If \(\begin{vmatrix}-a^2&ab&ac\\ba&-b^2&bc\\ac&bc&-c^2\end{vmatrix}=k.a^lb^mc^n\), then
Match List-I with List-II
| List-I | List-II | ||
| (A) | \(l=m=n=\) | (I) | 10 |
| (B) | \(k+l+m+n=\) | (II) | 6 |
| (C) | \(k^2 + l^2 + (m-n)^2=\) | (III) | 2 |
| (D) | \(l^2 + m^2 + (n-k)=\) | (IV) | 20 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) - (III), (B) - (I), (C) - (IV), (D) - (II)
Let \(\Delta\) be the determinant. Taking the common factors a from \(R_1\), b from \(R_2\) and c from \(R_3\):
\(\displaystyle \Delta = abc\begin{vmatrix}-a&b&c\\a&-b&c\\a&b&-c\end{vmatrix}\)
Now a from \(C_1\), b from \(C_2\) and c from \(C_3\):
\(\displaystyle \Delta = a^2b^2c^2\begin{vmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{vmatrix}\)
Expanding along the first row:
\(-1(1 - 1) - 1(-1 - 1) + 1(1 + 1) = 0 + 2 + 2 = 4\)
\(\Delta = 4a^2b^2c^2\), so \(k = 4\) and \(l = m = n = 2\)
(A) \(l=m=n=\)
This matches (III) 2.
(B) \(k+l+m+n=\)
This matches (I) 10:
\(4 + 2 + 2 + 2 = 10\)
(C) \(k^2 + l^2 + (m-n)^2=\)
This matches (IV) 20:
\(16 + 4 + 0 = 20\)
(D) \(l^2 + m^2 + (n-k)=\)
This matches (II) 6:
\(4 + 4 + (2 - 4) = 6\)
So the matching is (A) - (III), (B) - (I), (C) - (IV), (D) - (II).