1.

Evaluate :\(\begin{bmatrix} 102& 18 & 36 \\[0.3em] 1 & 3 & 4 \\[0.3em] 17 & 3 & 6 \end{bmatrix}\) [(102,18,36)(1,3,4)(17,3,6)]

Answer»

\(\begin{bmatrix} 102& 18 & 36 \\[0.3em] 1 & 3 & 4 \\[0.3em] 17 & 3 & 6 \end{bmatrix}\) = 6 x \(\begin{bmatrix} 17& 18 & 6 \\[0.3em] 1 & 6 & 4 \\[0.3em] 17 & 3 & 6 \end{bmatrix}\)[R1’ = R1/6]

Now, for any determinant, if at least two rows are identical, then the value of the determinant becomes zero. 

Here, the first and third rows are identical. 

So, the value of the above determinant evaluated = 0



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