Determinants — Class 12 Maths Solution

Let $A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&0\\0&{\cos \alpha }&{\sin \alpha }\\0&{\sin \alpha }&{ - \cos \alpha }\end{array}} \right],$

then
$|A| = 1( - {\cos ^2}\alpha - {\sin ^2}\alpha ) = - 1 \ne 0.$

$\therefore$ ${A^{ - 1}}exists.$

So, A is non-singular matrix and therefore, A is invertible.

Let ${A_{ij}}$be the cofactors of ${a_{ij}}$ in A.

Then the cofactors of elements of A are given by
For adjoint $A,{A_{ij}} = {( - 1)^{i + j}}{M_{ij}}$

${A_{11}} = {( - 1)^{1 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{\cos \alpha }&{\sin \alpha }\\{\sin \alpha }&{ - \cos \alpha }\end{array}} \right| = - {\cos ^2}\alpha - {\sin ^2}\alpha = - 1$

${A_{12}} = {( - 1)^{1 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&{\sin \alpha }\\0&{ - \cos \alpha }\end{array}} \right| = 0,$

${A_{13}} = {( - 1)^{1 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&{\cos \alpha }\\0&{\sin \alpha }\end{array}} \right| = 0$

${A_{21}} = {( - 1)^{2 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&0\\{\sin \alpha }&{ - \cos \alpha }\end{array}} \right| = 0$

${A_{22}} = {( - 1)^{2 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0\\0&{ - \cos \alpha }\end{array}} \right| = - \cos \alpha$

${A_{23}} = {( - 1)^{2 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0\\0&{\sin \alpha }\end{array}} \right| = - \sin \alpha$

${A_{31}} = {( - 1)^{3 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&0\\{\cos \alpha }&{\sin \alpha }\end{array}} \right| = 0$

${A_{32}} = {( - 1)^{3 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0\\0&{\sin \alpha }\end{array}} \right| = - \sin \alpha$

${A_{33}} = {( - 1)^{3 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0\\0&{\cos \alpha }\end{array}} \right| = \cos \alpha$

$\therefore$ $adj\;A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 1}&0&0\\0&{ - \cos \alpha }&{ - \sin \alpha }\\0&{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 1}&0&0\\0&{ - \cos \alpha }&{ - \sin \alpha }\\0&{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]$

Hence, ${A^{ - 1}} = \cfrac{1}{{|A|}}adjA = \cfrac{1}{{ - 1}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 1}&0&0\\0&{ - \cos \alpha }&{ - \sin \alpha }\\0&{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&0\\0&{\cos \alpha }&{\sin \alpha }\\0&{\sin \alpha }&{ - \cos \alpha }\end{array}} \right]$