Consider a determinant \( \Delta=\begin{vmatrix} 3a & -a+b & -a+c\\ -b+a & 3b & -b+c\\ -c+a & -c+b & 3c\end{vmatrix} \), then
Match the LIST-I with LIST-II
| LIST-I (Value of \( a,b,c \)) | LIST-II (Value of \( \Delta \)) | ||
| A. | \( a=0,b=0,c=1 \) | I. | 27 |
| B. | \( a=0,b=1,c=1 \) | II. | 0 |
| C. | \( a=0,b=2,c=-1 \) | III. | 6 |
| D. | \( a=1,b=1,c=1 \) | IV. | \( -6 \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-II, B-III, C-IV, D-I
In each case the values are put into \( \Delta \) and the determinant is expanded along the first row.
(A) a = 0, b = 0, c = 1
This matches (II) 0.
\( \displaystyle \Delta=\begin{vmatrix} 0 & 0 & 1\\ 0 & 0 & 1\\ -1 & -1 & 3\end{vmatrix}=0 \)
since the first two rows are identical.
(B) a = 0, b = 1, c = 1
This matches (III) 6.
\( \displaystyle \Delta=\begin{vmatrix} 0 & 1 & 1\\ -1 & 3 & 0\\ -1 & 0 & 3\end{vmatrix} \)
\( =0-1(-3-0)+1(0+3) \)
\( =3+3=6 \)
(C) a = 0, b = 2, c = −1
This matches (IV) −6.
\( \displaystyle \Delta=\begin{vmatrix} 0 & 2 & -1\\ -2 & 6 & -3\\ 1 & 3 & -3\end{vmatrix} \)
\( =0-2(6+3)+(-1)(-6-6) \)
\( =-18+12=-6 \)
(D) a = 1, b = 1, c = 1
This matches (I) 27.
\( \displaystyle \Delta=\begin{vmatrix} 3 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 3\end{vmatrix} \)
\( =3\times 3\times 3=27 \)
So the matching is (A)-(II), (B)-(III), (C)-(IV), (D)-(I).