Inverse of the matrix left[begin{array}{cc}0. | vedclass

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(A) Let $A = \left[\begin{array}{cc}0.8 & -0.6 \ 0.6 & 0.8\end{array}ight]$.The determinant of $A$ is given by $|A| = (0.8)(0.8) - (-0.6)(0.6) = 0.

Vedclass · Accepted Answer

$\left[\begin{array}{cc}0.8 & 0.6 \ -0.6 & 0.8\end{array}ight]$

Vedclass · Answer

(A) Let $A = \left[\begin{array}{cc}0.8 & -0.6 \ 0.6 & 0.8\end{array}ight]$.The determinant of $A$ is given by $|A| = (0.8)(0.8) - (-0.6)(0.6) = 0.64 + 0.36 = 1$.Since $|A| 
eq 0$,the inverse $A^{-1}$ exists.For a $2 	imes 2$ matrix $A = \left[\begin{array}{cc}a & b \ c & d\end{array}ight]$,the inverse is given by $A^{-1} = \frac{1}{|A|} \left[\begin{array}{cc}d & -b \ -c & a\end{array}ight]$.Substituting the values,we get $A^{-1} = \frac{1}{1} \left[\begin{array}{cc}0.8 & 0.6 \ -0.6 & 0.8\end{array}ight] = \left[\begin{array}{cc}0.8 & 0.6 \ -0.6 & 0.8\end{array}ight]$.

Inverse of the matrix $\left[\begin{array}{cc}0.8 & -0.6 \\ 0.6 & 0.8\end{array}\right]$ is

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