Consider the system of linear equations x+y+z=5, x+2 y+\lambda^2 z=9, x+3 y+\lambda z=\mu , where \lambda, \mu \in \mathbbR . Then, which of the following statement is NOT correct?

System has unique solution if \lambda eq 1 and \mu eq 13

System is consistent if \lambda eq 1 and \mu=13

System is inconsistent if \lambda=1 and \mu eq 13

System has infinite number of solutions if \lambda=1 and \mu=13

Matrices and Determinants PYQ — JEE Main 2024 · 30 Jan · Shift 2 · Q46 with Solution

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Correct answer: C

Write the system in matrix form

The system is

\begin{cases} x+y+z=5 \\ x+2y+\lambda^2 z=9 \\ x+3y+\lambda z=\mu \end{cases}

Coefficient matrix:

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

To determine whether the system has a unique solution, infinite solutions, or is inconsistent, compute $\det(A)$ .

Compute the determinant

Apply row operations:

R_2\to R_2-R_1, \qquad R_3\to R_3-R_1

Then

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

Expanding along the first column,

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

=\lambda-1-2\lambda^2+2 =-(2\lambda^2-\lambda-1)

=-(2\lambda+1)(\lambda-1)

So,

\det(A)=0 \iff \lambda=1 \text{ or } \lambda=-\frac12

Hence the system has a unique solution whenever

\lambda\neq 1 \quad \text{and} \quad \lambda\neq -\frac12.

Check the case $\lambda=1$

If $\lambda=1$ , the equations become

\begin{cases} x+y+z=5 \\ x+2y+z=9 \\ x+3y+z=\mu \end{cases}

Subtract first from second:

y=4

Subtract second from third:

y=\mu-9

Thus for consistency,

4=\mu-9 \implies \mu=13

If $\mu=13$ , then the third equation is dependent on the first two, so the system has infinitely many solutions.
If $\mu\neq 13$ , the system is inconsistent.

Therefore:

B is correct
D is correct

Check the case $\lambda\neq 1$ with $\mu=13$

Option A says: system is consistent if $\lambda\neq 1$ and $\mu=13$ .

This is certainly true whenever the system has either a unique solution or infinitely many solutions.

We already know:

If $\lambda\neq 1$ and also $\lambda\neq -\frac12$ , then $\det(A)\neq 0$ , so the system has a unique solution, hence it is consistent.
Now check $\lambda=-\frac12$ separately.

For $\lambda=-\frac12$ , the system is

\begin{cases} x+y+z=5 \\ x+2y+\frac14 z=9 \\ x+3y-\frac12 z=13 \end{cases}

Subtract the first equation from the second and third:

y-\frac34 z=4

2y-\frac32 z=8

The second derived equation is exactly twice the first, so the system is consistent and has infinitely many solutions.

Thus for $\lambda\neq 1$ and $\mu=13$ , the system is indeed consistent. So A is correct.

Check option C

Option C says: system has unique solution if $\lambda\neq 1$ and $\mu\neq 13$ .

This is not always true, because if

\lambda=-\frac12,

then $\det(A)=0$ , so unique solution is impossible.

Let us check consistency for $\lambda=-\frac12$ and general $\mu$ :

\begin{cases} x+y+z=5 \\ x+2y+\frac14 z=9 \\ x+3y-\frac12 z=\mu \end{cases}

From subtracting the first equation:

y-\frac34 z=4 \quad ...(1)

2y-\frac32 z=\mu-5 \quad ...(2)

But doubling (1) gives

2y-\frac32 z=8

So consistency requires

\mu-5=8 \implies \mu=13

If $\mu\neq 13$ , the system is inconsistent.

Therefore, for $\lambda=-\frac12$ and $\mu\neq 13$ , option C fails badly: the system does not have a unique solution.

Hence C is the NOT correct statement.

Final conclusion

The incorrect statement is

\boxed{\text{C}}

More from Matrices and Determinants

View written solutionFree

Correct answer: C

Write the system in matrix form

The system is

\begin{cases} x+y+z=5 \\ x+2y+\lambda^2 z=9 \\ x+3y+\lambda z=\mu \end{cases}

Coefficient matrix:

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

To determine whether the system has a unique solution, infinite solutions, or is inconsistent, compute $\det(A)$ .

Compute the determinant

Apply row operations:

R_2\to R_2-R_1, \qquad R_3\to R_3-R_1

Then

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

Expanding along the first column,

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

=\lambda-1-2\lambda^2+2 =-(2\lambda^2-\lambda-1)

=-(2\lambda+1)(\lambda-1)

So,

\det(A)=0 \iff \lambda=1 \text{ or } \lambda=-\frac12

Hence the system has a unique solution whenever

\lambda\neq 1 \quad \text{and} \quad \lambda\neq -\frac12.

Check the case $\lambda=1$

If $\lambda=1$ , the equations become

\begin{cases} x+y+z=5 \\ x+2y+z=9 \\ x+3y+z=\mu \end{cases}

Subtract first from second:

y=4

Subtract second from third:

y=\mu-9

Thus for consistency,

4=\mu-9 \implies \mu=13

If $\mu=13$ , then the third equation is dependent on the first two, so the system has infinitely many solutions.
If $\mu\neq 13$ , the system is inconsistent.

Therefore:

B is correct
D is correct

Check the case $\lambda\neq 1$ with $\mu=13$

Option A says: system is consistent if $\lambda\neq 1$ and $\mu=13$ .

This is certainly true whenever the system has either a unique solution or infinitely many solutions.

We already know:

If $\lambda\neq 1$ and also $\lambda\neq -\frac12$ , then $\det(A)\neq 0$ , so the system has a unique solution, hence it is consistent.
Now check $\lambda=-\frac12$ separately.

For $\lambda=-\frac12$ , the system is

\begin{cases} x+y+z=5 \\ x+2y+\frac14 z=9 \\ x+3y-\frac12 z=13 \end{cases}

Subtract the first equation from the second and third:

y-\frac34 z=4

2y-\frac32 z=8

The second derived equation is exactly twice the first, so the system is consistent and has infinitely many solutions.

Thus for $\lambda\neq 1$ and $\mu=13$ , the system is indeed consistent. So A is correct.

Check option C

Option C says: system has unique solution if $\lambda\neq 1$ and $\mu\neq 13$ .

This is not always true, because if

\lambda=-\frac12,

then $\det(A)=0$ , so unique solution is impossible.

Let us check consistency for $\lambda=-\frac12$ and general $\mu$ :

\begin{cases} x+y+z=5 \\ x+2y+\frac14 z=9 \\ x+3y-\frac12 z=\mu \end{cases}

From subtracting the first equation:

y-\frac34 z=4 \quad ...(1)

2y-\frac32 z=\mu-5 \quad ...(2)

But doubling (1) gives

2y-\frac32 z=8

So consistency requires

\mu-5=8 \implies \mu=13

If $\mu\neq 13$ , the system is inconsistent.

Therefore, for $\lambda=-\frac12$ and $\mu\neq 13$ , option C fails badly: the system does not have a unique solution.

Hence C is the NOT correct statement.

Final conclusion

The incorrect statement is

\boxed{\text{C}}

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