Consider the system of linear equations x+y+z=4 \mu, x+2 y+2 \lambda z=10 \mu, x+3 y+4 \lambda^2 z=\mu^2+15 where \lambda, \mu \in R . Which one of the following statements is NOT correct ?

The system is inconsistent if \lambda=\frac12 and \mu eq 1

The system has unique solution if \lambda eq \frac12 and \mu eq 1,15

The system has infinite number of solutions if \lambda=\frac12 and \mu=15

The system is consistent if \lambda eq \frac12

Matrices and Determinants PYQ — JEE Main 2024 · 30 Jan · Shift 1 · Q39 with Solution

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Correct answer: D

Write the system in matrix form

The system is

\begin{cases} x+y+z=4\mu \\ x+2y+2\lambda z=10\mu \\ x+3y+4\lambda^2 z=\mu^2+15 \end{cases}

with coefficient matrix

A=\begin{pmatrix} 1&1&1\\ 1&2&2\lambda\\ 1&3&4\lambda^2 \end{pmatrix}.

We must identify the statement which is NOT correct.

Check when the system has a unique solution

A unique solution exists when $\det(A)\neq 0$ .

Compute:

\det(A)= \begin{vmatrix} 1&1&1\\ 1&2&2\lambda\\ 1&3&4\lambda^2 \end{vmatrix}.

Apply row operations $R_2\to R_2-R_1$ , $R_3\to R_3-R_1$ :

\det(A)= \begin{vmatrix} 1&1&1\\ 0&1&2\lambda-1\\ 0&2&4\lambda^2-1 \end{vmatrix} = \begin{vmatrix} 1&2\lambda-1\\ 2&4\lambda^2-1 \end{vmatrix}.

So,

\det(A)=1(4\lambda^2-1)-2(2\lambda-1) =4\lambda^2-1-4\lambda+2 =4\lambda^2-4\lambda+1 =(2\lambda-1)^2.

Thus,

\det(A)\neq 0 \iff \lambda\neq \frac12.

Hence for $\lambda\neq \frac12$ , the system always has a unique solution for every real $\mu$ .

So statement A says unique solution if $\lambda\neq \frac12$ and $\mu\neq 1,15$ . This is true but unnecessarily restrictive, because unique solution already follows from $\lambda\neq \frac12$ . It is still not false.

Also statement C: if $\lambda\neq \frac12$ , unique solution exists, hence system is certainly consistent. So C is correct.

Now analyze the singular case $\lambda=\frac12$

Substitute $\lambda=\frac12$ into the equations:

\begin{cases} x+y+z=4\mu \\ x+2y+z=10\mu \\ x+3y+z=\mu^2+15 \end{cases}

since

2\lambda=1, \qquad 4\lambda^2=1.

Now subtract successively:

Equation 2 minus Equation 1:

y=10\mu-4\mu=6\mu.

Equation 3 minus Equation 2:

y=(\mu^2+15)-10\mu=\mu^2-10\mu+15.

For consistency, these two values of $y$ must be equal:

6\mu=\mu^2-10\mu+15.

So,

\mu^2-16\mu+15=0

(\mu-1)(\mu-15)=0.

Hence when $\lambda=\frac12$ , the system is consistent only for

\mu=1 \quad \text{or} \quad \mu=15.

For all other values of $\mu$ , the system is inconsistent.

Therefore statement D says: system is inconsistent if $\lambda=\frac12$ and $\mu\neq 1$ . This is not correct, because when $\mu=15$ , the system is also consistent.

So D is false.

Check statement B explicitly

Take $\lambda=\frac12$ and $\mu=15$ :

\begin{cases} x+y+z=60 \\ x+2y+z=150 \\ x+3y+z=240 \end{cases}

From differences,

y=90,

and then first equation gives

x+z=-30.

So one free variable remains, giving infinitely many solutions. Thus B is correct.

Similarly, if $\mu=1$ , we get

\begin{cases} x+y+z=4 \\ x+2y+z=10 \\ x+3y+z=16 \end{cases}

Then $y=6$ and $x+z=-2$ , again infinitely many solutions.

Conclusion

The only statement that is NOT correct is $\boxed{D}.$

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View written solutionFree

Correct answer: D

Write the system in matrix form

The system is

\begin{cases} x+y+z=4\mu \\ x+2y+2\lambda z=10\mu \\ x+3y+4\lambda^2 z=\mu^2+15 \end{cases}

with coefficient matrix

A=\begin{pmatrix} 1&1&1\\ 1&2&2\lambda\\ 1&3&4\lambda^2 \end{pmatrix}.

We must identify the statement which is NOT correct.

Check when the system has a unique solution

A unique solution exists when $\det(A)\neq 0$ .

Compute:

\det(A)= \begin{vmatrix} 1&1&1\\ 1&2&2\lambda\\ 1&3&4\lambda^2 \end{vmatrix}.

Apply row operations $R_2\to R_2-R_1$ , $R_3\to R_3-R_1$ :

\det(A)= \begin{vmatrix} 1&1&1\\ 0&1&2\lambda-1\\ 0&2&4\lambda^2-1 \end{vmatrix} = \begin{vmatrix} 1&2\lambda-1\\ 2&4\lambda^2-1 \end{vmatrix}.

So,

\det(A)=1(4\lambda^2-1)-2(2\lambda-1) =4\lambda^2-1-4\lambda+2 =4\lambda^2-4\lambda+1 =(2\lambda-1)^2.

Thus,

\det(A)\neq 0 \iff \lambda\neq \frac12.

Hence for $\lambda\neq \frac12$ , the system always has a unique solution for every real $\mu$ .

Also statement C: if $\lambda\neq \frac12$ , unique solution exists, hence system is certainly consistent. So C is correct.

Now analyze the singular case $\lambda=\frac12$

Substitute $\lambda=\frac12$ into the equations:

\begin{cases} x+y+z=4\mu \\ x+2y+z=10\mu \\ x+3y+z=\mu^2+15 \end{cases}

since

2\lambda=1, \qquad 4\lambda^2=1.

Now subtract successively:

Equation 2 minus Equation 1:

y=10\mu-4\mu=6\mu.

Equation 3 minus Equation 2:

y=(\mu^2+15)-10\mu=\mu^2-10\mu+15.

For consistency, these two values of $y$ must be equal:

6\mu=\mu^2-10\mu+15.

So,

\mu^2-16\mu+15=0

(\mu-1)(\mu-15)=0.

Hence when $\lambda=\frac12$ , the system is consistent only for

\mu=1 \quad \text{or} \quad \mu=15.

For all other values of $\mu$ , the system is inconsistent.

Therefore statement D says: system is inconsistent if $\lambda=\frac12$ and $\mu\neq 1$ . This is not correct, because when $\mu=15$ , the system is also consistent.

So D is false.

Check statement B explicitly

Take $\lambda=\frac12$ and $\mu=15$ :

\begin{cases} x+y+z=60 \\ x+2y+z=150 \\ x+3y+z=240 \end{cases}

From differences,

y=90,

and then first equation gives

x+z=-30.

So one free variable remains, giving infinitely many solutions. Thus B is correct.

Similarly, if $\mu=1$ , we get

\begin{cases} x+y+z=4 \\ x+2y+z=10 \\ x+3y+z=16 \end{cases}

Then $y=6$ and $x+z=-2$ , again infinitely many solutions.

Conclusion

The only statement that is NOT correct is $\boxed{D}.$

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