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Find the inverse of each of the matrices given below. \left[\begin{array}{cc} 2 & -3 \\ 4 & 6 \end{array}\right]
Adjoint and Inverse of a Matrix | CBSE | Class 12 | Exercise 7A | Maths | RS Aggarwal
Find the inverse of each of the matrices given below. \left[\begin{array}{cc} 2 & -3 \\ 4 & 6 \end{array}\right]

Solution:

Given matrix as A=\left(\begin{array}{cc}2 & -3 \\ 4 & 6\end{array}\right).
Now let’s calculate the determinant of the given matrix first.
|A|=\left|\begin{array}{cc} 2 & -3 \\ 4 & 6 \end{array}\right|=12+12=24 \neq 0
So the given matrix has inverse.
Find: the adjoint of the given matrix.

Step: 1 Find the minor matrix of A.
M_{A}=\left(\begin{array}{cc} |6| & |4| \\ |-3| & |2| \end{array}\right)=\left(\begin{array}{cc} 6 & 4 \\ -3 & 2 \end{array}\right)

Step: 2 Find the co-factor matrix of A.
C_{A}=\left(\begin{array}{cc} (-1)^{1+1}(6) & (-1)^{1+2}(4) \\ (-1)^{2+1}(-3) & (-1)^{2+2}(2) \end{array}\right)=\left(\begin{array}{cc} 6 & -4 \\ 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}6 & -4 \\ 3 & 2\end{array}\right)^{T}=\left(\begin{array}{cc}6 & 3 \\ -4 & 2\end{array}\right)
Finally the inverse of the matrix is
A^{-1}=\frac{1}{|A|} \text { adj } A=\frac{1}{24}\left(\begin{array}{cc} 6 & 3 \\ -4 & 2 \end{array}\right)

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