Question

If $A=\left[\begin{array}{c}45\\ 21\end{array}\right]$, then show that $A^{-1}=\frac{1}{6}\left(A-5I\right)$.

Answer 1

$A=\left[\begin{array}{cc}4& 5\\ 2& 1\end{array}\right]$

For characterictic equation of $A$

$|A-\lambda I|=0$

$\left[\begin{array}{cc}4-\lambda & 5\\ 2& 1-\lambda \end{array}\right]=0$

$(4-\lambda )(1-\lambda )-10=0$

${\lambda }^2-5\lambda +4-10=0$

${\lambda }^2=5\lambda -6=0$

Acc to cayley Hamiltm theorem, every square matrix satisfied its own characteristic equation

$A^2-5A-6I=0$

$-6I=5A-A^2$

Pre multiply by $A^{-1}$

$-6A-1=(5I-A)$

$A^{-1}=\frac{-1}{6}(5I-A)$

$A^{-1}=\frac{1}{6}(A-5I)$