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(A) Let $A = \left[\begin{array}{cc}1 & -1 \ 2 & 3\end{array}ight]$.First,we find the determinant of $A$:$|A| = (1)(3) - (-1)(2) = 3 + 2 = 5$.Since

Vedclass · Accepted Answer

$A^{-1}=\left[\begin{array}{cc}\frac{3}{5} & \frac{1}{5} \ -\frac{2}{5} & \frac{1}{5}\end{array}ight]$

Vedclass · Answer

(A) Let $A = \left[\begin{array}{cc}1 & -1 \ 2 & 3\end{array}ight]$.First,we find the determinant of $A$:$|A| = (1)(3) - (-1)(2) = 3 + 2 = 5$.Since $|A| 
eq 0$,the inverse $A^{-1}$ exists.We use the formula $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$.The matrix of cofactors is:$C_{11} = 3, C_{12} = -2, C_{21} = 1, C_{22} = 1$.Thus,$	ext{adj}(A) = \left[\begin{array}{cc}3 & 1 \ -2 & 1\end{array}ight]^T = \left[\begin{array}{cc}3 & -2 \ 1 & 1\end{array}ight]$.Wait,let's re-calculate the adjoint correctly:$C_{11} = 3, C_{12} = -2, C_{21} = -(-1) = 1, C_{22} = 1$.So,$	ext{adj}(A) = \left[\begin{array}{cc}3 & 1 \ -2 & 1\end{array}ight]$.Therefore,$A^{-1} = \frac{1}{5} \left[\begin{array}{cc}3 & 1 \ -2 & 1\end{array}ight] = \left[\begin{array}{cc}\frac{3}{5} & \frac{1}{5} \ -\frac{2}{5} & \frac{1}{5}\end{array}ight]$.

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

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