Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Equation | Value of x | ||
| (A) | \(\begin{vmatrix}x & 4\\ 3 & 6\end{vmatrix} = 6\) | (I) | 5 |
| (B) | \(\begin{vmatrix}2 & x\\ x & 8\end{vmatrix} = 0\), where \(x \gt 0\) | (II) | 6 |
| (C) | \(\begin{vmatrix}1 & 0 & 0\\ 2 & x & 0\\ 5 & 7 & 3\end{vmatrix} = 15\) | (III) | 3 |
| (D) | \(\begin{vmatrix}x + 1 & 2\\ 3 & 2\end{vmatrix} = 8\) | (IV) | 4 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
Expand each determinant and solve the resulting equation in x.
(A) \(\begin{vmatrix}x & 4\\ 3 & 6\end{vmatrix} = 6\)
This matches (III) 3.
\(6x - 12 = 6\)
\(x = 3\)
(B) \(\begin{vmatrix}2 & x\\ x & 8\end{vmatrix} = 0\), where \(x \gt 0\)
This matches (IV) 4.
\(16 - x^{2} = 0\)
\(x = \pm 4\)
With \(x \gt 0\), x = 4.
(C) \(\begin{vmatrix}1 & 0 & 0\\ 2 & x & 0\\ 5 & 7 & 3\end{vmatrix} = 15\)
This matches (I) 5.
The determinant is lower triangular, so it equals the product of the diagonal elements:
\(1 \cdot x \cdot 3 = 15\)
\(x = 5\)
(D) \(\begin{vmatrix}x + 1 & 2\\ 3 & 2\end{vmatrix} = 8\)
This matches (II) 6.
\(2(x + 1) - 6 = 8\)
\(2x - 4 = 8\)
\(x = 6\)