Question

Inverse of a diagonal non- singular matrix is
$
  (a){\text{ Scalar matrix}} \\
  (b){\text{ Skew - symmetric matrix}} \\
  (c){\text{ Zero matrix}} \\
  (d){\text{ Diagonal matrix}} \\
 $

Answer 1

Hint – We have to consider a diagonal non-singular matrix, diagonal matrices are those which have only diagonal elements while rest all are zero whereas non-singular means that the determinant must not be zero. Use this concept to write a diagonal non-singular matrix. Then use the concept of ${A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj\left( A \right)$ to get the inverse.

Complete step-by-step answer:
Consider a diagonal non-singular matrix
$A = \left[ {\begin{array}{*{20}{c}}
  a&0&0 \\
  0&b&0 \\
  0&0&c
\end{array}} \right]$ (Where a, b, c is any real number and not all equal to 1.)
Now we have to find out the inverse of this matrix.
Now as we know ${A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj\left( A \right)$
Where $adj\left( A \right) = {\left[ {\begin{array}{*{20}{c}}
  {{c_{11}}}&{{c_{12}}}&{{c_{13}}} \\
  {{c_{21}}}&{{c_{22}}}&{{c_{23}}} \\
  {{c_{31}}}&{{c_{32}}}&{{c_{33}}}
\end{array}} \right]^T}$
Where T is the transpose of matrix, so apply transpose of matrix
\[
   \Rightarrow adj\left( A \right) = \left[ {\begin{array}{*{20}{c}}
  {{c_{11}}}&{{c_{21}}}&{{c_{31}}} \\
  {{c_{12}}}&{{c_{22}}}&{{c_{32}}} \\
  {{c_{13}}}&{{c_{23}}}&{{c_{33}}}
\end{array}} \right] \\
  {A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj\left( A \right) = \dfrac{1}{{\left| A \right|}}\left[ {\begin{array}{*{20}{c}}
  {{c_{11}}}&{{c_{21}}}&{{c_{31}}} \\
  {{c_{12}}}&{{c_{22}}}&{{c_{32}}} \\
  {{c_{13}}}&{{c_{23}}}&{{c_{33}}}
\end{array}} \right]...........\left( 1 \right) \\
\]
Now, first calculate determinant of $A$
$ \Rightarrow \left| A \right| = \left| {\begin{array}{*{20}{c}}
  a&0&0 \\
  0&b&0 \\
  0&0&c
\end{array}} \right|$
Now, expand the determinant
$
   \Rightarrow \left| A \right| = \left| {\begin{array}{*{20}{c}}
  a&0&0 \\
  0&b&0 \\
  0&0&c
\end{array}} \right| = a\left| {\begin{array}{*{20}{c}}
  b&0 \\
  0&c
\end{array}} \right| - 0 + 0 \\
   = a\left( {bc - 0} \right) = abc \\
 $
Now calculate $adj\left( A \right)$
$ \Rightarrow adj\left( A \right) = \left[ {\begin{array}{*{20}{c}}
  {{c_{11}}}&{{c_{21}}}&{{c_{31}}} \\
  {{c_{12}}}&{{c_{22}}}&{{c_{32}}} \\
  {{c_{13}}}&{{c_{23}}}&{{c_{33}}}
\end{array}} \right]$
So, calculate its internal elements i.e. its cofactors
\[
  {c_{11}} = + 1\left| {\begin{array}{*{20}{c}}
  b&0 \\
  0&c
\end{array}} \right| = 1\left( {bc - 0} \right) = bc,{\text{ }}{{\text{c}}_{21}} = - 1\left| {\begin{array}{*{20}{c}}
  0&b \\
  0&0
\end{array}} \right| = - 1\left( {0 - 0} \right) = 0,{\text{ }}{{\text{c}}_{31}} = + 1\left| {\begin{array}{*{20}{c}}
  0&0 \\
  b&0
\end{array}} \right| = 1\left( {0 - 0} \right) = 0 \\
  {c_{12}} = - 1\left| {\begin{array}{*{20}{c}}
  0&0 \\
  0&c
\end{array}} \right| = - 1\left( {0 - 0} \right) = 0,{\text{ }}{{\text{c}}_{22}} = + 1\left| {\begin{array}{*{20}{c}}
  a&0 \\
  0&c
\end{array}} \right| = 1\left( {ac - 0} \right) = ac,{\text{ }}{{\text{c}}_{32}} = - 1\left| {\begin{array}{*{20}{c}}
  a&0 \\
  0&0
\end{array}} \right| = - 1\left( {0 - 0} \right) = 0 \\
  {c_{13}} = + 1\left| {\begin{array}{*{20}{c}}
  0&b \\
  0&0
\end{array}} \right| = 1\left( {0 - 0} \right) = 0,{\text{ }}{{\text{c}}_{23}} = - 1\left| {\begin{array}{*{20}{c}}
  a&0 \\
  0&0
\end{array}} \right| = - 1\left( {0 - 0} \right) = 0,{\text{ }}{{\text{c}}_{33}} = + 1\left| {\begin{array}{*{20}{c}}
  a&0 \\
  0&b
\end{array}} \right| = 1\left( {ab - 0} \right) = ab \\
\]$ \Rightarrow adj\left( A \right) = \left[ {\begin{array}{*{20}{c}}
  {bc}&0&0 \\
  0&{ca}&0 \\
  0&0&{ab}
\end{array}} \right]$
Now, from equation (1) we have,
${A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj\left( A \right) = \dfrac{1}{{abc}}\left[ {\begin{array}{*{20}{c}}
  {bc}&0&0 \\
  0&{ca}&0 \\
  0&0&{ab}
\end{array}} \right]$

$ \Rightarrow {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
  {\dfrac{{bc}}{{abc}}}&0&0 \\
  0&{\dfrac{{ca}}{{abc}}}&0 \\
  0&0&{\dfrac{{ab}}{{abc}}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {\dfrac{1}{a}}&0&0 \\
  0&{\dfrac{1}{b}}&0 \\
  0&0&{\dfrac{1}{c}}
\end{array}} \right]$
So the inverse of matrix A is also a diagonal non-singular matrix having diagonal elements inverted.
Hence, option (d) is correct.

Note – In order to face such types of problems the key concept is simply to have the understanding of basic definitions of scalar, skew-symmetric matrix, zero matrix and diagonal matrix. A scalar matrix is a special diagonal matrix in which all the diagonal elements are the same while the rest are zero. A skew symmetric matrix is one which when transposed gives exactly the same matrix however with a negative sign. Zero matrix is one in which all the elements and a diagonal matrix is being explained in the hint only. Use this concept along with the direct formula to find the inverse of a matrix to get the answer.