If \( P=\displaystyle\frac{1}{14}\begin{bmatrix}3 & 2\\-4 & 2\end{bmatrix} \) and \( P^{-1}=\begin{bmatrix}x & -2\\4 & y\end{bmatrix} \), then the value of \( x \) and \( y \) respectively are
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 2 and 3
Let \( P=\frac{1}{14}M \) with \( M=\begin{bmatrix}3 & 2\\-4 & 2\end{bmatrix} \). Then \( P^{-1}=14\,M^{-1} \).
\( \displaystyle |M|=3(2)-2(-4) \)
\( \displaystyle =6+8=14 \)
For a 2 × 2 matrix, adj M swaps the diagonal entries and changes the signs of the other two:
\( \displaystyle M^{-1}=\frac{1}{|M|}\,\mathrm{adj}\,M \)
\( \displaystyle =\frac{1}{14}\begin{bmatrix}2 & -2\\4 & 3\end{bmatrix} \)
\( \displaystyle P^{-1}=14\cdot\frac{1}{14}\begin{bmatrix}2 & -2\\4 & 3\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}2 & -2\\4 & 3\end{bmatrix} \)
Comparing with the given \( P^{-1} \): x = 2 and y = 3.
Shortcut: Use \( PP^{-1}=I \) and compare only two entries:
\( \displaystyle \frac{1}{14}\begin{bmatrix}3 & 2\\-4 & 2\end{bmatrix}\begin{bmatrix}x & -2\\4 & y\end{bmatrix}=\frac{1}{14}\begin{bmatrix}3x+8 & -6+2y\\-4x+8 & 8+2y\end{bmatrix} \)
\( \displaystyle 3x+8=14\ \Rightarrow\ x=2 \)
\( \displaystyle -6+2y=0\ \Rightarrow\ y=3 \)