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

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(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|

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

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(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

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