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Matrices and Determinants question

2023 · 6 Apr · Shift 2 · Q35
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Matrices and Determinants question

2023 · 6 Apr · Shift 2 · Q35

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
For the system of equations x+y+z=6x+2y+αz=10x+3y+5z=βx+y+z=6x+2 y+\alpha z=10x+3 y+5 z=\betax+y+z=6x+2y+αz=10x+3y+5z=β, which one of the following is NOT true?
  1. A
    System has a unique solution for α=3,βe14\alpha=3,\beta e14α=3,βe14.
  2. B
    System has infinitely many solutions for α=3,β=14\alpha=3, \beta=14α=3,β=14.
  3. C
    System has no solution for α=3,β=24\alpha=3, \beta=24α=3,β=24.
  4. D
    System has a unique solution for α=−3,β=14\alpha=-3, \beta=14α=−3,β=14.
View written solutionFree

Correct answer: B, C

  1. Interpret the system correctly

The given chain means the three equations are

{x+y+z=66x+2y+αz=10x+3y+5z=β\begin{cases} x+y+z=6 \\ 6x+2y+\alpha z=10 \\ x+3y+5z=\beta \end{cases}⎩⎨⎧​x+y+z=66x+2y+αz=10x+3y+5z=β​

We must determine which statement is NOT true.

  1. Coefficient matrix and its determinant

The coefficient matrix is

A=(11162α135)A=\begin{pmatrix} 1&1&1\\ 6&2&\alpha\\ 1&3&5 \end{pmatrix}A=​161​123​1α5​​

A unique solution exists iff det⁡(A)≠0\det(A)\neq 0det(A)=0.

Compute:

det⁡(A)=∣11162α135∣\det(A)= \begin{vmatrix} 1&1&1\\ 6&2&\alpha\\ 1&3&5 \end{vmatrix}det(A)=​161​123​1α5​​

Expanding along the first row,

det⁡(A)=1∣2α35∣−1∣6α15∣+1∣6213∣\det(A)=1\begin{vmatrix}2&\alpha\\3&5\end{vmatrix} -1\begin{vmatrix}6&\alpha\\1&5\end{vmatrix} +1\begin{vmatrix}6&2\\1&3\end{vmatrix}det(A)=1​23​α5​​−1​61​α5​​+1​61​23​​ =(10−3α)−(30−α)+(18−2)= (10-3\alpha) - (30-\alpha) + (18-2)=(10−3α)−(30−α)+(18−2) =10−3α−30+α+16=−4−2α=−2(α+2)=10-3\alpha-30+\alpha+16 = -4-2\alpha = -2(\alpha+2)=10−3α−30+α+16=−4−2α=−2(α+2)

So:

  • if α≠−2\alpha\neq -2α=−2, the system has a unique solution for every β\betaβ;
  • if α=−2\alpha=-2α=−2, the system may have infinitely many or no solution depending on consistency.
  1. Check each option

Option A: α=3, β=14\alpha=3,\ \beta=14α=3, β=14

Since α=3≠−2\alpha=3\neq -2α=3=−2,

det⁡(A)=−2(3+2)=−10≠0\det(A)=-2(3+2)=-10\neq 0det(A)=−2(3+2)=−10=0

Hence the system has a unique solution. So A is true.

Option B: α=3, β=14\alpha=3,\ \beta=14α=3, β=14

This is the same values as in option A. Since for α=3\alpha=3α=3 the determinant is nonzero, the system cannot have infinitely many solutions.

Therefore B is false.

Option C: α=3, β=24\alpha=3,\ \beta=24α=3, β=24

Again, α=3≠−2\alpha=3\neq -2α=3=−2, so determinant is nonzero. Hence the system has a unique solution, not “no solution”.

Therefore C is false.

Option D: α=−3, β=14\alpha=-3,\ \beta=14α=−3, β=14

Since α=−3≠−2\alpha=-3\neq -2α=−3=−2,

det⁡(A)=−2(−3+2)=2≠0\det(A)=-2(-3+2)=2\neq 0det(A)=−2(−3+2)=2=0

Hence the system has a unique solution. So D is true.

  1. Conclusion

The false statements are:

B and C\boxed{B \text{ and } C}B and C​

So the statement(s) that are NOT true are B and C.

Since this is labeled as MCQ but actually yields more than one false option, the question appears inconsistent as a single-correct MCQ.

  1. Compare with stored answer

Stored correct answer: A

But we found:

A is true, not false\boxed{A \text{ is true, not false}}A is true, not false​

Hence I disagree with the stored answer. The mathematically correct result is that B and C are not true.

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