Answer
Verified
415.2k+ views
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Which type of bond is stronger ionic or covalent class 12 chemistry CBSE
What organs are located on the left side of your body class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
When people say No pun intended what does that mea class 8 english CBSE