(S1): $[(B-A)(B+A)]^T = \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}$
and
(S2): $\det(\text{adj}(A+B)) = -5$,
Matrix A Determination
Given $A = \begin{bmatrix} 1 & 2 \\ 1 & \alpha \end{bmatrix}$ and the equation $A^2 - 4A + I = O$. The Cayley-Hamilton theorem states that $A^2 - (\text{tr}(A))A + (\det(A))I = O$. Comparing the given equation with the theorem:
- Trace: $\text{tr}(A) = 1 + \alpha = 4$, which implies $\alpha = 3$.
- Determinant: $\det(A) = (1)(\alpha) - (2)(1) = \alpha - 2 = 1$. Substituting $\alpha = 3$, we get $3 - 2 = 1$, confirming consistency.
Therefore, matrix $A = \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix}$.
Matrix B Determination
Given $B = \begin{bmatrix} 3 & 3 \\ \beta & 2 \end{bmatrix}$ and the equation $B^2 - 5B - 6I = O$. Applying the Cayley-Hamilton theorem ($B^2 - (\text{tr}(B))B + (\det(B))I = O$):
- Trace: $\text{tr}(B) = 3 + 2 = 5$. This matches the coefficient of $B$.
- Determinant: $\det(B) = (3)(2) - (3)(\beta) = 6 - 3\beta$. From the equation, $\det(B)$ must be $-6$.
- Equating determinants: $6 - 3\beta = -6 \implies 3\beta = 12 \implies \beta = 4$.
Therefore, matrix $B = \begin{bmatrix} 3 & 3 \\ 4 & 2 \end{bmatrix}$.
Statement (S1) Analysis
Statement (S1) claims that $[(B-A)(B+A)]^T = \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}$.
- Calculate $B-A$: $B-A = \begin{bmatrix} 3 & 3 \\ 4 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix}$.
- Calculate $B+A$: $B+A = \begin{bmatrix} 3 & 3 \\ 4 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 5 \\ 5 & 5 \end{bmatrix}$.
- Compute the product $(B-A)(B+A)$: $(B-A)(B+A) = \begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} 4 & 5 \\ 5 & 5 \end{bmatrix} = \begin{bmatrix} (8+5) & (10+5) \\ (12-5) & (15-5) \end{bmatrix} = \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}$.
- Compute the transpose: $[(B-A)(B+A)]^T = \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}^T = \begin{bmatrix} 13 & 7 \\ 15 & 10 \end{bmatrix}$.
The calculated transpose $\begin{bmatrix} 13 & 7 \\ 15 & 10 \end{bmatrix}$ does not match the matrix $\begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}$ in statement (S1). Hence, (S1) is incorrect.
Statement (S2) Analysis
Statement (S2) claims that $\det(\text{adj}(A+B)) = -5$.
- Calculate $A+B$: $A+B = \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix} + \begin{bmatrix} 3 & 3 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 4 & 5 \\ 5 & 5 \end{bmatrix}$.
- Calculate the determinant of $A+B$: $\det(A+B) = (4)(5) - (5)(5) = 20 - 25 = -5$.
- Apply the property $\det(\text{adj}(M)) = (\det(M))^{n-1}$ for an $n \times n$ matrix $M$. For the 2x2 matrix $A+B$: $\det(\text{adj}(A+B)) = (\det(A+B))^{2-1} = \det(A+B)$.
- Result: $\det(\text{adj}(A+B)) = -5$.
This matches the value in statement (S2). Hence, (S2) is correct.
Final Conclusion
Based on the analysis, statement (S1) is incorrect, while statement (S2) is correct. Therefore, only (S2) is correct.
Similar Questions
- Let A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
- Let ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, ..., p_{13}$, is _________
- If $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is a solution of the system of equations $AX = B$, where $\text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$, then $|x + y + z|$ is equal to :
- Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1 + x)^n$, $n \in \mathbb{N}, 0 \leq r \leq n$. If $P_n = C_0 - C_1 + \frac{2^2}{3} C_2 - \frac{2^3}{4} C_3 + \dots + \frac{(-2)^n}{n+1} C_n$, then the value of $\sum_{n=1}^{25} \frac{1}{P_{2n}}$ equals.
- The number of elements in the relation $R = \{(x, y) : 4x^2 + y^2 < 52, x, y \in \mathbb{Z}\}$ is
- Let $S = \{z \in \mathbb{C} : 4z^2 + \bar{z} = 0\}$. Then $\sum_{z \in S} |z|^2$ is equal to :
Let f and g be functions satisfying $f(x+y) = f(x)f(y), f(1) = 7$ and $g(x+y) = g(xy), g(1) = 1$, for all $x, y \in \mathbb{N}$. If $\sum_{x=1}^{n} \left(\frac{f(x)}{g(x)}\right) = 19607$, then n is equal to :
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
- Let $S = \frac{1}{25!} + \frac{1}{3!23!} + \frac{1}{5!21!} + \dots$ up to 13 terms. If $13S = \frac{2^k}{n!}, k \in \mathbb{N}$, then $n + k$ is equal to
Important Questions from Algebra
- Let A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
- Let ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, ..., p_{13}$, is _________
- If $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is a solution of the system of equations $AX = B$, where $\text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$, then $|x + y + z|$ is equal to :
- Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1 + x)^n$, $n \in \mathbb{N}, 0 \leq r \leq n$. If $P_n = C_0 - C_1 + \frac{2^2}{3} C_2 - \frac{2^3}{4} C_3 + \dots + \frac{(-2)^n}{n+1} C_n$, then the value of $\sum_{n=1}^{25} \frac{1}{P_{2n}}$ equals.
