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.