Let R=(\beginarraycccx & 0 & 0 0 & y & 0 0 & 0 & z\endarray) be a non-zero 3 \times 3 matrix, where x \sin \theta=y \sin (\theta+\frac2 \pi3)=z \sin (\theta+\frac4 \pi3) eq 0, \theta \in(0,2 \pi) . For a square matrix M , let trace (M) denote the sum of all the diagonal entries of M . Then, among the statements: (I) Trace (R)=0 (II) If trace (\operatornameadj(\operatornameadj(R))=0 , then R has exactly one non-zero entry.

Matrices and Determinants PYQ — JEE Main 2024 · 30 Jan · Shift 2 · Q37 with Solution

View written solutionFree

Correct answer: B

Given condition

Let

R=\begin{pmatrix} x&0&0\\ 0&y&0\\ 0&0&z \end{pmatrix}

with

x\sin\theta=y\sin\left(\theta+\frac{2\pi}{3}\right)=z\sin\left(\theta+\frac{4\pi}{3}\right)\neq 0.

Let the common non-zero value be $k$ . Then

x=\frac{k}{\sin\theta},\qquad y=\frac{k}{\sin\left(\theta+\frac{2\pi}{3}\right)},\qquad z=\frac{k}{\sin\left(\theta+\frac{4\pi}{3}\right)}.

Since the common value is non-zero, all three denominators are non-zero, hence $x,y,z$ are all non-zero.

Check statement (I): $\operatorname{Trace}(R)=0$

We need to evaluate

\operatorname{Trace}(R)=x+y+z.

Using the expressions above,

x+y+z=k\left(\frac{1}{\sin\theta}+\frac{1}{\sin\left(\theta+\frac{2\pi}{3}\right)}+\frac{1}{\sin\left(\theta+\frac{4\pi}{3}\right)}\right).

So it is enough to prove

\frac{1}{\sin\theta}+\frac{1}{\sin\left(\theta+\frac{2\pi}{3}\right)}+\frac{1}{\sin\left(\theta+\frac{4\pi}{3}\right)}=0.

Let

a=\sin\theta,\quad b=\sin\left(\theta+\frac{2\pi}{3}\right),\quad c=\sin\left(\theta+\frac{4\pi}{3}\right).

Then

a+b+c=0

because the three angles are equally spaced by $\frac{2\pi}{3}$ .

Now,

\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ab+bc+ca}{abc}.

So we need $ab+bc+ca=0$ .

Use

(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca).

Since $a+b+c=0$ ,

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

For these three sines spaced by $\frac{2\pi}{3}$ ,

a^2+b^2+c^2=\frac{3}{2}.

Thus this route does not give zero directly, so let us compute explicitly instead.

Using angle addition,

\sin\left(\theta+\frac{2\pi}{3}\right)=-\frac12\sin\theta+\frac{\sqrt3}{2}\cos\theta,

\sin\left(\theta+\frac{4\pi}{3}\right)=-\frac12\sin\theta-\frac{\sqrt3}{2}\cos\theta.

Let $s=\sin\theta$ , $c=\cos\theta$ . Then

\frac{1}{s}+\frac{1}{-\frac12 s+\frac{\sqrt3}{2}c}+\frac{1}{-\frac12 s-\frac{\sqrt3}{2}c}.

The last two terms add to

\frac{-s}{\left(-\frac12 s\right)^2-\left(\frac{\sqrt3}{2}c\right)^2} =\frac{-s}{\frac14 s^2-\frac34 c^2} =\frac{-4s}{s^2-3c^2}.

Hence total sum is

\frac1s-\frac{4s}{s^2-3c^2} =\frac{s^2-3c^2-4s^2}{s(s^2-3c^2)} =\frac{-3(s^2+c^2)}{s(s^2-3c^2)} =\frac{-3}{s(s^2-3c^2)}.

Now

s^2-3c^2=s^2-3(1-s^2)=4s^2-3.

So the sum becomes

\frac{-3}{s(4s^2-3)}.

This is not identically zero. Therefore statement (I) is false.

We can also verify by example: take $\theta=\frac{\pi}{2}$ . Then

\sin\theta=1,\quad \sin\left(\theta+\frac{2\pi}{3}\right)=-\frac12, \quad \sin\left(\theta+\frac{4\pi}{3}\right)=-\frac12.

If $k=1$ , then

x=1,\ y=-2,\ z=-2,

\operatorname{Trace}(R)=1-2-2=-3\neq 0.

