Let \(A = \begin{bmatrix}1&1&0\\0&1&1\\1&0&1\end{bmatrix}\) and \(B = \begin{bmatrix}1&-1&1\\1&1&-1\\-1&1&1\end{bmatrix}\). Then \(A^{-1}\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{1}{2}B\)
Multiplying each row of A by each column of B:
Row 1 of AB: \(\begin{bmatrix}1 + 1 + 0&-1 + 1 + 0&1 - 1 + 0\end{bmatrix} = \begin{bmatrix}2&0&0\end{bmatrix}\)
Row 2 of AB: \(\begin{bmatrix}0 + 1 - 1&0 + 1 + 1&0 - 1 + 1\end{bmatrix} = \begin{bmatrix}0&2&0\end{bmatrix}\)
Row 3 of AB: \(\begin{bmatrix}1 + 0 - 1&-1 + 0 + 1&1 + 0 + 1\end{bmatrix} = \begin{bmatrix}0&0&2\end{bmatrix}\)
So \(AB = 2I\), and in the same way \(BA = 2I\). Dividing by 2:
\( \displaystyle A\left(\frac{1}{2}B\right) = \left(\frac{1}{2}B\right)A = I \)
By the definition of the inverse,
\( \displaystyle A^{-1} = \frac{1}{2}B \)