The system of linear equations x + y + z = 2 2x + 3y + 2z = 5 2x + 3y + (a2 – 1) z = a + 1 then

has infinitely many solutions for a = 4

has a unique solution for |a| = \sqrt3

Matrices and Determinants PYQ — JEE Main 2019 · 9 Jan · Shift 1 · Q26 with Solution

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Correct answer: C

Write the system in matrix form

The equations are

\begin{aligned} x+y+z&=2 \\ 2x+3y+2z&=5 \\ 2x+3y+(a^2-1)z&=a+1 \end{aligned}

Coefficient matrix:

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

Check when the system has a unique solution

A linear system has a unique solution when $\det(A)\neq 0$ .

Compute the determinant:

\det(A)= \begin{vmatrix} 1&1&1\\ 2&3&2\\ 2&3&a^2-1 \end{vmatrix}

Apply the row operation $R_3 \to R_3-R_2$ :

\det(A)= \begin{vmatrix} 1&1&1\\ 2&3&2\\ 0&0&a^2-3 \end{vmatrix} =(a^2-3) \begin{vmatrix} 1&1\\ 2&3 \end{vmatrix}

Since

\begin{vmatrix} 1&1\\ 2&3 \end{vmatrix}=3-2=1,

we get

\det(A)=a^2-3.

So the system has a unique solution when

a^2-3\neq 0 \quad \Rightarrow \quad a\neq \pm \sqrt{3}.

Hence, for $|a|=\sqrt{3}$ , the system does not have a unique solution.

So Option B is false.

Check consistency when $a=\pm\sqrt{3}$

When $a^2=3$ , the third equation becomes

2x+3y+(3-1)z=a+1

that is,

2x+3y+2z=a+1.

But the second equation is

2x+3y+2z=5.

Thus for consistency, we must have

a+1=5 \Rightarrow a=4.

However, under the condition $a=\pm\sqrt{3}$ , we do not have $a=4$ . So the second and third equations have the same left-hand side but different right-hand sides.

Therefore, the system is inconsistent when

|a|=\sqrt{3}.

So Option C is true.

Check the case $a=4$

For $a=4$ ,

a^2-3=16-3=13\neq 0.

So $\det(A)\neq 0$ , hence the system has a unique solution.

Therefore:

it does not have infinitely many solutions,
it is not inconsistent.

So:

Option A is false,
Option D is false.

Final conclusion

Only Option C is correct.

$\boxed{\text{C}}$

More from Matrices and Determinants

View written solutionFree

Correct answer: C

Write the system in matrix form

The equations are

\begin{aligned} x+y+z&=2 \\ 2x+3y+2z&=5 \\ 2x+3y+(a^2-1)z&=a+1 \end{aligned}

Coefficient matrix:

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

Check when the system has a unique solution

A linear system has a unique solution when $\det(A)\neq 0$ .

Compute the determinant:

\det(A)= \begin{vmatrix} 1&1&1\\ 2&3&2\\ 2&3&a^2-1 \end{vmatrix}

Apply the row operation $R_3 \to R_3-R_2$ :

\det(A)= \begin{vmatrix} 1&1&1\\ 2&3&2\\ 0&0&a^2-3 \end{vmatrix} =(a^2-3) \begin{vmatrix} 1&1\\ 2&3 \end{vmatrix}

Since

\begin{vmatrix} 1&1\\ 2&3 \end{vmatrix}=3-2=1,

we get

\det(A)=a^2-3.

So the system has a unique solution when

a^2-3\neq 0 \quad \Rightarrow \quad a\neq \pm \sqrt{3}.

Hence, for $|a|=\sqrt{3}$ , the system does not have a unique solution.

So Option B is false.

Check consistency when $a=\pm\sqrt{3}$

When $a^2=3$ , the third equation becomes

2x+3y+(3-1)z=a+1

that is,

2x+3y+2z=a+1.

But the second equation is

2x+3y+2z=5.

Thus for consistency, we must have

a+1=5 \Rightarrow a=4.

However, under the condition $a=\pm\sqrt{3}$ , we do not have $a=4$ . So the second and third equations have the same left-hand side but different right-hand sides.

Therefore, the system is inconsistent when

|a|=\sqrt{3}.

So Option C is true.

Check the case $a=4$

For $a=4$ ,

a^2-3=16-3=13\neq 0.

So $\det(A)\neq 0$ , hence the system has a unique solution.

Therefore:

it does not have infinitely many solutions,
it is not inconsistent.

So:

Option A is false,
Option D is false.

Final conclusion

Only Option C is correct.

$\boxed{\text{C}}$

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