Inverse of the matrix begin{bmatrix} 1 & -2 | vedclass

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

Question

(A) Let $A = \begin{bmatrix} 1 & -2 \ 3 & 4 \end{bmatrix}$.First,calculate the determinant of $A$:$|A| = (1)(4) - (-2)(3) = 4 + 6 = 10$.Since $|A|

Vedclass · Accepted Answer

$\frac{1}{10} \begin{bmatrix} 4 & 2 \ -3 & 1 \end{bmatrix}$

Vedclass · Answer

(A) Let $A = \begin{bmatrix} 1 & -2 \ 3 & 4 \end{bmatrix}$.First,calculate the determinant of $A$:$|A| = (1)(4) - (-2)(3) = 4 + 6 = 10$.Since $|A| 
eq 0$,the inverse $A^{-1}$ exists.The adjoint of $A$,$adj(A)$,is obtained by interchanging the diagonal elements and changing the signs of the off-diagonal elements:$adj(A) = \begin{bmatrix} 4 & 2 \ -3 & 1 \end{bmatrix}$.The inverse is given by $A^{-1} = \frac{1}{|A|} adj(A)$:$A^{-1} = \frac{1}{10} \begin{bmatrix} 4 & 2 \ -3 & 1 \end{bmatrix}$.

Inverse of the matrix $\begin{bmatrix} 1 & -2 \\ 3 & 4 \end{bmatrix}$ is

Similar Questions

If $A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 1 \\ 1 & 2 & 1 \end{bmatrix}$, then $|\operatorname{Adj}(A^2)| = $

Find the inverse of the matrix,if it exists: $\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]$

If $A = \begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix}$,then $\text{adj}(3A^2 + 12A)$ is equal to

The element in the first row and third column of the inverse of the matrix $\begin{bmatrix} 1 & 2 & -3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix}$ is

For the matrix $A = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}$,find $A^{-1}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module