If A = begin{bmatrix} -2 & 2 -3 & 2 end{bmat | vedclass

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

Question

(A) We know that $(B^{-1} A^{-1})^{-1} = (A^{-1})^{-1} (B^{-1})^{-1} = AB$. Given $A = \begin{bmatrix} -2 & 2 \ -3 & 2 \end{bmatrix}$ and $B = \begin

Vedclass · Accepted Answer

$\begin{bmatrix} 2 & 2 \ 2 & 3 \end{bmatrix}$

Vedclass · Answer

(A) We know that $(B^{-1} A^{-1})^{-1} = (A^{-1})^{-1} (B^{-1})^{-1} = AB$. Given $A = \begin{bmatrix} -2 & 2 \ -3 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}$. Now,calculate the product $AB$: $AB = \begin{bmatrix} -2 & 2 \ -3 & 2 \end{bmatrix} \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}$ $AB = \begin{bmatrix} (-2)(0) + (2)(1) & (-2)(-1) + (2)(0) \ (-3)(0) + (2)(1) & (-3)(-1) + (2)(0) \end{bmatrix}$ $AB = \begin{bmatrix} 0 + 2 & 2 + 0 \ 0 + 2 & 3 + 0 \end{bmatrix} = \begin{bmatrix} 2 & 2 \ 2 & 3 \end{bmatrix}$.

If $A = \begin{bmatrix} -2 & 2 \\ -3 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$,then $(B^{-1} A^{-1})^{-1}$ is equal to

Similar Questions

$A$ disc has mass $m$ and radius $R$. How much tangential force should be applied to the rim of the disc so as to rotate it with angular velocity $\omega$ in time $t$?

The distance between the pair of lines $x^2+2 \sqrt{2} x y+2 y^2+4 x+4 \sqrt{2} y+1=0$ is:

Which of the following aqueous solutions will result in the maximum number of moles of metal deposited on passing the same amount of charge?

The direction ratios of the normal to the plane passing through $(0,0,1)$, $(0,1,2)$, and $(1,0,3)$ are

$A$ wire has a resistance of $12 \, \Omega$. It is bent in the form of an equilateral triangle. The effective resistance between any two corners of the triangle is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module