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If A=[0a1−1b1−1c0]A = \begin{bmatrix}0 & a & 1\\-1 & b & 1\\-1 & c & 0\end{bmatrix} is a skew symmetric matrix, then the value of (a+b−c)2(a+b-c)^2 is

Solution

✅ Correct Option: 1

For a skew-symmetric matrix, diagonal entries are zero and aij=−ajia_{ij}=-a_{ji}.

So b=0b=0; from a12=−a21a_{12}=-a_{21}, a=1a=1; from a23=−a32a_{23}=-a_{32}, 1=−c1=-c, so c=−1c=-1.

Then (a+b−c)2=(1+0+1)2=4(a+b-c)^2=(1+0+1)^2=4.

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