Let $\mathrm{A}=\left[\begin{array}{cc}1 & 2 \ -1 & 4\end{array} ight]$ 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

Question

Let $\mathrm{A}=\left[\begin{array}{cc}1 & 2 \ -1 & 4\end{array}ight]$ 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

Aicharya · Accepted Answer

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

Let $\mathrm{A}=\left[\begin{array}{cc}1 & 2 \\ -1 & 4\end{array}\right]$ and $\mathrm{A}^{-1}=\alpha…

Solution