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Find the adjoint of the given matrix and verify in each case that A. (\operatorname{adj} A)=(\operatorname{adj} A)=m|A|.I.\left[\begin{array}{cc} \operatorname{Cos} \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]
Adjoint and Inverse of a Matrix | CBSE | Class 12 | Maths | RS Aggarwal
Find the adjoint of the given matrix and verify in each case that A. (\operatorname{adj} A)=(\operatorname{adj} A)=m|A|.I.\left[\begin{array}{cc} \operatorname{Cos} \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]

Solution:

Given matrix as A=\left(\begin{array}{ll}\cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right).
Find: the adjoint of the matrix given.

Step: 1 Find the minor matrix of A.
M_{A}=\left(\begin{array}{ll} |\cos \alpha| & |\sin \alpha| \\ |\sin \alpha| & |\cos \alpha| \end{array}\right)=\left(\begin{array}{cc} \cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right)

Step: 2 Find the co-factor matrix of A.
C_{A}=\left(\begin{array}{cc} (-1)^{1+1}(\cos \alpha) & (-1)^{1+2}(\sin \alpha) \\ (-1)^{2+1}(\sin \alpha) & (-1)^{2+2}(\cos \alpha) \end{array}\right)=\left(\begin{array}{cc} \cos \alpha & -\overline{s i n} \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right)

Step: 3 By transpose of C_{A} we will have \operatorname{adj} A.
\operatorname{adj} A=C_{A}^{T}=\left(\begin{array}{cc}\cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right)^{T}=\left(\begin{array}{cc}\cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right)

Let’s now calculate:
(i) A(\operatorname{adj} A)=\left(\begin{array}{cc}\cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right)\left(\begin{array}{cc}\cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right) =\left(\begin{array}{cc}\cos \alpha(\cos \alpha)-\sin \alpha(\sin \alpha) & \cos \alpha(\sin \alpha)-\sin \alpha(\cos \alpha) \\ \sin \alpha(\cos \alpha)-\cos \alpha(\sin \alpha) & -\sin \alpha(\sin \alpha)+\cos \alpha(\cos \alpha)\end{array}\right) =\left(\begin{array}{cc}\cos ^{2} \alpha-\sin ^{2} \alpha & 0 \\ 0 & \cos ^{2} \alpha-\sin ^{2} \alpha\end{array}\right)
=\left(\cos ^{2} \alpha-\sin ^{2} \alpha\right) I_{2}
(ii) (\operatorname{adj} A) A=\left(\begin{array}{cc}\cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right)\left(\begin{array}{cc}\cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right) =\left(\begin{array}{cc}\cos \alpha(\cos \alpha)-\sin \alpha(\sin \alpha) & -\sin \alpha(\cos \alpha)+\cos \alpha(\sin \alpha) \\ -\cos \alpha(\sin \alpha)+\sin \alpha(\cos \alpha) & -\sin \alpha(\sin \alpha)+\cos \alpha(\cos \alpha)\end{array}\right) =\left(\begin{array}{cc}\cos ^{2} \alpha-\sin ^{2} \alpha & 0 \\ 0 & \cos ^{2} \alpha-\sin ^{2} \alpha\end{array}\right)
=\left(\cos ^{2} \alpha-\sin ^{2} \alpha\right) I_{2}

(iii) |A|=\left|\begin{array}{ll}\cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right|=\cos ^{2} \alpha-\sin ^{2} \alpha
As a result from above cases it can be written as A(\operatorname{adj} A)=(\operatorname{adj} A)=|A| I_{2} and adjoint of given matrix is \left(\begin{array}{cc}\cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right).

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