For the system of linear equations \alpha x+y+z=1,x+\alpha y+z=1,x+y+\alpha z=\beta , which one of the following statements is NOT correct?

It has infinitely many solutions if \alpha=2 and \beta=-1

It has infinitely many solutions if \alpha=1 and \beta=1

x+y+z=\frac34 if \alpha=2 and \beta=1

It has no solution if \alpha=-2 and \beta=1

Matrices and Determinants PYQ — JEE Main 2023 · 1 Feb · Shift 2 · Q28 with Solution

View written solutionFree

Correct answer: B

We are given the system

\begin{cases} \alpha x+y+z=1 \\ x+\alpha y+z=1 \\ x+y+\alpha z=\beta \end{cases}

and we must find which statement is NOT correct.

1. Write the coefficient matrix

The coefficient matrix is

A=\begin{pmatrix} \alpha & 1 & 1\\ 1 & \alpha & 1\\ 1 & 1 & \alpha \end{pmatrix}.

A useful observation is that

A=(\alpha-1)I+J,

where $J$ is the $3\times 3$ all-ones matrix.

The eigenvalues of $J$ are $3,0,0$ , so the eigenvalues of $A$ are

\alpha+2,\quad \alpha-1,\quad \alpha-1.

Hence

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

Therefore, the matrix is singular when

\alpha=1 \quad \text{or} \quad \alpha=-2.

So we must check the options carefully.

2. Check Option A: $\alpha=1,\ \beta=1$

Substitute $\alpha=1$ :

x+y+z=1,

x+y+z=1,

x+y+z=\beta=1.

So all three equations are identical:

x+y+z=1.

This represents a plane in three variables, so there are infinitely many solutions.

So Option A is correct.

3. Check Option B: $\alpha=2,\ \beta=-1$

Substitute $\alpha=2$ :

2x+y+z=1 \quad ...(1)

x+2y+z=1 \quad ...(2)

x+y+2z=-1 \quad ...(3)

Since $\det(A)$ at $\alpha=2$ is

(2+2)(2-1)^2=4\neq 0,

the system has a unique solution, not infinitely many solutions.

So Option B is not correct.

Let us verify by solving quickly.

Add (1), (2), (3):

4(x+y+z)=1+1-1=1

x+y+z=\frac14.

So indeed a unique solution exists.

Thus the claim "infinitely many solutions" is false.

4. Check Option C: $\alpha=2,\ \beta=1$

Now the system becomes

2x+y+z=1,

x+2y+z=1,

x+y+2z=1.

Add all three equations:

4(x+y+z)=3.

Hence

x+y+z=\frac34.

So Option C is correct.

5. Check Option D: $\alpha=-2,\ \beta=1$

Substitute $\alpha=-2$ :

-2x+y+z=1 \quad ...(1)

x-2y+z=1 \quad ...(2)

x+y-2z=1 \quad ...(3)

Add all three equations:

(-2x+x+x)+(y-2y+y)+(z+z-2z)=1+1+1

0=3,

which is impossible.

Therefore, the system has no solution.

So Option D is correct.

6. Final conclusion

Among the given statements, the only one that is NOT correct is:

\boxed{\text{B}}

This matches the stored correct answer.

More from Matrices and Determinants

View written solutionFree

Correct answer: B

We are given the system

\begin{cases} \alpha x+y+z=1 \\ x+\alpha y+z=1 \\ x+y+\alpha z=\beta \end{cases}

and we must find which statement is NOT correct.

1. Write the coefficient matrix

The coefficient matrix is

A=\begin{pmatrix} \alpha & 1 & 1\\ 1 & \alpha & 1\\ 1 & 1 & \alpha \end{pmatrix}.

A useful observation is that

A=(\alpha-1)I+J,

where $J$ is the $3\times 3$ all-ones matrix.

The eigenvalues of $J$ are $3,0,0$ , so the eigenvalues of $A$ are

\alpha+2,\quad \alpha-1,\quad \alpha-1.

Hence

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

Therefore, the matrix is singular when

\alpha=1 \quad \text{or} \quad \alpha=-2.

