1. Tardigrade
  2. Question
  3. Mathematics
  4. If D = diag (d1,d2, ...,dn) where d1 ≠ 0, for i = 1, 2, ..., n, then D-1 is equal to

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. If D = diag $\left(d_{1},d_{2}, ...,d_{n}\right)$ where $d_{1} \ne 0$, for i = 1, 2, ..., n, then $D^{-1}$ is equal to

VITEEEVITEEE 2008

Solution:

D = diag $\left(d_{1},d_{2}, ...d_{n}\right)$
$D=\begin{vmatrix}d_{1}&0&0&-&-&0\\ 0&d_{2}&0&-&-&0\\ 0&0&d_{3}&-&-&0\\ -&-&-&-&-&-\\ -&-&-&-&-&-\\ 0&0&0&0&-&0\end{vmatrix}$
$\left|D\right|=\begin{vmatrix}d_{1}&0&0&-&-&0\\ 0&d_{2}&0&-&-&0\\ 0&0&d_{3}&-&-&0\\ -&-&-&-&-&-\\ -&-&-&-&-&-\\ 0&0&0&0&-&0\end{vmatrix}$
$\quad\quad=d_{1}d_{2}d_{3}.......d_{n}$
$adj\left(D\right)=\begin{bmatrix}d_{2}d_{3}d_{4}...dn&0&0&-&-&0\\ 0&d_{1}d_{3}d_{4}...dn&0&-&-&0\\ 0&0&d_{1}d_{2}d_{4}...dn&-&-&0\\ -&-&-&-&-&-\\ -&-&-&-&-&-\\ 0&0&0&0&-&d_{1}d_{2}d_{3}...d_{n-1}\end{bmatrix}$
$adj\left(D\right)=\frac{1}{d_{1}d_{2}d_{3}....d_{n}}$
$\begin{bmatrix}d_{2}d_{3}d_{4}...dn&0&0&-&-&0\\ 0&d_{1}d_{3}d_{4}...dn&0&-&-&0\\ 0&0&d_{1}d_{2}d_{4}...dn&-&-&0\\ -&-&-&-&-&-\\ -&-&-&-&-&-\\ 0&0&0&0&-&d_{1}d_{2}d_{3}...d_{n-1}\end{bmatrix}$
$=\begin{bmatrix}\frac{1}{d_{1}}&0&0&-&-&0\\ 0&\frac{1}{d_{2}}&0&-&-&0\\ 0&0&\frac{1}{d_{3}}&-&-&0\\ -&-&-&-&-&-\\ -&-&-&-&-&-\\ 0&0&0&0&-&\frac{1}{d_{n}}\end{bmatrix}$
$=\begin{bmatrix}d^{ -1}_{1}&0&0&-&-&0\\ 0&d^{ -1}&0&-&-&0\\ 0&0&d^{ -1}_{3}&-&-&0\\ -&-&-&-&-&-\\ -&-&-&-&-&-\\ 0&0&0&0&-&d^{ -1}_{n}\end{bmatrix}$
$=diag\left(d^{ -1}_{1}, d^{ -1}_{2},....,d^{ -1}_{n}\right)$