The matrix X for which […] \ X= […] is

The correct answer is Option (3) → \( \begin{bmatrix}-3 & -14\\4 & 17\end{bmatrix} \)

Let \( P=\begin{bmatrix}5 & 4\\1 & 1\end{bmatrix} \) and \( Q=\begin{bmatrix}1 & -2\\1 & 3\end{bmatrix} \). Then PX = Q, so \( X=P^{-1}Q \) (P⁻¹ on the left, because P is on the left of X).

    \( \displaystyle |P|=5-4=1 \)

    \( \displaystyle P^{-1}=\frac{1}{|P|}\,\mathrm{adj}\,P=\begin{bmatrix}1 & -4\\-1 & 5\end{bmatrix} \)

    \( \displaystyle X=\begin{bmatrix}1 & -4\\-1 & 5\end{bmatrix}\begin{bmatrix}1 & -2\\1 & 3\end{bmatrix} \)

    \( \displaystyle =\begin{bmatrix}1-4 & -2-12\\-1+5 & 2+15\end{bmatrix} \)

    \( \displaystyle =\begin{bmatrix}-3 & -14\\4 & 17\end{bmatrix} \)

Shortcut: Multiply P by the options instead of finding the inverse: for Option (3), the first row of PX is (5(−3) + 4(4), 5(−14) + 4(17)) = (1, −2), and the second row is (−3 + 4, −14 + 17) = (1, 3). This is Q.

Why the other options are incorrect

• Option (1) \( \begin{bmatrix}1 & -4\\-1 & 5\end{bmatrix} \) — Incorrect. This is \( P^{-1} \) itself; it still has to be multiplied by Q.

• Option (2) \( \begin{bmatrix}-2 & 2\\-4 & 3\end{bmatrix} \) — Incorrect. Multiplying, \( P\begin{bmatrix}-2 & 2\\-4 & 3\end{bmatrix}=\begin{bmatrix}-26 & 22\\-6 & 5\end{bmatrix} \), which is not Q.

• Option (4) \( \begin{bmatrix}3 & 14\\-4 & -17\end{bmatrix} \) — Incorrect. This is the negative of X; the signs of all the entries are wrong.

Tip: Matrix multiplication is not commutative: from PX = Q, X = P⁻¹Q, but from XP = Q, X = QP⁻¹.