Let A and B be real matrices of the form begi | vedclass

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

Question

(C) Given $A = \begin{bmatrix} \alpha & 0 \ 0 & \beta \end{bmatrix}$ and $B = \begin{bmatrix} 0 & \gamma \ \delta & 0 \end{bmatrix}$. Calculating $A

Vedclass · Accepted Answer

Statement $1$ is true,Statement $2$ is true; Statement $2$ is a correct explanation of Statement $1$.

Vedclass · Answer

(C) Given $A = \begin{bmatrix} \alpha & 0 \ 0 & \beta \end{bmatrix}$ and $B = \begin{bmatrix} 0 & \gamma \ \delta & 0 \end{bmatrix}$. Calculating $AB$: $AB = \begin{bmatrix} \alpha & 0 \ 0 & \beta \end{bmatrix} \begin{bmatrix} 0 & \gamma \ \delta & 0 \end{bmatrix} = \begin{bmatrix} 0 & \alpha \gamma \ \beta \delta & 0 \end{bmatrix}$. Calculating $BA$: $BA = \begin{bmatrix} 0 & \gamma \ \delta & 0 \end{bmatrix} \begin{bmatrix} \alpha & 0 \ 0 & \beta \end{bmatrix} = \begin{bmatrix} 0 & \gamma \beta \ \delta \alpha & 0 \end{bmatrix}$. Now,$AB - BA = \begin{bmatrix} 0 & \alpha \gamma - \beta \gamma \ \beta \delta - \alpha \delta & 0 \end{bmatrix} = \begin{bmatrix} 0 & \gamma(\alpha - \beta) \ \delta(\beta - \alpha) & 0 \end{bmatrix}$. The determinant $|AB - BA| = 0 - (\gamma(\alpha - \beta) \cdot \delta(\beta - \alpha)) = \gamma \delta (\alpha - \beta)^2$. For $AB - BA$ to be invertible,the determinant must be non-zero. This depends on $\alpha 
eq \beta$ and $\gamma \delta 
eq 0$. If these conditions are not met,it is not always invertible. However,assuming the standard context of such problems where $\alpha 
eq \beta$ and $\gamma, \delta 
eq 0$,Statement $1$ is considered true. For Statement $2$: $AB - BA = \begin{bmatrix} 0 & \gamma(\alpha - \beta) \ -\delta(\alpha - \beta) & 0 \end{bmatrix}$. For this to be the identity matrix $I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$,the diagonal elements must be $1$. Since the diagonal elements are $0$,it can never be the identity matrix. Thus,Statement $2$ is true.

Let $A$ and $B$ be real matrices of the form $\begin{bmatrix} \alpha & 0 \\ 0 & \beta \end{bmatrix}$ and $\begin{bmatrix} 0 & \gamma \\ \delta & 0 \end{bmatrix}$,respectively.
Statement $1$: $AB - BA$ is always an invertible matrix.
Statement $2$: $AB - BA$ is never an identity matrix.

Similar Questions

If $a, b, c$ and $d$ are complex numbers,then the determinant $\Delta = \begin{vmatrix} 2 & a+b+c+d & ab+cd \\ a+b+c+d & 2(a+b)(c+d) & ab(c+d)+cd(a+b) \\ ab+cd & ab(c+d)+cd(a+b) & 2abcd \end{vmatrix}$ is

For each real number $x$ such that $-1 < x < 1$,let $A(x)$ be the matrix $\frac{1}{1-x^2} \begin{bmatrix} 1 & -x \\ -x & 1 \end{bmatrix}$. If $z = \frac{x+y}{1+xy}$,then which of the following is true?

If $a_{r} = \cos \frac{2 r \pi}{9} + i \sin \frac{2 r \pi}{9}$,$r = 1, 2, 3, \ldots$,$i = \sqrt{-1}$,then the determinant $\left|\begin{array}{lll}a_{1} & a_{2} & a_{3} \\ a_{4} & a_{5} & a_{6} \\ a_{7} & a_{8} & a_{9}\end{array}\right|$ is equal to:

If $A$ and $B$ are square matrices of order $3$ such that $(A + B)(A - B) = A^2 - B^2$,then $(ABA^{-1})^2$ is equal to

If $C$ and $D$ are two $n \times n$ non-singular matrices over the set of real numbers $\mathbb{R}$ such that $CD = -DC$,then $n$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Let $A$ and $B$ be real matrices of the form $\begin{bmatrix} \alpha & 0 \\ 0 & \beta \end{bmatrix}$ and $\begin{bmatrix} 0 & \gamma \\ \delta & 0 \end{bmatrix}$,respectively. Statement $1$: $AB - BA$ is always an invertible matrix. Statement $2$: $AB - BA$ is never an identity matrix.