The value of the determinant |1 a a²-bc; 1 b b²-ca; 1 c c²-ab| is .........

Q: Wait, why can't we just expand the determinant directly? It's a $3 \times 3$ matrix, so it shouldn't be that hard, right?

You are absolutely right that you *could* expand it directly. However, in JEE Advanced, time is your most precious resource. Expanding directly involves cubic terms that are prone to algebraic errors and take much longer to simplify. Using row operations to create zeros is the 'pro' way to solve this, turning a 5-minute calculation into a 30-second observation.

Q: Why did we choose $R_1 - R_2$ and $R_2 - R_3$? Could we have done something else?

Great question! You could have used $R_2 - R_1$ or $R_3 - R_2$. The goal is simply to create zeros in the first column. As long as you perform operations that create at least two zeros in a row or column, you are on the right track. The specific choice of rows just depends on which subtraction makes the algebra in the third column easier to factor.

Q: What if the rows didn't end up being identical? How would I know when to stop?

If the rows don't become identical, you continue the simplification until you have a row or column with at least two zeros. Once you have that, you expand along that row or column. The goal is always to reduce the order of the determinant or to make the expansion trivial. If you see a pattern, follow it; if you don't, simplify until you do!

Question

The value of the determinant $\begin{vmatrix} 1 & a & a^2 - bc \ 1 & b & b^2 - ca \ 1 & c & c^2 - ab \end{vmatrix}$ is .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a massive, intimidating $3 	imes 3$ matrix. The problem asks you to find the value of the determinant:

$$\Delta = \begin{vmatrix} 1 & a & a^2 - bc \ 1 & b & b^2 - ca \ 1 & c & c^2 - ab \end{vmatrix}$$

At first glance, your instinct might be to dive straight into the expansion. You might start calculating the minors, multiplying terms, and bracing yourself for a storm of cubic algebra. But stop. Take a breath. In the world of JEE Advanced, the most elegant solutions are rarely the ones that require the most brute force.

### The Power of the First Column

Look closely at the first column. It is a sequence of ones: $[1, 1, 1]^T$. This is not a coincidence; it is a gift. Whenever you see a column (or row) of ones in a determinant problem, your brain should immediately light up.

This is a golden opportunity to create zeros. Why do we want zeros? Because zeros are the best friends of a mathematician—they make expansion trivial. If we can turn two of those ones into zeros, the entire determinant collapses into a much simpler $2 	imes 2$ form.

Our strategy is simple: we will perform row operations. We will subtract the second row from the first ($R_1 ightarrow R_1 - R_2$) and the third row from the second ($R_2 ightarrow R_2 - R_3$).

### The Algebraic Dance

Let us execute the first operation, $R_1 ightarrow R_1 - R_2$. The first column becomes $1 - 1 = 0$. The second column becomes $a - b$.

Now, the third column is where the magic happens. We subtract $(b^2 - ca)$ from $(a^2 - bc)$, giving us $(a^2 - b^2) + (ca - bc)$.

This looks complex, but let us group the terms: $(a^2 - b^2) + c(a - b)$. Notice the structure? $a^2 - b^2$ is a difference of squares, which factors into $(a - b)(a + b)$.

So, our expression becomes $(a - b)(a + b) + c(a - b)$. If we factor out the $(a - b)$, we get $(a - b)(a + b + c)$.

We repeat this for the second row ($R_2 ightarrow R_2 - R_3$). The first column becomes $0$, the second becomes $b - c$, and the third becomes $(b^2 - c^2) + a(b - c)$, which factors into $(b - c)(b + c) + a(b - c) = (b - c)(a + b + c)$.

### The 'Aha!' Moment

Now, look at what we have created. Our determinant looks like this:

$$\Delta = \begin{vmatrix} 0 & a - b & (a - b)(a + b + c) \ 0 & b - c & (b - c)(a + b + c) \ 1 & c & c^2 - ab \end{vmatrix}$$

Notice that in the first row, every element has a common factor of $(a - b)$. In the second row, every element has a common factor of $(b - c)$. We can pull these factors out of the determinant entirely:

$$\Delta = (a - b)(b - c) \begin{vmatrix} 0 & 1 & a + b + c \ 0 & 1 & a + b + c \ 1 & c & c^2 - ab \end{vmatrix}$$

Do you see it? The first row and the second row are now identical! They are both $[0, 1, a + b + c]$.

### The Final Revelation

One of the most fundamental properties of determinants is that if any two rows (or columns) are identical, the value of the determinant is zero. It does not matter what $a$, $b$, or $c$ are. The symmetry we uncovered forces the entire expression to vanish.

$$\Delta = (a - b)(b - c) 	imes 0 = 0$$

This is the beauty of mathematics. We didn't need to expand a single cubic term. We didn't need to get lost in a forest of variables. We simply looked for the pattern, used the properties of determinants to our advantage, and watched as the complexity dissolved into nothingness.

Remember this: always look for symmetry, always look for the 'ones,' and always trust the properties of the system. You've got this! The final answer is **0**.

Animated Solution for Mathematics - Matrices and Determinants: The value of the determinant $111 a b c a^{2} - b c b^{2} - c a c^{2} - ab$ is .........

Enter Numerical Value:

Visualized Solution

The Given Determinant $Δ$

Planning Row Operations

Transforming Row $1$ ( $R_{1} \to R_{1} - R_{2}$ )

Transforming Row $2$ ( $R_{2} \to R_{2} - R_{3}$ )

Factoring Column $3$

Extracting Common Factors

Identical Rows Property

Final Answer & Key Takeaway

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Power of the First Column

The Algebraic Dance

The 'Aha!' Moment

The Final Revelation

Similar Questions

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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; } The value of the determinant ​111​abc​a2−bcb2−cac2−ab​​ is .........