If A = begin{bmatrix} 2 & 3 1 & 2 end{bmatri | vedclass

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

Question

(A) We know that $(AB)^{-1} = B^{-1} A^{-1}$.First,we find the product $AB$:$AB = \begin{bmatrix} 2 & 3 \ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \

Vedclass · Accepted Answer

$\begin{bmatrix} 2 & -3 \ -7 & 11 \end{bmatrix}$

Vedclass · Answer

(A) We know that $(AB)^{-1} = B^{-1} A^{-1}$.First,we find the product $AB$:$AB = \begin{bmatrix} 2 & 3 \ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \ 3 & 1 \end{bmatrix} = \begin{bmatrix} (2)(1) + (3)(3) & (2)(0) + (3)(1) \ (1)(1) + (2)(3) & (1)(0) + (2)(1) \end{bmatrix} = \begin{bmatrix} 11 & 3 \ 7 & 2 \end{bmatrix}$.Now,we find $(AB)^{-1} = \frac{1}{|AB|} 	ext{adj}(AB)$.$|AB| = (11)(2) - (3)(7) = 22 - 21 = 1$.$	ext{adj}(AB) = \begin{bmatrix} 2 & -3 \ -7 & 11 \end{bmatrix}$.Therefore,$B^{-1} A^{-1} = (AB)^{-1} = \frac{1}{1} \begin{bmatrix} 2 & -3 \ -7 & 11 \end{bmatrix} = \begin{bmatrix} 2 & -3 \ -7 & 11 \end{bmatrix}$.

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

Similar Questions

If $A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}$ such that $A^2 - 4A + 3I = 0$,where $I$ is a unit matrix of order $2$,then $A^{-1}$ is

Let $A = (a_{ij})_{1 \leq i, j \leq 3}$ be a $3 \times 3$ invertible matrix where each $a_{ij}$ is a real number. Denote the inverse of the matrix $A$ by $A^{-1}$. If $\sum_{j=1}^3 a_{ij} = 1$ for $1 \leq i \leq 3$,then:

The inverse of the matrix $A = \begin{bmatrix} 1 + pq & p & 0 \\ q & 1 + pq & p \\ 0 & q & 1 \end{bmatrix}$ is

If $A = \begin{bmatrix} 3 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 7 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{bmatrix}$, then find $(2A + 3B)^{-1}$.

If $A = \begin{bmatrix} \cos \theta & -\sin \theta \\ -\sin \theta & -\cos \theta \end{bmatrix}$,then $A^{-1} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module