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Choose the correct option.
If |a b c; x+a y+2a z+3a; 1 2 3| = |a b c; x y z; 1 2 3| + |B|, then the value of |B| is equal to
If $\begin{vmatrix} a & b & c \\ x+a & y+2a & z+3a \\ 1 & 2 & 3 \end{vmatrix} = \begin{vmatrix} a & b & c \\ x & y & z \\ 1 & 2 & 3 \end{vmatrix} + |B|$, then the value of $|B|$ is equal to
- (a)$a + 2b + 3c$
- (b)$a - b + 2c$
- (c)$0$
- (d)$3a$
Answer
Answer
Correct option: (c)
Answer: (c) $0$
The second row splits as $(x, y, z) + (a, 2a, 3a)$, so $|B|$ has rows $(a, b, c)$, $a(1, 2, 3)$ and $(1, 2, 3)$.
Rows 2 and 3 are proportional, so $|B| = 0$.