If \( XA=(I-A)^2 \) where \( X,A \) and \( I \) are \( 2\times 2 \) matrices and \( A=\begin{bmatrix}1 & 2\\1 & 4\end{bmatrix} \), then \( |X|= \)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{4}{|A|} \)
Taking determinants on both sides, with \( |PQ|=|P||Q| \):
\( \displaystyle |X|\,|A|=|I-A|^2 \)
\( \displaystyle I-A=\begin{bmatrix}1-1 & 0-2\\0-1 & 1-4\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}0 & -2\\-1 & -3\end{bmatrix} \)
\( |I-A|=0-2=-2 \)
\( |I-A|^2=4 \)
\( \displaystyle |X|=\frac{4}{|A|} \)
Here \( |A|=4-2=2 \), so |X| = 2. With |A| = 2 the options give \( 2|A|=4 \), \( \frac{4}{|A|}=2 \), \( |A|^2=4 \) and \( \frac{2}{|A|}=1 \); only Option (2) gives 2.