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Matrices and Determinants question

2022 · 26 Jun · Shift 2 · Q26
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Matrices and Determinants question

2022 · 26 Jun · Shift 2 · Q26

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
If the system of equations α\alphaα x + y + z = 5, x + 2y + 3z = 4, x + 3y + 5z = β\betaβ has infinitely many solutions, then the ordered pair (α\alphaα, β\betaβ) is equal to :
  1. A
    (1, −-− 3)
  2. B
    (−-− 1, 3)
  3. C
    (1, 3)
  4. D
    (−-− 1, −-− 3)
View written solutionFree

Correct answer: C

  1. Write the system in matrix form:
{αx+y+z=5x+2y+3z=4x+3y+5z=β\begin{cases} \alpha x + y + z = 5 \\ x + 2y + 3z = 4 \\ x + 3y + 5z = \beta \end{cases}⎩⎨⎧​αx+y+z=5x+2y+3z=4x+3y+5z=β​

Coefficient matrix:

A=(α11123135)A=\begin{pmatrix} \alpha & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 5 \end{pmatrix}A=​α11​123​135​​

For the system to have infinitely many solutions, we need:

  • det⁡(A)=0\det(A)=0det(A)=0 and
  • the system must be consistent, i.e. ranks of coefficient and augmented matrices are equal and less than 333.
  1. Compute det⁡(A)\det(A)det(A):
det⁡(A)=α∣2335∣−1∣1315∣+1∣1213∣\det(A)=\alpha\begin{vmatrix}2&3\\3&5\end{vmatrix}-1\begin{vmatrix}1&3\\1&5\end{vmatrix}+1\begin{vmatrix}1&2\\1&3\end{vmatrix}det(A)=α​23​35​​−1​11​35​​+1​11​23​​ =α(10−9)−(5−3)+(3−2)=\alpha(10-9)-(5-3)+(3-2)=α(10−9)−(5−3)+(3−2) =α−2+1=α−1=\alpha-2+1=\alpha-1=α−2+1=α−1

For infinitely many solutions,

α−1=0⇒α=1\alpha-1=0 \Rightarrow \alpha=1α−1=0⇒α=1
  1. Now substitute α=1\alpha=1α=1:
{x+y+z=5x+2y+3z=4x+3y+5z=β\begin{cases} x+y+z=5 \\ x+2y+3z=4 \\ x+3y+5z=\beta \end{cases}⎩⎨⎧​x+y+z=5x+2y+3z=4x+3y+5z=β​

Notice that

2(x+2y+3z)−(x+y+z)=x+3y+5z2(x+2y+3z)-(x+y+z)=x+3y+5z2(x+2y+3z)−(x+y+z)=x+3y+5z

So the left-hand side of the third equation is a linear combination of the first two. For consistency, the right-hand side must satisfy the same relation:

β=2⋅4−5=8−5=3\beta=2\cdot 4-5=8-5=3β=2⋅4−5=8−5=3

Thus,

(α,β)=(1,3)(\alpha,\beta)=(1,3)(α,β)=(1,3)
  1. Check options:
  • A: (1,−3)(1,-3)(1,−3) — wrong
  • B: (−1,3)(-1,3)(−1,3) — wrong
  • C: (1,3)(1,3)(1,3) — correct
  • D: (−1,−3)(-1,-3)(−1,−3) — wrong

Hence the correct option is C.

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