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)
|
List-I |
List-II |
|
(A) $l=m=n=$ |
(III) 2 |
|
(B) $k+l+m+n=$ |
(I) 10 |
|
(C) $k^2 + l^2 + (m-n)^2=$ |
(IV) 20 |
|
(D) $l^2 + m^2 + (n-k)=$ |
(II) 6 |
Given:
$\left|\begin{array}{ccc} -a^2 & ab & ac \\ ba & -b^2 & bc \\ ac & bc & -c^2 \end{array}\right| = k \cdot a^l b^m c^n$
Factor terms row-wise:
- Row 1: factor out $a$ → $a(-a, b, c)$
- Row 2: factor out $b$ → $b(a, -b, c)$
- Row 3: factor out $c$ → $c(a, b, -c)$
$abc \cdot \left|\begin{array}{ccc} -a & b & c \\ a & -b & c \\ a & b & -c \end{array}\right|$
Now evaluate the inner $3 \times 3$ determinant:
Expand along first row:
$= -a \cdot \left| \begin{array}{cc} -b & c \\ b & -c \end{array} \right| - b \cdot \left| \begin{array}{cc} a & c \\ a & -c \end{array} \right| + c \cdot \left| \begin{array}{cc} a & -b \\ a & b \end{array} \right|$
$= -a(0) - b(-2ac) + c(2ab) = 0 + 2abc + 2abc = 4abc$
Total determinant:
$abc \cdot 4abc = 4a^2b^2c^2$
So: $k = 4$, $l = m = n = 2$
Now match List-I with List-II:
(A) $l = m = n = 2$ ⟹ (A) → (III)
(B) $k + l + m + n = 4 + 2 + 2 + 2 = 10$ ⟹ (B) → (I)
(C) $k^2 + l^2 + (m - n)^2 = 4^2 + 2^2 + (2 - 2)^2 = 16 + 4 + 0 = 20$ ⟹ (C) → (IV)
(D) $l^2 + m^2 + (n - k) = 2^2 + 2^2 + (2 - 4) = 4 + 4 - 2 = 6$ ⟹ (D)→ (II)