Let \(\mathrm{A}=\left[\begin{array}{cc}1 & 2 \\ -1 & 4\end{array}\right]\) and \(\mathrm{A}^{-1}=\alpha \mathrm{I}+\beta \mathrm{A}, \alpha, \beta \in \mathbb{R}\), I is the identity matrix of order 2 , then \(4(\alpha-\beta)\) is

\(|A|=\left|\begin{array}{cc}
1 & 2 \\
-1 & 4
\end{array}\right|=4+2=6 \neq 0\)
If \(A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\) and \(\mathrm{ad}-\mathrm{bc} \neq 0\), then
\(\begin{aligned}
A^{-1} & =\frac{1}{a d-b c}\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right] \\
\therefore \quad A^{-1} & =\frac{1}{6}\left[\begin{array}{cc}
4 & -2 \\
1 & 1
\end{array}\right] .
\end{aligned}\)
\(\begin{aligned} & A^{-1}=\alpha I+\beta A \\ & \Rightarrow \frac{1}{6}\left[\begin{array}{cc}4 & -2 \\ 1 & 1\end{array}\right]=\left[\begin{array}{cc}\alpha & 0 \\ 0 & \alpha\end{array}\right]+\left[\begin{array}{cc}\beta & 2 \beta \\ -\beta & 4 \beta\end{array}\right] \\ & \Rightarrow\left[\begin{array}{cc}\frac{2}{3} & -\frac{1}{3} \\ \frac{1}{6} & \frac{1}{6}\end{array}\right]=\left[\begin{array}{cc}\alpha+\beta & 2 \beta \\ -\beta & \alpha+4 \beta\end{array}\right]\end{aligned}\)
\(\therefore \quad\) By the equality of matrices,
\(\begin{aligned}
& \frac{1}{6}=-\beta \text { and } \alpha+\beta=\frac{2}{3} \\
& \Rightarrow \beta=\frac{-1}{6} \text { and } \alpha-\frac{1}{6}=\frac{2}{3} \\
& \Rightarrow \beta=\frac{-1}{6} \text { and } \alpha=\frac{5}{6} \\
\therefore \quad & 4(\alpha-\beta)=4
\end{aligned}\)