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If the matrix [zx−y+2x+4y−3x+y0a−32x−y+3a−10]\begin{bmatrix} z & x-y+2 & x+4y-3 \\ x+y & 0 & a-3 \\ 2x-y+3 & a-1 & 0 \end{bmatrix} is skew-symmetric, then the value of (a+x+y+z)(a+x+y+z) is:

Solution

✅ Correct Option: 3

Skew-symmetry forces zero diagonal, so z=0z=0. Comparing a12=−a21a_{12}=-a_{21} gives x−y+2=−(x+y)x-y+2=-(x+y), so x=−1x=-1. Comparing a13=−a31a_{13}=-a_{31} gives 3x+3y=03x+3y=0, so y=1y=1. Comparing a23=−a32a_{23}=-a_{32} gives a−3=−(a−1)a-3=-(a-1), so a=2a=2. Hence a+x+y+z=2a+x+y+z=2.

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