If A is an invertible matrix such that \(A^{-1} = \begin{bmatrix}2&-1\\4&3\end{bmatrix}\), then \((3A')^{-1}\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{1}{3}\begin{bmatrix}2&4\\-1&3\end{bmatrix}\)
Two results are used: \((kM)^{-1} = \displaystyle\frac{1}{k}M^{-1}\) for \(k \ne 0\), and \((A')^{-1} = (A^{-1})'\). So
\( \displaystyle (3A')^{-1} = \frac{1}{3}(A')^{-1} = \frac{1}{3}(A^{-1})' \)
The transpose of \(A^{-1}\) turns its rows into columns:
\( \displaystyle (A^{-1})' = \begin{bmatrix}2&4\\-1&3\end{bmatrix} \)
Hence
\( \displaystyle (3A')^{-1} = \frac{1}{3}\begin{bmatrix}2&4\\-1&3\end{bmatrix} \)
Why the other options are incorrect
Option (1) \(3\begin{bmatrix}2&4\\-1&3\end{bmatrix}\) — Incorrect. The scalar 3 must be replaced by \(\displaystyle\frac{1}{3}\) in the inverse.
Option (2) \(\displaystyle\frac{1}{3}\begin{bmatrix}2&-1\\4&3\end{bmatrix}\) — Incorrect. This is \((3A)^{-1}\); the transpose has been left out.
Option (3) \(\displaystyle\frac{1}{3}\begin{bmatrix}-2&-4\\1&-3\end{bmatrix}\) — Incorrect. It is the negative of the correct matrix.