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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| \cdot I. \left[\begin{array}{ll}2 & 3 \\ 5 & 9\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| \cdot I. \left[\begin{array}{ll}2 & 3 \\ 5 & 9\end{array}\right]

Solution:

Given matrix as A=\left(\begin{array}{ll}2 & 3 \\ 5 & 9\end{array}\right).
Find: the adjoint of the matrix given.

Step: 1 Find the minor matrix of A.
M_{A}=\left(\begin{array}{ll} |9| & |5| \\ |3| & |2| \end{array}\right)=\left(\begin{array}{ll} 9 & 5 \\ 3 & 2 \end{array}\right)

Step: 2 Find the co-factor matrix of A.
C_{A}=\left(\begin{array}{ll} (-1)^{1+1} 9 & (-1)^{1+2} 5 \\ (-1)^{2+1} 3 & (-1)^{2+2} 2 \end{array}\right)=\left(\begin{array}{cc} 9 & -5 \\ -3 & 2 \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}9 & -5 \\ -3 & 2\end{array}\right)^{T}=\left(\begin{array}{cc}9 & -3 \\ -5 & 2\end{array}\right) .

Let’s now calculate:

(i) A(\operatorname{adj} A)=\left(\begin{array}{ll}2 & 3 \\ 5 & 9\end{array}\right)\left(\begin{array}{cc}9 & -3 \\ -5 & 2\end{array}\right)=
\left(\begin{array}{cc}2(9)+3(-5) & 2(-3)+3(2) \\ 5(9)+9(2) & 5(-3)+9(2)\end{array}\right)=\left(\begin{array}{cc}3 & 0 \\ 0 & 3\end{array}\right)=3 I_{2}

(ii) (\operatorname{adj} A) A=\left(\begin{array}{cc}9 & -3 \\ -5 & 2\end{array}\right)\left(\begin{array}{ll}2 & 3 \\ 5 & 9\end{array}\right)=
\left(\begin{array}{ll}9(2)+(-3)(5) & 9(3)+(-3)(9) \\ (-5)(2)+2(5) & (-5)(3)+2(9)\end{array}\right)=\left(\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right)=3 I_{2}

(iii) |A|=\left|\begin{array}{ll}2 & 3 \\ 5 & 9\end{array}\right|=2(9)-5(3)=18-15=3
As a result from the 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}9 & -3 \\ -5 & 2\end{array}\right).

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