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Matrices and Determinants question

2023 · 10 Apr · Shift 1 · Q34
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Matrices and Determinants question

2023 · 10 Apr · Shift 1 · Q34

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
For the system of linear equations 2x−y+3z=53x+2y−z=74x+5y+αz=β2x - y + 3z = 53x + 2y - z = 74x + 5y + \alpha z = \beta2x−y+3z=53x+2y−z=74x+5y+αz=β, which of the following is NOT correct?
  1. A
    The system has infinitely many solutions for α=−6\alpha=-6α=−6 and β=9\beta=9β=9
  2. B
    The system has a unique solution for αe−5\alpha e -5αe−5 and β=8\beta=8β=8
  3. C
    The system is inconsistent for α=−5\alpha=-5α=−5 and β=8\beta=8β=8
  4. D
    The system has infinitely many solutions for α=−5\alpha=-5α=−5 and β=9\beta=9β=9
View written solutionFree

Correct answer: A

  1. Interpret the system correctly

The given system is

{2x−y+3z=53x+2y−z=74x+5y+αz=β\begin{cases} 2x-y+3z=5\\ 3x+2y-z=7\\ 4x+5y+\alpha z=\beta \end{cases}⎩⎨⎧​2x−y+3z=53x+2y−z=74x+5y+αz=β​

We must determine which option is NOT correct.

  1. Form the coefficient matrix

The coefficient matrix is

A=(2−1332−145α)A=\begin{pmatrix} 2 & -1 & 3\\ 3 & 2 & -1\\ 4 & 5 & \alpha \end{pmatrix}A=​234​−125​3−1α​​

Its determinant decides whether the system has a unique solution.

∣A∣=∣2−1332−145α∣|A|= \begin{vmatrix} 2 & -1 & 3\\ 3 & 2 & -1\\ 4 & 5 & \alpha \end{vmatrix}∣A∣=​234​−125​3−1α​​

Expanding along the first row,

∣A∣=2∣2−15α∣−(−1)∣3−14α∣+3∣3245∣|A|=2\begin{vmatrix}2 & -1\\5 & \alpha\end{vmatrix} -(-1)\begin{vmatrix}3 & -1\\4 & \alpha\end{vmatrix} +3\begin{vmatrix}3 & 2\\4 & 5\end{vmatrix}∣A∣=2​25​−1α​​−(−1)​34​−1α​​+3​34​25​​ =2(2α+5)+(3α+4)+3(15−8)=2(2\alpha+5)+(3\alpha+4)+3(15-8)=2(2α+5)+(3α+4)+3(15−8) =4α+10+3α+4+21=4\alpha+10+3\alpha+4+21=4α+10+3α+4+21 =7α+35=7(α+5)=7\alpha+35=7(\alpha+5)=7α+35=7(α+5)

So,

  • if α≠−5\alpha\neq -5α=−5, then ∣A∣≠0|A|\neq 0∣A∣=0 and the system has a unique solution;
  • if α=−5\alpha=-5α=−5, then ∣A∣=0|A|=0∣A∣=0 and we must check consistency.
  1. Check option B

Option B says: unique solution for α≠−5\alpha\ne -5α=−5 and β=8\beta=8β=8.

Since unique solution depends only on the coefficient determinant,

α≠−5  ⟹  ∣A∣≠0\alpha\ne -5 \implies |A|\ne 0α=−5⟹∣A∣=0

so the system has a unique solution for any β\betaβ.

Hence option B is correct.

  1. Analyze the case α=−5\alpha=-5α=−5

Then the system becomes

{2x−y+3z=53x+2y−z=74x+5y−5z=β\begin{cases} 2x-y+3z=5\\ 3x+2y-z=7\\ 4x+5y-5z=\beta \end{cases}⎩⎨⎧​2x−y+3z=53x+2y−z=74x+5y−5z=β​

Now check whether the third equation is a linear combination of the first two.

Let

a(2x−y+3z)+b(3x+2y−z)=4x+5y−5za(2x-y+3z)+b(3x+2y-z)=4x+5y-5za(2x−y+3z)+b(3x+2y−z)=4x+5y−5z

Then,

2a+3b=4,2a+3b=4,2a+3b=4, −a+2b=5,-a+2b=5,−a+2b=5, 3a−b=−53a-b=-53a−b=−5

From

−a+2b=5  ⟹  a=2b−5-a+2b=5 \implies a=2b-5−a+2b=5⟹a=2b−5

Substitute into

2a+3b=4:2a+3b=4:2a+3b=4: 2(2b−5)+3b=42(2b-5)+3b=42(2b−5)+3b=4 7b=14  ⟹  b=27b=14 \implies b=27b=14⟹b=2

Thus

a=2(2)−5=−1a=2(2)-5=-1a=2(2)−5=−1

Check third coefficient:

3(−1)−2=−53(-1)-2=-53(−1)−2=−5

which is true.

So,

(4,5,−5)=−1(2,−1,3)+2(3,2,−1)(4,5,-5)=-1(2,-1,3)+2(3,2,-1)(4,5,−5)=−1(2,−1,3)+2(3,2,−1)

Hence the third equation is

−1×(first equation)+2×(second equation)-1\times(\text{first equation})+2\times(\text{second equation})−1×(first equation)+2×(second equation)

Therefore for consistency, the RHS must also satisfy

β=−1⋅5+2⋅7=9\beta=-1\cdot 5+2\cdot 7=9β=−1⋅5+2⋅7=9

So when α=−5\alpha=-5α=−5:

  • if β=9\beta=9β=9, system is consistent dependent ⇒\Rightarrow⇒ infinitely many solutions;
  • if β≠9\beta\ne 9β=9, system is inconsistent.
  1. Check options C and D
  • Option C: inconsistent for α=−5\alpha=-5α=−5 and β=8\beta=8β=8.

Since for α=−5\alpha=-5α=−5, consistency requires β=9\beta=9β=9, the value β=8\beta=8β=8 makes the system inconsistent. So C is correct.

  • Option D: infinitely many solutions for α=−5\alpha=-5α=−5 and β=9\beta=9β=9.

This matches the condition found above. So D is correct.

  1. Check option A

Option A says: infinitely many solutions for α=−6\alpha=-6α=−6 and β=9\beta=9β=9.

But for α=−6\alpha=-6α=−6,

∣A∣=7(α+5)=7(−6+5)=−7≠0|A|=7(\alpha+5)=7(-6+5)=-7\ne 0∣A∣=7(α+5)=7(−6+5)=−7=0

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

Hence A is NOT correct.

  1. Final conclusion

The incorrect statement is

A\boxed{\text{A}}A​
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