Question

The roots of the equation $\begin{vmatrix} 2 & -2 & 4 \\ -5 & x+2 & -10 \\ -1 & 1 & x+1 \end{vmatrix} = 0$, are

• A. 3, -3
• B. 0, 5
• C. 6, -6
• D. 5, -5
• E. 0, -5

Hint:

Look for relationships between columns. Here $C_1$ and $C_2$ were very similar ($2, -2$ and $-1, 1$). Adding them creates zeros, which drastically simplifies the expansion.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We need to find values of $x$ that satisfy the determinantal equation. Row/column operations can simplify the determinant before expansion.

Step 2: Detailed Explanation:
1. Perform Column Operation $C_2 \to C_2 + C_1$:
\[ \begin{vmatrix} 2 & 0 & 4 \\ -5 & x-3 & -10 \\ -1 & 0 & x+1 \end{vmatrix} = 0 \]
2. Expand the determinant along the second column:
\[ (x-3) \begin{vmatrix} 2 & 4 \\ -1 & x+1 \end{vmatrix} = 0 \]
(Note: the sign for the $(2,2)$ element in a $3\times3$ expansion is positive).
3. Solve the resulting equation:
\[ (x-3) [2(x+1) - (-4)] = 0 \]
\[ (x-3) [2x + 2 + 4] = 0 \]
\[ (x-3) (2x + 6) = 0 \]
4. Find the roots:
- $x - 3 = 0 \implies x = 3$
- $2x + 6 = 0 \implies x = -3$

Step 3: Final Answer:
The roots are 3 and -3.