So we must check the options carefully.

2. Check Option A: $\alpha=1,\ \beta=1$

Substitute $\alpha=1$ :

x+y+z=1,

x+y+z=1,

x+y+z=\beta=1.

So all three equations are identical:

x+y+z=1.

This represents a plane in three variables, so there are infinitely many solutions.

So Option A is correct.

3. Check Option B: $\alpha=2,\ \beta=-1$

Substitute $\alpha=2$ :

2x+y+z=1 \quad ...(1)

x+2y+z=1 \quad ...(2)

x+y+2z=-1 \quad ...(3)

Since $\det(A)$ at $\alpha=2$ is

(2+2)(2-1)^2=4\neq 0,

the system has a unique solution, not infinitely many solutions.

So Option B is not correct.

Let us verify by solving quickly.

Add (1), (2), (3):

4(x+y+z)=1+1-1=1

x+y+z=\frac14.

So indeed a unique solution exists.

Thus the claim "infinitely many solutions" is false.

4. Check Option C: $\alpha=2,\ \beta=1$

Now the system becomes

2x+y+z=1,

x+2y+z=1,

x+y+2z=1.

Add all three equations:

4(x+y+z)=3.

Hence

x+y+z=\frac34.

So Option C is correct.

5. Check Option D: $\alpha=-2,\ \beta=1$

Substitute $\alpha=-2$ :

-2x+y+z=1 \quad ...(1)

x-2y+z=1 \quad ...(2)

x+y-2z=1 \quad ...(3)

Add all three equations:

(-2x+x+x)+(y-2y+y)+(z+z-2z)=1+1+1

0=3,

which is impossible.

Therefore, the system has no solution.

So Option D is correct.

6. Final conclusion

Among the given statements, the only one that is NOT correct is:

\boxed{\text{B}}

This matches the stored correct answer.

Matrices and Determinants question

Matrices and Determinants question

1. Write the coefficient matrix

2. Check Option A: $\alpha=1,\ \beta=1$

3. Check Option B: $\alpha=2,\ \beta=-1$

4. Check Option C: $\alpha=2,\ \beta=1$

5. Check Option D: $\alpha=-2,\ \beta=1$

6. Final conclusion

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Matrices and Determinants question

Matrices and Determinants question

1. Write the coefficient matrix

2. Check Option A: $\alpha=1,\ \beta=1$

3. Check Option B: $\alpha=2,\ \beta=-1$

4. Check Option C: $\alpha=2,\ \beta=1$

5. Check Option D: $\alpha=-2,\ \beta=1$

6. Final conclusion

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1. Write the coefficient matrix

2. Check Option A: α=1, β=1\alpha=1,\ \beta=1α=1, β=1

3. Check Option B: α=2, β=−1\alpha=2,\ \beta=-1α=2, β=−1

4. Check Option C: α=2, β=1\alpha=2,\ \beta=1α=2, β=1

5. Check Option D: α=−2, β=1\alpha=-2,\ \beta=1α=−2, β=1

6. Final conclusion

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1. Write the coefficient matrix

2. Check Option A: α=1, β=1\alpha=1,\ \beta=1α=1, β=1

3. Check Option B: α=2, β=−1\alpha=2,\ \beta=-1α=2, β=−1

4. Check Option C: α=2, β=1\alpha=2,\ \beta=1α=2, β=1

5. Check Option D: α=−2, β=1\alpha=-2,\ \beta=1α=−2, β=1

6. Final conclusion

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2. Check Option A: $\alpha=1,\ \beta=1$

3. Check Option B: $\alpha=2,\ \beta=-1$

4. Check Option C: $\alpha=2,\ \beta=1$

5. Check Option D: $\alpha=-2,\ \beta=1$

2. Check Option A: $\alpha=1,\ \beta=1$

3. Check Option B: $\alpha=2,\ \beta=-1$

4. Check Option C: $\alpha=2,\ \beta=1$

5. Check Option D: $\alpha=-2,\ \beta=1$