NIMCET 2015 — Mathematics PYQ

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Question

NIMCET 2015 — Mathematics PYQ

NIMCET | Mathematics | 2015

If $A = \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix}$ , where $a, b$ and $c$ are real positive numbers such that $abc = 1$ and $A^T A = I$ then the equation that holds true among the following is:

Choose the correct answer:

Explanation

1. Condition for Orthogonal Matrix

The given condition $A^T A = I$ implies that $A$ is an orthogonal matrix. For an orthogonal matrix, the determinant $|A|$ must be $\pm 1$ .

2. Calculate the Determinant of A

The matrix $A$ is a circulant matrix. Its determinant is given by:

|A| = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}

Expanding the determinant:

|A| = a(cb - a^2) - b(b^2 - ac) + c(ba - c^2)

|A| = abc - a^3 - b^3 + abc + abc - c^3

|A| = 3abc - (a^3 + b^3 + c^3)

3. Use the Orthogonal Property

Since $A^T A = I$ , then $|A^T A| = |I|$ , which means $|A|^2 = 1$ .

Therefore, $|A| = 1$ or $|A| = -1$ .

For a circulant matrix where $a, b, c > 0$ , the expression $a^3 + b^3 + c^3 - 3abc$ is usually positive (unless $a=b=c$ ). Looking at the options, we use the property:

(a^3 + b^3 + c^3 - 3abc)^2 = 1

However, a simpler way is to look at the result of $A^T A = I$ directly. The product of the first row of $A^T$ (which is the first column of $A$ ) and the first column of $A$ must be $1$ :

a^2 + b^2 + c^2 = 1

4. Relate to the Options

We are given $abc = 1$ . Let's test the value of the determinant expression:

|A| = -(a^3 + b^3 + c^3 - 3abc)

We know the algebraic identity:

a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)

From $A^T A = I$ , the dot product of any two distinct columns must be $0$ :

ab + bc + ca = 0

But since $a, b, c$ are real positive numbers, their sum and products cannot be zero.

Wait, let's re-examine $A^T A = I$ :

\begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix} \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

(Note: For this symmetric matrix, $A^T = A$ ).

$a^2 + b^2 + c^2 = 1$
$ab + bc + ca = 0$

Since $a, b, c > 0$ , $ab + bc + ca$ cannot be $0$ . This implies the question likely contains a typo and should be interpreted via the determinant $|A| = \pm 1$ .

If $|A| = -1$ :

3abc - (a^3 + b^3 + c^3) = -1

Substitute $abc = 1$ :

3(1) - (a^3 + b^3 + c^3) = -1

3 - (a^3 + b^3 + c^3) = -1

a^3 + b^3 + c^3 = 4

Conclusion

Based on the calculation $a^3 + b^3 + c^3 = 4$ , the relationship holds true.

Correct Option: (d)