Thus (I) is false.

Check statement (II)

We need to analyze:

If $\operatorname{Trace}(\operatorname{adj}(\operatorname{adj}(R)))=0$ , then $R$ has exactly one non-zero entry.

Since $R=\operatorname{diag}(x,y,z)$ , first find its adjugate:

\operatorname{adj}(R)=\operatorname{diag}(yz,zx,xy).

Then

\operatorname{adj}(\operatorname{adj}(R))= \operatorname{diag}((zx)(xy),(xy)(yz),(yz)(zx)) = \operatorname{diag}(x^2yz,xy^2z,xyz^2).

\operatorname{Trace}(\operatorname{adj}(\operatorname{adj}(R))) =x^2yz+xy^2z+xyz^2 =xyz(x+y+z).

From the given condition, $x,y,z$ are all non-zero. Hence

\operatorname{Trace}(\operatorname{adj}(\operatorname{adj}(R)))=0 \iff x+y+z=0.

But from part (I), $x+y+z$ is not always zero; however it can be zero for some choices. If it is zero, then all three entries can still be non-zero.

Take an explicit example with all non-zero entries and trace zero: Choose $x=1,y=1,z=-2$ . Then we need to check whether such $x,y,z$ can satisfy the given relation. Since

x:y:z=\frac1{\sin\theta}:\frac1{\sin(\theta+2\pi/3)}:\frac1{\sin(\theta+4\pi/3)},

this means we want

\sin\theta=\sin\left(\theta+\frac{2\pi}{3}\right)=-\frac12\sin\left(\theta+\frac{4\pi}{3}\right).

A simpler direct choice is to solve $x+y+z=0$ using the formula found above:

x+y+z=k\cdot \frac{-3}{\sin\theta(4\sin^2\theta-3)}.

This can never be zero because denominator is finite and numerator is $-3k\neq 0$ . Thus under the given condition,

\operatorname{Trace}(\operatorname{adj}(\operatorname{adj}(R)))=xyz(x+y+z) eq 0

always.

So the hypothesis in statement (II) is never satisfied.

A statement of the form "If P, then Q" is considered true when $P$ is never true (vacuously true). Therefore (II) is true.

Conclusion

(I) is false.
(II) is true.

Hence the correct option is

\boxed{\text{B: Only (II) is true}}

Comparison with stored answer

Stored correct answer: $\boxed{\text{D}}$

My derived answer is $\boxed{\text{B}}$ , so I disagree with the stored answer.

More from Matrices and Determinants

View written solutionFree

Correct answer: B

Given condition

Let

R=\begin{pmatrix} x&0&0\\ 0&y&0\\ 0&0&z \end{pmatrix}

with

x\sin\theta=y\sin\left(\theta+\frac{2\pi}{3}\right)=z\sin\left(\theta+\frac{4\pi}{3}\right)\neq 0.

Let the common non-zero value be $k$ . Then

x=\frac{k}{\sin\theta},\qquad y=\frac{k}{\sin\left(\theta+\frac{2\pi}{3}\right)},\qquad z=\frac{k}{\sin\left(\theta+\frac{4\pi}{3}\right)}.

Since the common value is non-zero, all three denominators are non-zero, hence $x,y,z$ are all non-zero.

Check statement (I): $\operatorname{Trace}(R)=0$

We need to evaluate

\operatorname{Trace}(R)=x+y+z.

Using the expressions above,

x+y+z=k\left(\frac{1}{\sin\theta}+\frac{1}{\sin\left(\theta+\frac{2\pi}{3}\right)}+\frac{1}{\sin\left(\theta+\frac{4\pi}{3}\right)}\right).

So it is enough to prove

\frac{1}{\sin\theta}+\frac{1}{\sin\left(\theta+\frac{2\pi}{3}\right)}+\frac{1}{\sin\left(\theta+\frac{4\pi}{3}\right)}=0.

Let

a=\sin\theta,\quad b=\sin\left(\theta+\frac{2\pi}{3}\right),\quad c=\sin\left(\theta+\frac{4\pi}{3}\right).

Then

a+b+c=0

because the three angles are equally spaced by $\frac{2\pi}{3}$ .

Now,

\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ab+bc+ca}{abc}.

So we need $ab+bc+ca=0$ .

Use

(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca).

