NIMCET 2015 — Mathematics PYQ
NIMCET | Mathematics | 2015If , where and are real positive numbers such that and then the equation that holds true among the following is:
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If A=abcamp;bamp;camp;aamp;camp;aamp;b, where a,b and c are real positive numbers such that abc=1 and ATA=I then the equation that holds true among the following is:
a+b+c=1
a2+b2+c2=1
ab+bc+ca=0
a3+b3+c3=4
(Correct Answer)a3+b3+c3=4
The given condition ATA=I implies that A is an orthogonal matrix. For an orthogonal matrix, the determinant ∣A∣ must be ±1.
The matrix A is a circulant matrix. Its determinant is given by:
Expanding the determinant:
Since ATA=I, then ∣ATA∣=∣I∣, which means ∣A∣2=1.
Therefore, ∣A∣=1 or ∣A∣=−1.
For a circulant matrix where a, b, c > 0, the expression a3+b3+c3−3abc is usually positive (unless a=b=c). Looking at the options, we use the property:
However, a simpler way is to look at the result of ATA=I directly. The product of the first row of AT (which is the first column of A) and the first column of A must be 1:
We are given abc=1. Let's test the value of the determinant expression:
We know the algebraic identity:
From ATA=I, the dot product of any two distinct columns must be 0:
But since a,b,c are real positive numbers, their sum and products cannot be zero.
Wait, let's re-examine ATA=I:
(Note: For this symmetric matrix, AT=A).
a2+b2+c2=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∣=±1.
If ∣A∣=−1:
Substitute abc=1:
Based on the calculation a3+b3+c3=4, the relationship holds true.
Correct Option: (d)
The given condition ATA=I implies that A is an orthogonal matrix. For an orthogonal matrix, the determinant ∣A∣ must be ±1.
The matrix A is a circulant matrix. Its determinant is given by:
Expanding the determinant:
Since ATA=I, then ∣ATA∣=∣I∣, which means ∣A∣2=1.
Therefore, ∣A∣=1 or ∣A∣=−1.
For a circulant matrix where a, b, c > 0, the expression a3+b3+c3−3abc is usually positive (unless a=b=c). Looking at the options, we use the property:
However, a simpler way is to look at the result of ATA=I directly. The product of the first row of AT (which is the first column of A) and the first column of A must be 1:
We are given abc=1. Let's test the value of the determinant expression:
We know the algebraic identity:
From ATA=I, the dot product of any two distinct columns must be 0:
But since a,b,c are real positive numbers, their sum and products cannot be zero.
Wait, let's re-examine ATA=I:
(Note: For this symmetric matrix, AT=A).
a2+b2+c2=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∣=±1.
If ∣A∣=−1:
Substitute abc=1:
Based on the calculation a3+b3+c3=4, the relationship holds true.
Correct Option: (d)