduce the fourth power term Substitute the expression for mathrmA^3$\mathrm{A}^3$ into our formula for mathrmA^4$\mathrm{A}^4$: mathrmA^4 = 2left( 2mathrmA^2 + 4mathrmA - 4mathrmI right) + 4mathrmA^2 - 4mathrmA$$\mathrm{A}^4 = 2\left( 2\mathrm{A}^2 + 4\mathrm{A} - 4\mathrm{I} \right) + 4\mathrm{A}^2 - 4\mathrm{A}$$ mathrmA^4 = 4mathrmA^2 + 8mathrmA - 8mathrmI + 4mathrmA^2 - 4mathrmA = 8mathrmA^2 + 4mathrmA - 8mathrmI$$\mathrm{A}^4 = 4\mathrm{A}^2 + 8\mathrm{A} - 8\mathrm{I} + 4\mathrm{A}^2 - 4\mathrm{A} = 8\mathrm{A}^2 + 4\mathrm{A} - 8\mathrm{I}$$ Multiply by matrix A$A$ to find the fifth power: mathrmA^5 = 8mathrmA^3 + 4mathrmA^2 - 8mathrmA$$\mathrm{A}^5 = 8\mathrm{A}^3 + 4\mathrm{A}^2 - 8\mathrm{A}$$ ### Step 3: Reduce the fifth power term and solve Substitute the expression for mathrmA^3$\mathrm{A}^3$ again: mathrmA^5 = 8left( 2mathrmA^2 + 4mathrmA - 4mathrmI right) + 4mathrmA^2 - 8mathrmA$$\mathrm{A}^5 = 8\left( 2\mathrm{A}^2 + 4\mathrm{A} - 4\mathrm{I} \right) + 4\mathrm{A}^2 - 8\mathrm{A}$$ mathrmA^5 = 16mathrmA^2 + 32mathrmA - 32mathrmI + 4mathrmA^2 - 8mathrmA = 20mathrmA^2 + 24mathrmA - 32mathrmI$$\mathrm{A}^5 = 16\mathrm{A}^2 + 32\mathrm{A} - 32\mathrm{I} + 4\mathrm{A}^2 - 8\mathrm{A} = 20\mathrm{A}^2 + 24\mathrm{A} - 32\mathrm{I}$$ Comparing this with mathrmA^5 = alpha mathrmA^2 + beta mathrmA + gamma mathrmI$\mathrm{A}^5 = \alpha \mathrm{A}^2 + \beta \mathrm{A} + \gamma \mathrm{I}$, we find: - alpha = 20$\alpha = 20$ - beta = 24$\beta = 24$ - gamma = -32$\gamma = -32$ Sum the coefficients: alpha + beta + gamma = 20 + 24 - 32 = 12$$\alpha + \beta + \gamma = 20 + 24 - 32 = 12$$ ### Pattern Recognition Cayley-Hamilton reduction: For any polynomial equation of a matrix, higher powers A^k$A^k$ can always be reduced down to polynomials of order less than the degree of the characteristic equation by recursive substitution. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants

### Related Formula For a matrix satisfying A^2 = A$A^2 = A$ (Idempotent Matrix): (I+A)^n = I + (2^n - 1)A$$(I+A)^n = I + (2^n - 1)A$$ ### Core Logic Given det(A) = alphabeta + 6 = 0 implies alphabeta = -6$\det(A) = \alpha\beta + 6 = 0 \implies \alpha\beta = -6$ and alpha + beta = 1$\alpha + \beta = 1$. Solving these gives alpha = 3, beta = -2$\alpha = 3, \beta = -2$ (since alpha > 0$\alpha > 0$). ### Step 1: Check Powers of A Substitute values into A$A$: A = beginbmatrix 3 & -1 \\ 6 & -2 endbmatrix$$A = \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix}$$ Compute A^2$A^2$: A^2 = beginbmatrix 3 & -1 \\ 6 & -2 endbmatrix beginbmatrix 3 & -1 \\ 6 & -2 endbmatrix = beginbmatrix 9-6 & -3+2 \\ 18-12 & -6+4 endbmatrix = beginbmatrix 3 & -1 \\ 6 & -2 endbmatrix = A$$A^2 = \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix} \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 9-6 & -3+2 \\ 18-12 & -6+4 \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix} = A$$ ### Step 2: Expand Matrix Expression Since A^2 = A$A^2 = A$, it follows that A^n = A$A^n = A$ for all integers n ge 1$n \ge 1$. (I + A)^8 = I + sum_k=1^8 binom8k A^k = I + A sum_k=1^8 binom8k = I + (2^8 - 1)A = I + 255A$$(I + A)^8 = I + \sum_{k=1}^{8} \binom{8}{k} A^k = I + A \sum_{k=1}^{8} \binom{8}{k} = I + (2^8 - 1)A = I + 255A$$ ### Step 3: Construct the Final Matrix (I + A)^8 = beginbmatrix 1 & 0 \\ 0 & 1 endbmatrix + 255 beginbmatrix 3 & -1 \\ 6 & -2 endbmatrix = beginbmatrix 1 + 765 & -255 \\ 1530 & 1 - 510 