The inverse of the matrix begin{bmatrix} 3 & | vedclass

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(A) Let $A = \begin{bmatrix} 3 & -2 \ 1 & 4 \end{bmatrix}$.The determinant of $A$ is $|A| = (3)(4) - (-2)(1) = 12 + 2 = 14$.The adjoint of $A$ is obt

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

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(A) Let $A = \begin{bmatrix} 3 & -2 \ 1 & 4 \end{bmatrix}$.The determinant of $A$ is $|A| = (3)(4) - (-2)(1) = 12 + 2 = 14$.The adjoint of $A$ is obtained by swapping the diagonal elements and changing the signs of the off-diagonal elements: $adj(A) = \begin{bmatrix} 4 & 2 \ -1 & 3 \end{bmatrix}$.The inverse of the matrix is given by $A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{14} \begin{bmatrix} 4 & 2 \ -1 & 3 \end{bmatrix} = \begin{bmatrix} \frac{4}{14} & \frac{2}{14} \ \frac{-1}{14} & \frac{3}{14} \end{bmatrix}$.

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

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