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_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.