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The values of x,y,zx, y, z and ww such that [x−y2z+w2x−y2x+w]=[531215]\begin{bmatrix} x-y & 2z+w \\ 2x-y & 2x+w \end{bmatrix} = \begin{bmatrix} 5 & 3 \\ 12 & 15 \end{bmatrix} are given by

Solution

✅ Correct Option: 3

Equating entries: x−y=5x-y=5 and 2x−y=122x-y=12 give x=7x=7, y=2y=2. Then 2x+w=152x+w=15 gives w=1w=1, and 2z+w=32z+w=3 gives z=1z=1.

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