Question

If \( \begin{vmatrix} 3i & -9i & 1 \\ 2 & 9i & -1 \\ 10 & 9 & i \end{vmatrix} = x + iy \), then:

• A. \( x = 1, y = 1 \)
• B. \( x = 0, y = 1 \)
• C. \( x = 1, y = 0 \)
• D. \( x = 0, y = 0 \)
• E. \( x = -1, y = 0 \)

Hint:

When a determinant contains many \( i \) terms, check if expansion leads to terms that cancel out. Often, these problems are designed to result in \( 0 \) or a purely real/imaginary number.

Answer 1

The Correct Option is D

Concept: To find \( x \) and \( y \), we evaluate the determinant. We can simplify the columns or rows by factoring out common terms like \( i \) or \( 9 \). If any two rows or columns become proportional, the determinant will be zero.

Step 1: Simplifying the determinant columns.
Let's factor out \( 3i \) from the first column and \( 9 \) from the second column: \[ \Delta = (3i)(9) \begin{vmatrix} 1 & -i & 1 \\ \frac{2}{3i} & i & -1 \\ \frac{10}{3i} & 1 & i \end{vmatrix} \] Alternatively, notice Row 1 and Row 2. If we add Row 1 and Row 2: \( R_1 + R_2 = (3i+2, 0, 0) \). This doesn't immediately show zero. Let's expand directly: \[ \Delta = 3i(9i^2 - (-9)) - (-9i)(2i - (-10)) + 1(18 - 90i) \] \[ = 3i(-9 + 9) + 9i(2i + 10) + 18 - 90i \]

Step 2: Calculating the final value.
\[ \Delta = 3i(0) + 18i^2 + 90i + 18 - 90i \] \[ \Delta = 0 - 18 + 90i + 18 - 90i = 0 \] Since \( x + iy = 0 \), comparing real and imaginary parts gives \( x = 0 \) and \( y = 0 \).