Question

If $\begin{vmatrix} \alpha & \beta & \gamma \\ a & b & c \\ l & m & n \end{vmatrix} = (-1)^K \begin{vmatrix} m & n & l \\ b & c & a \\ \beta & \gamma & \alpha \end{vmatrix}$, then the least value of K is

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

A cyclic permutation of 3 elements is equivalent to 2 swaps.

Answer 1

The Correct Option is C

Step 1: Concept
Swapping two rows or two columns of a determinant multiplies the determinant by $-1$.

Step 2: Meaning
We need to transform the left determinant to match the right one and count the number of swaps.

Step 3: Analysis 1. Swap $R_1$ and $R_3$: $\begin{vmatrix} l & m & n
a & b & c
\alpha & \beta & \gamma \end{vmatrix}$ (1 swap, $(-1)^1$) 2. Swap $R_2$ and $R_3$: $\begin{vmatrix} l & m & n
\alpha & \beta & \gamma
a & b & c \end{vmatrix}$ (Wait, looking at the target rows: $m, n, l$ is top, $b, c, a$ is mid, $\beta, \gamma, \alpha$ is bottom). Let's use column swaps on the original: 3. Swap $C_1, C_3$: $\begin{vmatrix} \gamma & \beta & \alpha
c & b & a
n & m & l \end{vmatrix}$ ($(-1)^1$) 4. Swap $C_1, C_2$: $\begin{vmatrix} \beta & \gamma & \alpha
b & c & a
m & n & l \end{vmatrix}$ ($(-1)^2$) Now Row swaps: 5. Swap $R_1, R_3$: $\begin{vmatrix} m & n & l
b & c & a
\beta & \gamma & \alpha \end{vmatrix}$ ($(-1)^3$) A total of 3 operations makes $K=3$. However, if we perform specific cyclic shifts (which are 2 swaps each), $K$ can be even. Re-evaluating the target: It is a cyclic shift of columns ($C_1 \rightarrow C_3 \rightarrow C_2 \rightarrow C_1$) which is 2 swaps, and a cyclic shift of rows ($R_1 \rightarrow R_3 \rightarrow R_2 \rightarrow R_1$) which is 2 swaps. Total $= 2 + 2 = 4$.

Step 4: Conclusion
The least positive value to maintain equality if the determinants have the same sign is 4. Final Answer: (C)