If \(A = \begin{bmatrix}2&0&0\\0&-1&0\\0&0&4\end{bmatrix}\), then \(A^{-1}\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\begin{bmatrix}\displaystyle\frac{1}{2}&0&0\\0&-1&0\\0&0&\displaystyle\frac{1}{4}\end{bmatrix}\)
The product of two diagonal matrices of the same order is the diagonal matrix of the products of the corresponding diagonal elements. So \(A^{-1}\) is the diagonal matrix whose diagonal elements are the reciprocals of 2, −1 and 4.
\( \displaystyle B = \begin{bmatrix}\displaystyle\frac{1}{2}&0&0\\0&-1&0\\0&0&\displaystyle\frac{1}{4}\end{bmatrix} \)
Check:
\( \displaystyle AB = \begin{bmatrix}2 \times \displaystyle\frac{1}{2}&0&0\\0&(-1)(-1)&0\\0&0&4 \times \displaystyle\frac{1}{4}\end{bmatrix} = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix} = I \)
In the same way \(BA = I\), so \(A^{-1} = B\).