The values of \( x,y,z \) and \( w \) such that \( \begin{bmatrix}x-y & 2z+w\\2x-y & 2x+w\end{bmatrix}=\begin{bmatrix}5 & 3\\12 & 15\end{bmatrix} \) are given by
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( x=7,y=2,\ z=1 \) and \( w=1 \)
Two matrices are equal when their corresponding entries are equal:
\( x-y=5 \) and \( 2x-y=12 \). Subtracting, x = 7, and then y = 2.
\( 2x+w=15 \), so 14 + w = 15 and w = 1.
\( 2z+w=3 \), so 2z = 2 and z = 1.
Shortcut: Only Options (2), (3) and (4) have x = 7. Of these, x − y = 5 needs y = 2, leaving (2) and (3); and 2x + w = 15 needs w = 1, which leaves Option (3).