If \( \begin{bmatrix}3 & 1\\2 & 1\end{bmatrix}P\begin{bmatrix}2 & 1\\3 & 2\end{bmatrix}=\begin{bmatrix}1 & -1\\0 & 1\end{bmatrix} \) then matrix \( P \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \begin{bmatrix}8 & -5\\-19 & 12\end{bmatrix} \)
Let \( M=\begin{bmatrix}3 & 1\\2 & 1\end{bmatrix} \), \( N=\begin{bmatrix}2 & 1\\3 & 2\end{bmatrix} \) and \( C=\begin{bmatrix}1 & -1\\0 & 1\end{bmatrix} \). Then \( MPN=C \), so \( P=M^{-1}CN^{-1} \), with the inverses on the correct sides.
\( \displaystyle |M|=3-2=1 \)
\( \displaystyle M^{-1}=\begin{bmatrix}1 & -1\\-2 & 3\end{bmatrix} \)
\( \displaystyle |N|=4-3=1 \)
\( \displaystyle N^{-1}=\begin{bmatrix}2 & -1\\-3 & 2\end{bmatrix} \)
\( \displaystyle M^{-1}C=\begin{bmatrix}1 & -1\\-2 & 3\end{bmatrix}\begin{bmatrix}1 & -1\\0 & 1\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}1 & -2\\-2 & 5\end{bmatrix} \)
\( \displaystyle P=\begin{bmatrix}1 & -2\\-2 & 5\end{bmatrix}\begin{bmatrix}2 & -1\\-3 & 2\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}2+6 & -1-4\\-4-15 & 2+10\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}8 & -5\\-19 & 12\end{bmatrix} \)
Shortcut: Taking determinants of both sides, \( |M|\,|P|\,|N|=|C| \) gives \( 1\cdot|P|\cdot 1=1 \), so |P| = 1. Option (1) has determinant −96 − 95 = −191, Option (2) −95 + 96 = 1, Option (3) 96 − 95 = 1 and Option (4) 95 − 96 = −1. Only Options (2) and (3) remain; the (1, 1) entry of \( M^{-1}CN^{-1} \) is \( 1\cdot 2+(-2)(-3)=8 \), which picks Option (3).