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 $M$ be a $3 \times 3$ invertible matrix with real entries and let $I$ denote the $3 \times 3$ identity matrix. If $M^{-1} = \operatorname{adj}(\operatorname{adj} M)$,then which of the following statement$(s)$ is/are $ALWAYS \text{ } TRUE$?

For $\alpha, \beta \in R$ and a natural number $n$,let $A_r = \begin{vmatrix} r & 1 & \frac{n^2}{2} + \alpha \\ 2r & 2 & n^2 - \beta \\ 3r - 2 & 3 & \frac{n(3n - 1)}{2} \end{vmatrix}$. Then $2A_{10} - A_8$ is equal to:

If $A$ is a $2 \times 2$ matrix such that $\operatorname{det} A = -21$ and $\operatorname{trace}(A^3) = 2024$, then the trace of $A$ is

If the matrix $M_r$ is given by $M_r = \begin{bmatrix} r & r-1 \\ r-1 & r \end{bmatrix}$ for $r = 1, 2, 3, \ldots$, then $\det(M_1) + \det(M_2) + \ldots + \det(M_{2008}) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module