Since $a+b+c=0$ ,

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

For these three sines spaced by $\frac{2\pi}{3}$ ,

a^2+b^2+c^2=\frac{3}{2}.

Thus this route does not give zero directly, so let us compute explicitly instead.

Using angle addition,

\sin\left(\theta+\frac{2\pi}{3}\right)=-\frac12\sin\theta+\frac{\sqrt3}{2}\cos\theta,

\sin\left(\theta+\frac{4\pi}{3}\right)=-\frac12\sin\theta-\frac{\sqrt3}{2}\cos\theta.

Let $s=\sin\theta$ , $c=\cos\theta$ . Then

\frac{1}{s}+\frac{1}{-\frac12 s+\frac{\sqrt3}{2}c}+\frac{1}{-\frac12 s-\frac{\sqrt3}{2}c}.

The last two terms add to

\frac{-s}{\left(-\frac12 s\right)^2-\left(\frac{\sqrt3}{2}c\right)^2} =\frac{-s}{\frac14 s^2-\frac34 c^2} =\frac{-4s}{s^2-3c^2}.

Hence total sum is

\frac1s-\frac{4s}{s^2-3c^2} =\frac{s^2-3c^2-4s^2}{s(s^2-3c^2)} =\frac{-3(s^2+c^2)}{s(s^2-3c^2)} =\frac{-3}{s(s^2-3c^2)}.

Now

s^2-3c^2=s^2-3(1-s^2)=4s^2-3.

So the sum becomes

\frac{-3}{s(4s^2-3)}.

This is not identically zero. Therefore statement (I) is false.

We can also verify by example: take $\theta=\frac{\pi}{2}$ . Then

\sin\theta=1,\quad \sin\left(\theta+\frac{2\pi}{3}\right)=-\frac12, \quad \sin\left(\theta+\frac{4\pi}{3}\right)=-\frac12.

If $k=1$ , then

x=1,\ y=-2,\ z=-2,

\operatorname{Trace}(R)=1-2-2=-3\neq 0.

Thus (I) is false.

Check statement (II)

We need to analyze:

If $\operatorname{Trace}(\operatorname{adj}(\operatorname{adj}(R)))=0$ , then $R$ has exactly one non-zero entry.

Since $R=\operatorname{diag}(x,y,z)$ , first find its adjugate:

\operatorname{adj}(R)=\operatorname{diag}(yz,zx,xy).

Then

\operatorname{adj}(\operatorname{adj}(R))= \operatorname{diag}((zx)(xy),(xy)(yz),(yz)(zx)) = \operatorname{diag}(x^2yz,xy^2z,xyz^2).

\operatorname{Trace}(\operatorname{adj}(\operatorname{adj}(R))) =x^2yz+xy^2z+xyz^2 =xyz(x+y+z).

From the given condition, $x,y,z$ are all non-zero. Hence

\operatorname{Trace}(\operatorname{adj}(\operatorname{adj}(R)))=0 \iff x+y+z=0.

But from part (I), $x+y+z$ is not always zero; however it can be zero for some choices. If it is zero, then all three entries can still be non-zero.

Take an explicit example with all non-zero entries and trace zero: Choose $x=1,y=1,z=-2$ . Then we need to check whether such $x,y,z$ can satisfy the given relation. Since

x:y:z=\frac1{\sin\theta}:\frac1{\sin(\theta+2\pi/3)}:\frac1{\sin(\theta+4\pi/3)},

this means we want

\sin\theta=\sin\left(\theta+\frac{2\pi}{3}\right)=-\frac12\sin\left(\theta+\frac{4\pi}{3}\right).

A simpler direct choice is to solve $x+y+z=0$ using the formula found above:

x+y+z=k\cdot \frac{-3}{\sin\theta(4\sin^2\theta-3)}.

This can never be zero because denominator is finite and numerator is $-3k\neq 0$ . Thus under the given condition,

\operatorname{Trace}(\operatorname{adj}(\operatorname{adj}(R)))=xyz(x+y+z) eq 0

always.

So the hypothesis in statement (II) is never satisfied.

A statement of the form "If P, then Q" is considered true when $P$ is never true (vacuously true). Therefore (II) is true.

Conclusion

(I) is false.
(II) is true.

Hence the correct option is

\boxed{\text{B: Only (II) is true}}

Comparison with stored answer

Stored correct answer: $\boxed{\text{D}}$

My derived answer is $\boxed{\text{B}}$ , so I disagree with the stored answer.

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