The inverse of begin{bmatrix} 2 & -3 -4 & 2 | vedclass

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Question

(A) Let $A = \begin{bmatrix} 2 & -3 \ -4 & 2 \end{bmatrix}$.First,we find the determinant $|A| = (2)(2) - (-3)(-4) = 4 - 12 = -8$.Since $|A| 
eq 0$,

Vedclass · Accepted Answer

$-\frac{1}{8} \begin{bmatrix} 2 & 3 \ 4 & 2 \end{bmatrix}$

Vedclass · Answer

(A) Let $A = \begin{bmatrix} 2 & -3 \ -4 & 2 \end{bmatrix}$.First,we find the determinant $|A| = (2)(2) - (-3)(-4) = 4 - 12 = -8$.Since $|A| 
eq 0$,the inverse exists.The adjoint of a $2 	imes 2$ matrix $\begin{bmatrix} a & b \ c & d \end{bmatrix}$ is $\begin{bmatrix} d & -b \ -c & a \end{bmatrix}$.Thus,$adj(A) = \begin{bmatrix} 2 & 3 \ 4 & 2 \end{bmatrix}$.The inverse is given by $A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{-8} \begin{bmatrix} 2 & 3 \ 4 & 2 \end{bmatrix} = -\frac{1}{8} \begin{bmatrix} 2 & 3 \ 4 & 2 \end{bmatrix}$.

The inverse of $\begin{bmatrix} 2 & -3 \\ -4 & 2 \end{bmatrix}$ is

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