Answer

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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.

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.

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