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Matrices and Determinants question

2022 · 27 Jun · Shift 1 · Q21
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  5. /2022 · 27 Jun · Shift 1 · Q21

Matrices and Determinants question

2022 · 27 Jun · Shift 1 · Q21

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let the system of linear equations x+2y+z=2x + 2y + z = 2x+2y+z=2, αx+3y−z=α\alpha x + 3y - z = \alphaαx+3y−z=α, −αx+y+2z=−α- \alpha x + y + 2z = - \alpha−αx+y+2z=−α be inconsistent. Then α\alphaα is equal to :
  1. A
    52{5 \over 2}25​
  2. B
    −52-{5 \over 2}−25​
  3. C
    72{7 \over 2}27​
  4. D
    −72-{7 \over 2}−27​
View written solutionFree

Correct answer: D

  1. Write the system in matrix form:
{x+2y+z=2αx+3y−z=α−αx+y+2z=−α\begin{cases} x+2y+z=2\\ \alpha x+3y-z=\alpha\\ -\alpha x+y+2z=-\alpha \end{cases}⎩⎨⎧​x+2y+z=2αx+3y−z=α−αx+y+2z=−α​

The coefficient matrix is

A=(121α3−1−α12)A=\begin{pmatrix} 1&2&1\\ \alpha&3&-1\\ -\alpha&1&2 \end{pmatrix}A=​1α−α​231​1−12​​

For the system to be inconsistent, we need:

  • det⁡(A)=0\det(A)=0det(A)=0, and
  • the augmented system must have rank greater than the coefficient matrix rank.
  1. Compute det⁡(A)\det(A)det(A):
det⁡(A)=∣121α3−1−α12∣\det(A)= \begin{vmatrix} 1&2&1\\ \alpha&3&-1\\ -\alpha&1&2 \end{vmatrix}det(A)=​1α−α​231​1−12​​

Expanding along the first row:

det⁡(A)=1∣3−112∣−2∣α−1−α2∣+1∣α3−α1∣\det(A)=1\begin{vmatrix}3&-1\\1&2\end{vmatrix} -2\begin{vmatrix}\alpha&-1\\-\alpha&2\end{vmatrix} +1\begin{vmatrix}\alpha&3\\-\alpha&1\end{vmatrix}det(A)=1​31​−12​​−2​α−α​−12​​+1​α−α​31​​

Now,

∣3−112∣=3⋅2−(−1)⋅1=7\begin{vmatrix}3&-1\\1&2\end{vmatrix}=3\cdot2-(-1)\cdot1=7​31​−12​​=3⋅2−(−1)⋅1=7 ∣α−1−α2∣=α⋅2−(−1)(−α)=2α−α=α\begin{vmatrix}\alpha&-1\\-\alpha&2\end{vmatrix}=\alpha\cdot2-(-1)(-\alpha)=2\alpha-\alpha=\alpha​α−α​−12​​=α⋅2−(−1)(−α)=2α−α=α ∣α3−α1∣=α⋅1−3(−α)=α+3α=4α\begin{vmatrix}\alpha&3\\-\alpha&1\end{vmatrix}=\alpha\cdot1-3(-\alpha)=\alpha+3\alpha=4\alpha​α−α​31​​=α⋅1−3(−α)=α+3α=4α

So,

det⁡(A)=7−2(α)+4α=7+2α\det(A)=7-2(\alpha)+4\alpha=7+2\alphadet(A)=7−2(α)+4α=7+2α

For inconsistency, set

7+2α=07+2\alpha=07+2α=0 α=−72\alpha=-\frac{7}{2}α=−27​
  1. Check whether the system is actually inconsistent for α=−72\alpha=-\frac72α=−27​.

Substitute α=−72\alpha=-\frac72α=−27​ into the equations:

x+2y+z=2...(1)x+2y+z=2 \quad ...(1)x+2y+z=2...(1) −72x+3y−z=−72...(2)-\frac72 x+3y-z=-\frac72 \quad ...(2)−27​x+3y−z=−27​...(2) 72x+y+2z=72...(3)\frac72 x+y+2z=\frac72 \quad ...(3)27​x+y+2z=27​...(3)

Multiply all equations by 222 to simplify:

2x+4y+2z=4...(1′)2x+4y+2z=4 \quad ...(1')2x+4y+2z=4...(1′) −7x+6y−2z=−7...(2′)-7x+6y-2z=-7 \quad ...(2')−7x+6y−2z=−7...(2′) 7x+2y+4z=7...(3′)7x+2y+4z=7 \quad ...(3')7x+2y+4z=7...(3′)

Now add (2′)(2')(2′) and (3′)(3')(3′):

(−7x+6y−2z)+(7x+2y+4z)=−7+7(-7x+6y-2z)+(7x+2y+4z)= -7+7(−7x+6y−2z)+(7x+2y+4z)=−7+7 8y+2z=08y+2z=08y+2z=0 4y+z=0...(4)4y+z=0 \quad ...(4)4y+z=0...(4)

From (1′)(1')(1′):

2x+4y+2z=42x+4y+2z=42x+4y+2z=4

Using z=−4yz=-4yz=−4y from (4):

2x+4y+2(−4y)=42x+4y+2(-4y)=42x+4y+2(−4y)=4 2x−4y=42x-4y=42x−4y=4 x−2y=2...(5)x-2y=2 \quad ...(5)x−2y=2...(5)

Now use (3′)(3')(3′):

7x+2y+4z=77x+2y+4z=77x+2y+4z=7

Substitute z=−4yz=-4yz=−4y:

7x+2y−16y=77x+2y-16y=77x+2y−16y=7 7x−14y=77x-14y=77x−14y=7 x−2y=1x-2y=1x−2y=1

But from (5),

x−2y=2x-2y=2x−2y=2

This is a contradiction. Hence the system is inconsistent.

  1. Therefore,
α=−72\boxed{\alpha=-\frac72}α=−27​​

This corresponds to Option D.

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