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Choose the correct option.
If Δ = |a b c; x y z; p q r|, then |ka kb kc; kx ky kz; kp kq kr| is:
If $\Delta = \begin{vmatrix} a & b & c \\ x & y & z \\ p & q & r \end{vmatrix}$, then $\begin{vmatrix} ka & kb & kc \\ kx & ky & kz \\ kp & kq & kr \end{vmatrix}$ is:
- (a)$\Delta$
- (b)$k\Delta$
- (c)$3k\Delta$
- (d)$k^3\Delta$
Answer
Answer
Correct option: (d)
Answer: (d) $k^3\Delta$
Each of the three rows has the common factor $k$; taking it out of every row gives $k\cdot k\cdot k\,\Delta=k^3\Delta$.