If\[A = \left[ {\begin{array}{*{20}{c}}
  9&0&0 \\
  0&{10}&0 \\
  0&0&8
\end{array}} \right]\] , then ${A^{ - 1}}$ ,
$
  A)\left[ {\begin{array}{*{20}{c}}
  {\dfrac{1}{9}}&0&0 \\
  0&{10}&0 \\
  0&0&{\dfrac{1}{8}}
\end{array}} \right] \\
  B)\left[ {\begin{array}{*{20}{c}}
  9&0&0 \\
  0&{\dfrac{1}{{10}}}&0 \\
  0&0&8
\end{array}} \right] \\
  C)\left[ {\begin{array}{*{20}{c}}
  9&0&0 \\
  0&{10}&0 \\
  0&0&{\dfrac{1}{8}}
\end{array}} \right] \\
  D)\left[ {\begin{array}{*{20}{c}}
  {\dfrac{1}{9}}&0&0 \\
  0&{\dfrac{1}{{10}}}&0 \\
  0&0&{\dfrac{1}{8}}
\end{array}} \right] \\
$

Hint: In order to find out the inverse of a matrix $A$, we have to use the formula,${A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj(A){\text{ and }}\left| A \right| \ne 0$
Where, ${A^{ - 1}}$ = Inverse of matrix A
$\left| A \right|$= Determinant of Matrix A
And, $Adj(A)$= Adjoint of Matrix A

Complete step by step answer:
Now Firstly, we have to find $\left| A \right|$
That is,
$
  \left| A \right| = 9[(10 \times 8) - (0 \times 0)] - 0[(0 \times 8) - (0 \times 0)] + 0[(0 \times 0) - (10 \times 0)] \\
   = 9[80 - 0] - 0[0 - 0] + 0[0 - 0] \\
   = 9[80] \\
   \Rightarrow \left| A \right| = 720 \\
  \therefore \left| A \right| \ne 0 \\
$
Hence, Inverse exists as it is non-singular matrix
Now, we need to find the Adjoint of Matrix A.
In order to find the Adjoint of Matrix A, we will need to find the cofactors of each of the elements of Matrix A, and then find the transpose of that resultant Matrix.
And cofactor of an element of a matrix can be found using the formula ${C_{ij}} = {( - 1)^{i + j}}\left| {{M_{ij}}} \right|$, where M is the minor of the corresponding element.
Therefore, Cofactor of first element that is 9,
Minor is the determinant of what is left after excluding its corresponding row and column
$
   = {( - 1)^{1 + 1}}\left| {\begin{array}{*{20}{c}}
  {10}&0 \\
  0&8
\end{array}} \right| \\
   = {( - 1)^2}[(10 \times 8) - (0 \times 0)] \\
   = 1 \times (80 - 0) \\
   = 80 \\
$
Co-factor of 0 =
$
   = {( - 1)^{1 + 2}}\left| {\begin{array}{*{20}{c}}
  0&0 \\
  0&8
\end{array}} \right| \\
   = {( - 1)^3}[(0 \times 8) - (0 \times 0)] \\
   = - 1 \times (0 - 0) \\
   = 0 \\
$

Co-factor of 0
$
   = {( - 1)^{1 + 2}}\left| {\begin{array}{*{20}{c}}
  0&{10} \\
  0&0
\end{array}} \right| \\
   = {( - 1)^4}[(0 \times 0) - (0 \times 10)] \\
   = 1 \times (0 - 0) \\
   = 0 \\
$

Co-factor of 0
$
   = {( - 1)^{2 + 1}}\left| {\begin{array}{*{20}{c}}
  0&0 \\
  0&8
\end{array}} \right| \\
   = {( - 1)^3}[(0 \times 8) - (0 \times 0)] \\
   = - 1 \times (0 - 0) \\
   = 0 \\
$
Co-factor of 10 =
$
   = {( - 1)^{2 + 2}}\left| 9 \right| \\
   = {( - 1)^4}[(9 \times 8) - (0 \times 0)] \\
   = 1 \times (72 - 0) \\
   = 72 \\
$
Co-factor of 0 =
$
   = {( - 1)^{2 + 3}}\left| {\begin{array}{*{20}{c}}
  9&0 \\
  0&0
\end{array}} \right| \\
   = {( - 1)^5}[(9 \times 0) - (0 \times 0)] \\
   = - 1 \times (0 - 0) \\
   = 0 \\
$
Co-factor of 0 =
$
   = {( - 1)^{3 + 1}}\left| {\begin{array}{*{20}{c}}
  6&0 \\
  {10}&0
\end{array}} \right| \\
   = {( - 1)^4}[(6 \times 0) - (10 \times 0)] \\
   = 1 \times (0 - 0) \\
   = 0 \\
$
Co-factor of 0 =
$
   = {( - 1)^{3 + 2}}\left| {\begin{array}{*{20}{c}}
  9&0 \\
  0&0
\end{array}} \right| \\
   = {( - 1)^5}[(9 \times 0) - (0 \times 0)] \\
   = - 1 \times (0 - 0) \\
   = 0 \\
$
Co-factor of 8 =
$
   = {( - 1)^{3 + 3}}\left| {\begin{array}{*{20}{c}}
  9&0 \\
  0&{10}
\end{array}} \right| \\
   = {( - 1)^6}[(9 \times 10) - (0 \times 0)] \\
   = 1 \times (90 - 0) \\
   = 90 \\
$
Therefore, the required matrix =$\left[ {\begin{array}{*{20}{c}}
  {80}&0&0 \\
  0&{72}&0 \\
  0&0&{90}
\end{array}} \right]$
This was the cofactor matrix , Now we need to find the transpose of this matrix in order to get the adjoint matrix.
 Therefore, Transpose matrix or Adjoint of A =$Adj(A)$ =$\left[ {\begin{array}{*{20}{c}}
  {80}&0&0 \\
  0&{72}&0 \\
  0&0&{90}
\end{array}} \right]$
Now , we know ${A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj(A)$
Substituting the values , we get
${A^{ - 1}} = \dfrac{1}{{720}}\left[ {\begin{array}{*{20}{c}}
  {80}&0&0 \\
  0&{72}&0 \\
  0&0&{90}
\end{array}} \right]$
Dividing by 720 all through the elements , we get
${A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
  {\dfrac{1}{9}}&0&0 \\
  0&{\dfrac{1}{{10}}}&0 \\
  0&0&{\dfrac{1}{8}}
\end{array}} \right]$

So, the correct answer is “Option D”.

Note: In this particular question, we will try to find out the determinant of the matrix by using “minors” and “cofactors”. Cofactor of an element of a square matrix is the minor of the element with appropriate sign. By using these basic steps one can easily find the inverse of a matrix.