If \( \Delta=\begin{vmatrix}1+a&1&1\\1&1+b&1\\1&1&1+c\end{vmatrix} \), Then
Match List-I with List-II
| List-I | List-II | ||
| Values of \( a,b \) & \( c \) | value of \( \Delta \) | ||
| (A). | \( a=b=c=1 \) | (I). | 2 |
| (B). | \( a=-1,b=1,c=1 \) | (II). | 0 |
| (C). | \( a=-1,b=-1,c=1 \) | (III). | −2 |
| (D). | \( a=-1,b=-1,c=-1 \) | (IV). | 4 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) - (IV), (B) - (III), (C) - (II), (D) - (I)
Expanding along the first row,
\( \Delta=(1+a)\left[(1+b)(1+c)-1\right]-\left[(1+c)-1\right]+\left[1-(1+b)\right] \)
\( =(1+a)(b+c+bc)-c-b \)
\( =abc+ab+bc+ca \)
(A) \( a=b=c=1 \)
This matches (IV) 4: \( \Delta=1+1+1+1=4 \).
(B) \( a=-1,b=1,c=1 \)
This matches (III) −2: \( \Delta=-1-1+1-1=-2 \).
(C) \( a=-1,b=-1,c=1 \)
This matches (II) 0: \( \Delta=1+1-1-1=0 \).
(D) \( a=-1,b=-1,c=-1 \)
This matches (I) 2: \( \Delta=-1+1+1+1=2 \).
So the matching is (A)-(IV), (B)-(III), (C)-(II), (D)-(I).
Tip: The result \( \Delta=abc+ab+bc+ca \) for this determinant is worth remembering; for non-zero a, b, c it is often written as \( abc\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \).