Login/Sign Up
Home » Find the inverse of each of the matrices given below.
Find the inverse of each of the matrices given below. \left[\begin{array}{ccc}8 & -4 & 1 \\ 10 & 0 & 6 \\ 8 & 1 & 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}{ccc}8 & -4 & 1 \\ 10 & 0 & 6 \\ 8 & 1 & 6\end{array}\right]

Solution:

Given matrix as A=\left(\begin{array}{ccc}8 & -4 & 1 \\ 10 & 0 & 6 \\ 8 & 1 & 6\end{array}\right)
Now let’s calculate the determinant of the given matrix first.
\begin{array}{l} |A|=\left|\begin{array}{ccc} 8 & -4 & 1 \\ 10 & 0 & 6 \\ 8 & 1 & 6 \end{array}\right|=8(0-6)-4(60-48)+1(10+0)=10 \\ \neq 0 \end{array}
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|cc} \left|\begin{array}{cc} 0 & 6 \\ 1 & 6 \end{array}\right| & \left|\begin{array}{cc} 10 & 6 \\ 8 & 6 \end{array}\right| & \left|\begin{array}{cc} 10 & 0 \\ 8 & 1 \end{array}\right| \\ \left|\begin{array}{cc} -4 & 1 \\ 1 & 6 \end{array}\right| & \left|\begin{array}{cc} 8 & 1 \\ 8 & 6 \end{array}\right| & \left|\begin{array}{cc} 8 & -4 \\ 8 & 1 \end{array}\right| \\ \left|\begin{array}{cc} -4 & 1 \\ 0 & 6 \end{array}\right| & \left|\begin{array}{cc} 8 & 1 \\ 10 & 6 \end{array}\right| & \left|\begin{array}{cc} 8 & -4 \\ 10 & 0 \end{array}\right| \end{array}\right)=\left(\begin{array}{ccc} -6 & 12 & 10 \\ -25 & 40 & 40 \\ -24 & 38 & 40 \end{array}\right)

Step: 2 Find the co-factor matrix of A.
\begin{array}{l} C_{A}=\left(\begin{array}{ccc} (-1)^{1+1}(-6) & (-1)^{1+2}(12) & (-1)^{1+3}(10) \\ (-1)^{2+1}(-25) & (-1)^{2+2}(40) & (-1)^{2+3}(40) \\ (-1)^{3+1}(-24) & (-1)^{3+2}(38) & (-1)^{3+3}(40) \end{array}\right) \\ =\left(\begin{array}{ccc} -6 & -12 & 10 \\ 25 & 40 & -40 \\ -24 & -38 & 40 \end{array}\right) \end{array}

Step: 3 By transpose of C_{A} we will have \operatorname{adj} A.
\operatorname{adj} A=C_{A}^{T}=\left(\begin{array}{ccc}-6 & -12 & 10 \\ 25 & 40 & -40 \\ -24 & -38 & 40\end{array}\right)^{T}=\left(\begin{array}{ccc}-6 & 25 & -24 \\ -12 & 40 & -38 \\ 10 & -40 & 40\end{array}\right)
Finally the inverse of the matrix is
A^{-1}=\frac{1}{|A|} \operatorname{adj} A=\frac{1}{10}\left(\begin{array}{ccc} -6 & 25 & -24 \\ -12 & 40 & -38 \\ 10 & -40 & 40 \end{array}\right)

i

More Material