If A = begin{bmatrix} 2 & -2 4 & 3 end{bmatr | vedclass

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

Question

(B) Given $A = \begin{bmatrix} 2 & -2 \ 4 & 3 \end{bmatrix}$.First,we find the determinant of $A$:$|A| = (2)(3) - (-2)(4) = 6 + 8 = 14$.Since $|A|

Vedclass · Accepted Answer

$\frac{1}{14} \begin{bmatrix} 3 & 2 \ -4 & 2 \end{bmatrix}$

Vedclass · Answer

(B) Given $A = \begin{bmatrix} 2 & -2 \ 4 & 3 \end{bmatrix}$.First,we find the determinant of $A$:$|A| = (2)(3) - (-2)(4) = 6 + 8 = 14$.Since $|A| 
eq 0$,$A^{-1}$ exists.The adjoint of $A$ is obtained by swapping the diagonal elements and changing the signs of the off-diagonal elements:$	ext{Adj } A = \begin{bmatrix} 3 & 2 \ -4 & 2 \end{bmatrix}$.Using the formula $A^{-1} = \frac{1}{|A|} 	ext{Adj } A$:$A^{-1} = \frac{1}{14} \begin{bmatrix} 3 & 2 \ -4 & 2 \end{bmatrix}$.

If $A = \begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}$,then $A^{-1} =$

Similar Questions

If $B$ is the inverse of a third order matrix $A$ and $\det B = k$,then $(\operatorname{adj}(\operatorname{adj} A))^{-1} =$

${\left[ {\begin{array}{*{20}{c}}{ - 6}&5\\{ - 7}&6\end{array}} \right]^{ - 1}}$ =

If $A$ is a square matrix such that $A(\operatorname{adj} A) = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}$, then $\operatorname{det}(\operatorname{adj} A)$ is equal to

If $A$ is an invertible matrix of order $n$, then the determinant of $\operatorname{adj} A$ is equal to :

Find the inverse of the matrix,if it exists: $\left[\begin{array}{cc}3 & 10 \\ 2 & 7\end{array}\right]$

Vedclass Test Series

Exam Paper Generator

Online Exam Module