Complex Numbers
Complex Number in Iota Form
Complex Numbers_PYQ
Grade 11

Question:

If $\begin{vmatrix} 6i & -3i & 1 \\ 4 & 3i & -1 \\ 20 & 3 & i \end{vmatrix} = x + iy$, then
$x = 3,\; y = 1$
$x = 1,\; y = 3$
$x = 0,\; y = 3$
$x = 0,\; y = 0$

Step-by-Step Solution

Key Concept: In a determinant with complex entries, one minor being zero (here $M_{11}=0$) can simplify computation dramatically; always evaluate minors before expanding.
**Step 1: Expand along Row 1** $$D = 6i\begin{vmatrix}3i & -1\\3 & i\end{vmatrix} - (-3i)\begin{vmatrix}4 & -1\\20 & i\end{vmatrix} + 1\begin{vmatrix}4 & 3i\\20 & 3\end{vmatrix}$$ **Step 2: Evaluate each 2×2 minor** $M_{11} = 3i^2 + 3 = -3+3 = 0$; $M_{12} = 4i + 20$; $M_{13} = 12 - 60i$. **Step 3: Assemble and simplify** $D = 6i(0) + 3i(4i+20) + (12-60i) = 0 + 12i^2 + 60i + 12 - 60i = -12 + 12 = 0$. So $x = 0,\; y = 0$.
Correct Answer: 4

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