Question

Let \[D = diag\left( {{d_1},{d_2},{d_3},.....{d_n}} \right)\] where\[{d_i} \ne 0\nabla i\], then \[{D^{ - 1}}\] equals
A.D
B\[diag\left( {d_1^n,d_2^n,.....,d_n^n} \right)\]
C.\[{I_n}\]
D.\[diag\left( {d_1^{ - 1},d_2^{ - 1},d_3^{ - 1}.....,d_n^{ - 1}} \right)\]

Answer 1

Hint: In this question, we need to determine the expression of \[{D^{ - 1}}\] such that \[D = diag\left( {{d_1},{d_2},{d_3},.....{d_n}} \right)\]. For this, we will follow the property of the matrix along with the inverse matrix and general algebraic calculations.

Complete step-by-step answer:
Given
\[D = diag\left( {{d_1},{d_2},{d_3},.....{d_n}} \right)\]
Let a matrix D with n numbers of rows and columns be given as
\[D = \left[ {\begin{array}{*{20}{c}}
  {{d_1}}&0&0&0&{...}&0 \\
  0&{{d_2}}&0&0&{...}&0 \\
  0&0&{{d_3}}&{...}&{...}&0 \\
  {...}&{...}&{...}&{...}&{...}&{...} \\
  0&0&0&{...}&{...}&{{d_n}}
\end{array}} \right]\]
So the product of the diagonal elements of the matrix D will be
\[\left| D \right| = {d_1}{d_2}......{d_n} - - (i)\]
We know that the cofactor of a matrix is the number which is obtained when the column and the row of a designated element in a matrix are removed.
Hence we can write the cofactor of diagonal elements of matrix D as
\[{D_{11}} = {d_2}{d_3}....{d_n}\]
Similarly the cofactor of \[{D_{22}} = {d_1}{d_3}....{d_n}\]
Similarly we can write \[{D_{ij}} = 0\nabla i \ne j\]
We know the inverse of a matrix is given by the formula \[{A^{ - 1}} = \dfrac{{\left( {adjA} \right)}}{{\left| A \right|}}\], hence the inverse of the matrix D will be
\[{D^{ - 1}} = \dfrac{{\left( {adjD} \right)}}{{\left| D \right|}} - - (ii)\]
Hence by substituting the values of formula (ii) from (i), we get
 \[\Rightarrow {D^{ - 1}} = \dfrac{{\left( {adjD} \right)}}{{\left| D \right|}} = \dfrac{1}{{{d_1}{d_2}...{d_n}}}\left[ {\begin{array}{*{20}{c}}
  {{d_2}{d_3}...{d_n}}&0&0&0 \\
  0&{{d_1}{d_3}...{d_n}}&0&0 \\
  {...}&{...}&{...}&{...} \\
  {...}&{...}&{...}&{...} \\
  0&0&0&{{d_1}{d_2}{d_3}...{d_{n - 1}}}
\end{array}} \right]\]
This can be further written as
\[{D^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
  {\dfrac{1}{{{d_1}}}}&0&0&0 \\
  0&{\dfrac{1}{{{d_2}}}}&0&0 \\
  {...}&{...}&{...}&{...} \\
  {...}&{...}&{...}&{...} \\
  0&0&0&{\dfrac{1}{{{d_n}}}}
\end{array}} \right]\]
Now we know the product of the diagonal elements of the matrix is written as
\[\Rightarrow {D^{ - 1}} = diag\left( {\dfrac{1}{{{d_1}}}.\dfrac{1}{{{d_2}}}......\dfrac{1}{{{d_n}}}} \right)\]
This can be also written as
\[{D^{ - 1}} = diag\left( {d_1^{ - 1}.d_2^{ - 1}.d_3^{ - 1}.......d_n^{ - 1}} \right)\]
Therefore we can say \[{D^{ - 1}}\]equal\[ = diag\left( {d_1^{ - 1}.d_2^{ - 1}.d_3^{ - 1}.......d_n^{ - 1}} \right)\]
So, the correct answer is “Option D”.

Note: Matrix is an array of numbers aligned in a proper manner in rows and columns. They are expressed as $\left( {n \times m} \right)$ where ‘n’ is the number of rows and ‘m’ is the number of columns. Students must note that the determinant of a triangular matrix is equal to the product of its diagonal elements.