Let a, b, c ∈ R be all non-zero and satisfy a³ + b³ + c³ = 2. If the matrix A = [[a, b, c], [b, c, a

Q: Wait, why is $AA^T = I$ the same as saying the rows are orthonormal?

Because $AA^T = I$ is the definition of an orthogonal matrix. If $A = [r_1, r_2, r_3]^T$, then $AA^T$ is a matrix where the $(i,j)$-th entry is the dot product of row $i$ and row $j$. Since $AA^T = I$, the diagonal entries (dot product of a row with itself) are 1, and off-diagonal entries (dot product of distinct rows) are 0.

Q: Why do we get two cases for $a+b+c$?

Because $(a+b+c)^2 = 1$. When you take the square root, you must consider both positive and negative roots. Ignoring the negative root is a common trap in JEE Advanced problems.

Q: Is there a faster way to solve this without the cubic identity?

While you could try to solve for $a, b, c$ individually, that would be a nightmare of algebra. The cubic identity is the most elegant and efficient path. Always look for symmetric identities when dealing with cyclic expressions.

Question

Let $a, b, c \in R$ be all non-zero and satisfy $a^3 + b^3 + c^3 = 2$. If the matrix $A = \begin{pmatrix} a & b & c \ b & c & a \ c & a & b \end{pmatrix}$ satisfies $AA^T = I$, then a value of $abc$ can be :

Options:

(A) $3$

(B) $\frac{1}{3}$

(C) $-\frac{1}{3}$

(D) $\frac{2}{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a matrix $A = \begin{pmatrix} a & b & c \ b & c & a \ c & a & b \end{pmatrix}$. At first glance, it looks like a simple arrangement of variables.

In the world of JEE Advanced, this is not just a matrix; it is a structure of hidden symmetry. The problem states that $AA^T = I$, which is the golden ticket.

This condition implies that $A$ is an **orthogonal matrix**. In the language of linear algebra, this means the rows of $A$ are orthonormal vectors; they are mutually perpendicular and each has a magnitude of one.

### Decoding the Orthogonality

Let us look at the first row: $(a, b, c)$. Since it is a unit vector, its dot product with itself must be one.

This gives us our first equation:
$$a^2 + b^2 + c^2 = 1$$

Now, consider the first and second rows: $(a, b, c)$ and $(b, c, a)$. Because they are perpendicular, their dot product must be zero.

This yields:
$$ab + bc + ca = 0$$

These two equations are the foundation of our solution. They are simple, yet they contain the entire geometry of the matrix.

### The Algebraic Bridge

We are given $a^3 + b^3 + c^3 = 2$ and we need to find $abc$. We utilize the classic algebraic identity:
$$a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - (ab + bc + ca))$$

We already know $a^2 + b^2 + c^2 = 1$ and $ab + bc + ca = 0$. To find $(a+b+c)$, we use the expansion:
$$(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$$

Substituting our known values:
$$(a+b+c)^2 = 1 + 2(0) = 1$$
This implies that $a+b+c = \pm 1$.

### Final Calculation

Now we evaluate the two possible cases using our master identity.

**Case 1:** $a+b+c = 1$
Substituting into the identity:
$$2 - 3abc = (1)(1 - 0)$$
$$2 - 3abc = 1 \implies 3abc = 1 \implies abc = \frac{1}{3}$$

**Case 2:** $a+b+c = -1$
Substituting into the identity:
$$2 - 3abc = (-1)(1 - 0)$$
$$2 - 3abc = -1 \implies 3abc = 3 \implies abc = 1$$

We have two possible values for $abc$: $\frac{1}{3}$ and $1$. Given the standard options for such problems, the result is **$\frac{1}{3}$**.

Animated Solution for Mathematics - Matrices and Determinants: Let $a, b, c \in R$ be all non-zero and satisfy $a^{3} + b^{3} + c^{3} = 2$ . If the matrix $A = a b c b c a c a b$ satisfies $A A^{T} = I$ , then a value of $ab c$ can be :

Select Answer:

Visualized Solution

Introduction to Matrix $A$

The Orthogonality Condition $A A^{T} = I$

Deriving Equations from Orthogonality

Calculating $(a + b + c)^{2}$

Finding $a + b + c$

The Cubic Algebraic Identity

Substitution into the Identity

Solving Case 1: $a + b + c = 1$

Solving Case 2: $a + b + c = - 1$

Final Conclusion and Summary

The Sigma Insight: Types of Matrices

Analyzing the Setup

Decoding the Orthogonality

The Algebraic Bridge

Final Calculation

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Visualized Solution

Introduction to Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

The Orthogonality Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AAT=I

Deriving Equations from Orthogonality

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (a+b+c)2

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