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Mathematics Question on Determinants

Let A be a 3×33 \times 3 matrix such that A[123 023 011]=[001 100 010]A\begin{bmatrix}1&2&3\\\ 0&2&3\\\ 0&1&1\end{bmatrix} = \begin{bmatrix}0&0&1\\\ 1&0&0\\\ 0&1&0\end{bmatrix} Then A−1A^{-1} is :

A

[312 302 101]\begin{bmatrix}3&1&2\\\ 3&0&2\\\ 1&0&1\end{bmatrix}

B

[321 320 110]\begin{bmatrix}3&2&1\\\ 3 &2&0\\\ 1 &1&0\end{bmatrix}

C

[013 023 111]\begin{bmatrix}0&1&3\\\ 0 &2&3\\\ 1 &1&1\end{bmatrix}

D

[123 011 023] \begin{bmatrix}1&2&3\\\ 0 &1&1\\\ 0 &2&3\end{bmatrix}

Answer

[312 302 101]\begin{bmatrix}3&1&2\\\ 3&0&2\\\ 1&0&1\end{bmatrix}

Explanation

Solution

Given A=[123 023 011]=[001 100 010]A = \begin{bmatrix}1&2&3\\\ 0 &2&3\\\ 0 &1&1\end{bmatrix} = \begin{bmatrix}0&0&1\\\ 1 &0&0\\\ 0 &1&0\end{bmatrix} Applying C1↔C3C_{1} \leftrightarrow C_{3} A[321 320 110]=[100 001 010]A \begin{bmatrix}3&2&1\\\ 3&2&0\\\ 1 &1&0\end{bmatrix} = \begin{bmatrix}1&0&0\\\ 0&0&1\\\ 0 &1&0\end{bmatrix} Again Applying C2↔C3C_{2}\leftrightarrow C_{3} A[312 302 101]=[100 010 001]A \begin{bmatrix}3&1&2\\\ 3 &0&2\\\ 1 &0&1\end{bmatrix} = \begin{bmatrix}1&0&0\\\ 0 &1&0\\\ 0 &0&1\end{bmatrix} pre-multiplying both sides by A−1A^{-1} A−1A[312 302 101]=A−1[100 010 001]A^{-1} A \begin{bmatrix}3&1&2\\\ 3 &0&2\\\ 1&0&1\end{bmatrix} = A^{-1} \begin{bmatrix}1&0&0\\\ 0 &1&0\\\ 0 &0&1\end{bmatrix} I[312 302 101]=A−1I=A−1I \begin{bmatrix}3&1&2\\\ 3 &0&2\\\ 1 &0&1\end{bmatrix} = A^{-1} I = A^{-1} (∵A−1A=I (\because A^{-1} A = I and I = Identity matrix) Hence, A−1=[312 302 101] A^{-1}= \begin{bmatrix}3&1&2\\\ 3 &0&2\\\ 1 &0&1\end{bmatrix}