Let A = begin{bmatrix} 1 & 2 1 & alpha end{b | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $A^2 - 4A + I = O$. By Cayley-Hamilton theorem, the characteristic equation of $A$ is $\lambda^2 - 	ext{tr}(A)\lambda + \det(A) = 0$. Here

Vedclass · Accepted Answer

only (S2) is correct

Vedclass · Answer

(B) Given $A^2 - 4A + I = O$. By Cayley-Hamilton theorem, the characteristic equation of $A$ is $\lambda^2 - 	ext{tr}(A)\lambda + \det(A) = 0$. Here $	ext{tr}(A) = 1 + \alpha$ and $\det(A) = \alpha - 2$. So, $\lambda^2 - (1 + \alpha)\lambda + (\alpha - 2) = 0$. Comparing with $A^2 - 4A + I = O$, we get $1 + \alpha = 4 \Rightarrow \alpha = 3$ and $\alpha - 2 = 1 \Rightarrow \alpha = 3$. Thus, $A = \begin{bmatrix} 1 & 2 \ 1 & 3 \end{bmatrix}$. For $B^2 - 5B - 6I = O$, the characteristic equation is $\lambda^2 - 	ext{tr}(B)\lambda + \det(B) = 0$. Here $	ext{tr}(B) = 3 + 2 = 5$ and $\det(B) = 6 - 3\beta$. Comparing with $B^2 - 5B - 6I = O$, we get $	ext{tr}(B) = 5$ (which is $5=5$) and $\det(B) = -6$. So, $6 - 3\beta = -6 \Rightarrow 3\beta = 12 \Rightarrow \beta = 4$. Thus, $B = \begin{bmatrix} 3 & 3 \ 4 & 2 \end{bmatrix}$. Now, $A + B = \begin{bmatrix} 1+3 & 2+3 \ 1+4 & 3+2 \end{bmatrix} = \begin{bmatrix} 4 & 5 \ 5 & 5 \end{bmatrix}$. $\det(A + B) = (4)(5) - (5)(5) = 20 - 25 = -5$. For (S2): $\det(	ext{adj}(A + B)) = (\det(A + B))^{n-1} = (-5)^{2-1} = -5$. Thus, (S2) is correct. For (S1): $B - A = \begin{bmatrix} 2 & 1 \ 3 & -1 \end{bmatrix}$, $B + A = \begin{bmatrix} 4 & 5 \ 5 & 5 \end{bmatrix}$. $(B - A)(B + A) = \begin{bmatrix} 2 & 1 \ 3 & -1 \end{bmatrix} \begin{bmatrix} 4 & 5 \ 5 & 5 \end{bmatrix} = \begin{bmatrix} 8+5 & 10+5 \ 12-5 & 15-5 \end{bmatrix} = \begin{bmatrix} 13 & 15 \ 7 & 10 \end{bmatrix}$. Transpose $[(B - A)(B + A)]^T = \begin{bmatrix} 13 & 7 \ 15 & 10 \end{bmatrix}$. This does not match the given matrix in (S1). Thus, (S1) is wrong.

Similar Questions

Let $\alpha$ be a root of the equation $x^{2}+x+1=0$ and the matrix $A=\frac{1}{\sqrt{3}}\begin{bmatrix} 1 & 1 & 1 \\ 1 & \alpha & \alpha^{2} \\ 1 & \alpha^{2} & \alpha^{4} \end{bmatrix}$,then the matrix $A^{31}$ is equal to

The number of singular matrices of order $2 \times 2$,whose elements are from the set $\{2, 3, 6, 9\}$ is

If $A = \begin{bmatrix} -1 & x & -3 \\ 2 & 4 & z \\ y & 5 & -6 \end{bmatrix}$ is a symmetric matrix and $B = \begin{bmatrix} 0 & 2 & q \\ p & 0 & -4 \\ -3 & r & s \end{bmatrix}$ is a skew-symmetric matrix,then $|A| + |B| - |AB| = $

Let for some real numbers $\alpha$ and $\beta$,$a = \alpha - i \beta$. If the system of equations $4ix + (1 + i)y = 0$ and $8(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3})x + \bar{a}y = 0$ has more than one solution,then $\frac{\alpha}{\beta}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module