MHT CET · Maths · Matrices
The element in the third row and first column of the inverse of the matrix \(\left[\begin{array}{ccc}1 & -3 & 2 \\ -3 & 3 & -1 \\ 2 & -1 & 0\end{array}\right]\) is
- A \(-3\)
- B \(x=4\)
- C \(3\)
- D \(2\)
Answer & Solution
Correct Answer
(C) \(3\)
Step-by-step Solution
Detailed explanation
We have \(A=\left[\begin{array}{ccc}-1 & -3 & 2 \\ -3 & 3 & -1 \\ 2 & -1 & 0\end{array}\right] \Rightarrow \begin{aligned}|A| &=\left|\begin{array}{ccc}-1 & -3 & 2 \\ -3 & 3 & -1 \\ 2 & -1 & 0\end{array}\right| \\ &=-(-1)+3(2)+2(3-6)=1+6-6=1 \end{aligned}\)
We have to find \(a_{31}\) in \(A^{-1}\). So we will find \(A_{13}\) in \(A\).
Cofactor of \(2=(-1)^{1+3}\left|\begin{array}{cc}-3 & 3 \\ 2 & -1\end{array}\right|=3-6=-3\)
Hence \(a_{31}\) in \(A^{-1}=\frac{-3}{|A|}=\frac{-3}{1}=-3\)
We have to find \(a_{31}\) in \(A^{-1}\). So we will find \(A_{13}\) in \(A\).
Cofactor of \(2=(-1)^{1+3}\left|\begin{array}{cc}-3 & 3 \\ 2 & -1\end{array}\right|=3-6=-3\)
Hence \(a_{31}\) in \(A^{-1}=\frac{-3}{|A|}=\frac{-3}{1}=-3\)
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