Question

Let A be a \(3\times3\) matrix such that \(A\begin{bmatrix}1&2&3\\0&2&3\\0&1&1\end{bmatrix}=\begin{bmatrix}0&0&1\\1&0&0\\0&1&0\end{bmatrix}\). Then \(A^{-1}\) is:

• A. \(\begin{bmatrix}3&1&2\\3&0&2\\1&0&1\end{bmatrix}\)
• B. \(\begin{bmatrix}3&2&1\\3&2&0\\1&1&0\end{bmatrix}\)
• C. \(\begin{bmatrix}0&1&3\\0&2&3\\1&1&1\end{bmatrix}\)
• D. \(\begin{bmatrix}1&2&3\\0&1&1\\0&2&3\end{bmatrix}\)

Hint:

Write the equation as AM = P; then \(A^{-1} = M P^{-1}\) and P is a permutation matrix, so \(P^{-1} = P^T\).

Answer 1

The Correct Option is A

Let \(M = \begin{bmatrix}1&2&3\\0&2&3\\0&1&1\end{bmatrix}\) and \(P = \begin{bmatrix}0&0&1\\1&0&0\\0&1&0\end{bmatrix}\), so AM = P.
Then \(M = A^{-1}P\), hence \(A^{-1} = M P^{-1}\).
P is a permutation matrix so \(P^{-1} = P^T = \begin{bmatrix}0&1&0\\0&0&1\\1&0&0\end{bmatrix}\).
Multiplying M by \(P^T\): column 1 = (third column of M) = (3, 3, 1), column 2 = (first column of M) = (1, 0, 0), column 3 = (second column of M) = (2, 2, 1).
So \(A^{-1} = \begin{bmatrix}3&1&2\\3&0&2\\1&0&1\end{bmatrix}\).
Answer: (1) \(\begin{bmatrix}3&1&2\\3&0&2\\1&0&1\end{bmatrix}\)