If det |6i -3i 1; 4 3i -1; 20 3 i| = x + iy, then (A) x = 3, y = 2 (B) x = 1, y = 3 (C) x = 0, y = 3

Q: Why can't I just expand the determinant directly? It seems straightforward.

While direct expansion is mathematically valid, it is a trap. In JEE Advanced, time is your most precious resource. Direct expansion of a $3 \times 3$ matrix with complex numbers is a breeding ground for sign errors. Always look for row or column operations to create zeros first.

Q: Does adding one row to another change the value of the determinant?

Absolutely not! The property states that $\det(A) = \det(A')$ where $A'$ is obtained by $R_i \to R_i + kR_j$. It is one of the most elegant properties of linear algebra, allowing us to simplify matrices without altering their fundamental value.

Q: How do I handle the complex numbers $i$ during the calculation?

Treat $i$ as a constant throughout the algebraic manipulation. Only when you reach the final step, use the fundamental definition $i^2 = -1$ to simplify the expression. Don't let the presence of $i$ intimidate you; it follows the same algebraic rules as any other variable.

Question

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

Options:

(A) x = 3, y = 2

(B) x = 1, y = 3

(C) x = 0, y = 3

(D) x = 0, y = 0

Sigma Percentile · Accepted Answer

### The Art of the Elegant Solution

Welcome, future engineer. Today, we are going to look at a problem that might seem like a tedious exercise in arithmetic, but it is actually a masterclass in the beauty of linear algebra.

When you first see a determinant like

$$
\begin{vmatrix} 6i & -3i & 1 \ 4 & 3i & -1 \ 20 & 3 & i \end{vmatrix} = x + iy
$$

your instinct might be to jump straight into the expansion. But wait! In the world of JEE Advanced, the fastest way to the answer is rarely the brute-force path. It is the path of observation.

### The Power of Observation

Before we touch a single variable, let us look at the structure. Look at the first and second rows, and the second and third columns.

In the second column, we have $-3i$ and $3i$. In the third column, we have $1$ and $-1$. These are additive inverses!

This is not a coincidence; it is an invitation to simplify. If we add the second row to the first, the magic happens. We apply the operation $R_1 	o R_1 + R_2$.

This operation is perfectly safe because it does not change the value of the determinant. It is like looking at the same object from a slightly different, more convenient angle.

### The Transformation

Let us perform the addition. For the first column, we have $6i + 4$. For the second column, we have $-3i + 3i = 0$. For the third column, we have $1 + (-1) = 0$.

Our determinant now looks like this:

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

Look at that first row! Two zeros! This is the moment where the problem collapses. Expanding along a row with the maximum number of zeros is the hallmark of a seasoned problem solver.

### The Final Collapse

Now, we expand along the first row. Because two elements are zero, we only need to consider the first term:

$$
(6i + 4) 	imes \begin{vmatrix} 3i & -1 \ 3 & i \end{vmatrix}
$$

Now, we evaluate the $2 	imes 2$ minor. The product of the diagonal elements is $(3i)(i) = 3i^2$. The product of the off-diagonal elements is $(3)(-1) = -3$.

Subtracting these gives us:

$$
3i^2 - (-3) = 3i^2 + 3
$$

Since $i^2 = -1$, this becomes $3(-1) + 3 = -3 + 3 = 0$. The entire minor is zero!

Therefore, the whole determinant is $(6i + 4) 	imes 0 = 0$. We are left with $0 = x + iy$.

By comparing the real and imaginary parts, we conclude that **$x = 0$** and **$y = 0$**. It is elegant, it is swift, and it is the kind of insight that wins exams. Keep looking for these symmetries, and you will find that even the most intimidating problems have a simple, beautiful core.

Animated Solution for Mathematics - Matrices and Determinants: If $6 i 420 - 3 i 3 i 3 1 - 1 i = x + i y$ , then

Select Answer:

Visualized Solution

Analyze the Determinant

Observe Row Elements

Apply Row Operation

Calculate New Elements

Simplified Determinant

Expand along $R_{1}$

Identify the Minor

Evaluate the $2 \times 2$ Determinant

Simplify using $i^{2} = - 1$

Final Determinant Value

Equate to RHS

Compare and Conclude

The Sigma Insight: Properties of Determinants

The Art of the Elegant Solution

The Power of Observation

The Transformation

The Final Collapse

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Animated Solution for Mathematics - Matrices and Determinants: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ​6i420​−3i3i3​1−1i​​=x+iy, then