If \( A=\begin{bmatrix}4 & 1\\2 & 3\end{bmatrix} \) and \( B=\begin{bmatrix}-2 & 3\\1 & 2\end{bmatrix} \) such that \( 2B-3A+X=0 \) then \( X \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \begin{bmatrix}16 & -3\\4 & 5\end{bmatrix} \)
Here 0 on the right is the zero matrix O of order 2 × 2. Making X the subject:
\( \displaystyle X=3A-2B \)
\( \displaystyle 3A=\begin{bmatrix}12 & 3\\6 & 9\end{bmatrix} \)
\( \displaystyle 2B=\begin{bmatrix}-4 & 6\\2 & 4\end{bmatrix} \)
\( \displaystyle X=\begin{bmatrix}12-(-4) & 3-6\\6-2 & 9-4\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}16 & -3\\4 & 5\end{bmatrix} \)
Why the other options are incorrect
Option (1) \( \begin{bmatrix}-16 & 3\\-4 & -5\end{bmatrix} \) — Incorrect. This is \( 2B-3A=-X \); the sign is lost while moving the terms to the other side.
Option (3) \( \begin{bmatrix}16 & 4\\-3 & 5\end{bmatrix} \) — Incorrect. This is the transpose of X (rows and columns interchanged).
Option (4) \( \begin{bmatrix}-16 & -4\\3 & -5\end{bmatrix} \) — Incorrect. This is the negative of the transpose of X.