endbmatrix = beginbmatrix 766 & -255 \\ 1530 & -509 endbmatrix$$(I + A)^8 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + 255 \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 1 + 765 & -255 \\ 1530 & 1 - 510 \end{bmatrix} = \begin{bmatrix} 766 & -255 \\ 1530 & -509 \end{bmatrix}$$ ### Pattern Recognition Whenever texttr(A) = 1$\text{tr}(A) = 1$ and det(A) = 0$\det(A) = 0$ for a 2 times 2$2 \times 2$ matrix, Cayley-Hamilton theorem gives A^2 - texttr(A)A + det(A)I = 0 implies A^2 = A$A^2 - \text{tr}(A)A + \det(A)I = 0 \implies A^2 = A$. Thus A$A$ is idempotent, simplifying polynomial expansions exponentially. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants

### Related Formula Cramer's Rule for infinite solutions specifies that the main determinant and all component determinants must vanish: Delta = 0 quad textand quad Delta_1 = Delta_2 = Delta_3 = 0$$\Delta = 0 \quad \text{and} \quad \Delta_1 = \Delta_2 = \Delta_3 = 0$$ ### Core Logic Set the key system determinants to zero to form equations linking alpha$\alpha$ and beta$\beta$, then isolate the constants. ### Step 1: Set Main Determinant to Zero Delta = beginvmatrix 3 & 1 & beta \\ 2 & alpha & -1 \\ 1 & 2 & 1 endvmatrix = 0$$\Delta = \begin{vmatrix} 3 & 1 & \beta \\ 2 & \alpha & -1 \\ 1 & 2 & 1 \end{vmatrix} = 0$$ Expand along the first row: 3(alpha + 2) - 1(2 + 1) + beta(4 - alpha) = 0$$3(\alpha + 2) - 1(2 + 1) + \beta(4 - \alpha) = 0$$ 3alpha + 6 - 3 + 4beta - alphabeta = 0 implies 3alpha + 4beta - alphabeta + 3 = 0 quad dots (1)$$3\alpha + 6 - 3 + 4\beta - \alpha\beta = 0 \implies 3\alpha + 4\beta - \alpha\beta + 3 = 0 \quad \dots (1)$$ ### Step 2: Set Subsidiary Determinant to Zero Using Delta_3 = 0$\Delta_3 = 0$ by substituting the constants vector into the third column: Delta_3 = beginvmatrix 3 & 1 & 3 \\ 2 & alpha & -3 \\ 1 & 2 & 4 endvmatrix = 0$$\Delta_3 = \begin{vmatrix} 3 & 1 & 3 \\ 2 & \alpha & -3 \\ 1 & 2 & 4 \end{vmatrix} = 0$$ Expand along the first row: 3(4alpha + 6) - 1(8 + 3) + 3(4 - alpha) = 0$$3(4\alpha + 6) - 1(8 + 3) + 3(4 - \alpha) = 0$$ 12alpha + 18 - 11 + 12 - 3alpha = 0 implies 9alpha + 19 = 0 implies alpha = -frac199$$12\alpha + 18 - 11 + 12 - 3\alpha = 0 \implies 9\alpha + 19 = 0 \implies \alpha = -\frac{19}{9}$$ ### Step 3: Solve for Beta and Final Expression Substitute alpha = -frac199$\alpha = -\frac{19}{9}$ into equation (1): 3left(-frac199right) + 4beta - left(-frac199right)beta + 3 = 0$$3\left(-\frac{19}{9}\right) + 4\beta - \left(-\frac{19}{9}\right)\beta + 3 = 0$$ -frac193 + 3 + betaleft(4 + frac199right) = 0 implies -frac103 + betaleft(frac559right) = 0$$-\frac{19}{3} + 3 + \beta\left(4 + \frac{19}{9}\right) = 0 \implies -\frac{10}{3} + \beta\left(\frac{55}{9}\right) = 0$$ frac559beta = frac103 implies beta = frac103 cdot frac955 = frac611$$\frac{55}{9}\beta = \frac{10}{3} \implies \beta = \frac{10}{3} \cdot \frac{9}{55} = \frac{6}{11}$$ Now compute 22beta - 9alpha$22\beta - 9\alpha$: 22left(frac611right) - 9left(-frac199 ight) = 12 + 19 = 31$$22\left(\frac{6}{11}\right) - 9\left(-\frac{19}{9} ight) = 12 + 19 = 31$$ ### Pattern Recognition Choosing Delta_3$\Delta_3$ over Delta_1$\Delta_1$ or Delta_2$\Delta_2$ eliminates beta$\beta$ entirely because the variable parameters are localized in specific positions. This yields alpha$\alpha$ directly without requiring a coupled system solution. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Matrices and Determinants