Question

If A = diag (a, b, c) = $\left[\begin{array}{ccc}a& 0& 0\\ 0& b& 0\\ 0& 0& c\end{array}\right]$ such that $abc\ne 0$
then $A^{-1}=diag(\frac{1}{a},\frac{1}{b},\frac{1}{c})=\left[\begin{array}{ccc}\frac{1}{a}& 0& 0\\ 0& \frac{1}{b}& 0\\ 0& 0& \frac{1}{c}\end{array}\right]$

Answer 1

|A| = abc $\ne$ 0 and hence A is non-singular whose inverse will exit. Now $A^{-1}=\frac{adj.\left(A\right)}{|A|}$
If C be the matrix formed by cofactors of the elements of A then C = $\left[\begin{array}{ccc}bc& 0& 0\\ 0& ca& 0\\ 0& 0& ab\end{array}\right]$
$\therefore$ adj. A = Transpose of C = $\left[\begin{array}{ccc}bc& 0& 0\\ 0& ca& 0\\ 0& 0& ab\end{array}\right]$
$Rows\to colandCol\to rows$
$\therefore A^{-1}=\frac{1}{|A|}Adj.A=\frac{1}{abc}\left[\begin{array}{ccc}bc& 0& 0\\ 0& ca& 0\\ 0& 0& ab\end{array}\right]$
=$\left[\begin{array}{ccc}\frac{1}{a}& 0& 0\\ 0& \frac{1}{b}& 0\\ 0& 0& \frac{1}{c}\end{array}\right]$