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Matrices and Determinants question

2021 · 27 Aug · Shift 1 · Q38
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Matrices and Determinants question

2021 · 27 Aug · Shift 1 · Q38

JEE MainMathematicsMatrices and DeterminantsNumerical+4 / −1
If the system of linear equations 2x + y −-− z = 3 x −-− y −-− z =α\alphaα 3x + 3y +β\betaβ z = 3 has infinitely many solution, then α\alphaα+β−αβ\beta-\alpha\betaβ−αβ is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 5

  1. Write the system in matrix form:
{2x+y−z=3x−y−z=α3x+3y+βz=3\begin{cases} 2x+y-z=3 \\ x-y-z=\alpha \\ 3x+3y+\beta z=3 \end{cases}⎩⎨⎧​2x+y−z=3x−y−z=α3x+3y+βz=3​

For infinitely many solutions, the three equations must be dependent, i.e. the third equation must be a linear combination of the first two, and the system must be consistent.

  1. Let the third equation be obtained as:
k(2x+y−z)+ℓ(x−y−z)=3x+3y+βzk(2x+y-z)+\ell(x-y-z)=3x+3y+\beta zk(2x+y−z)+ℓ(x−y−z)=3x+3y+βz

Comparing coefficients of x,y,zx,y,zx,y,z:

2k+ℓ=32k+\ell=32k+ℓ=3 k−ℓ=3k-\ell=3k−ℓ=3 −k−ℓ=β-k-\ell=\beta−k−ℓ=β

Also, comparing constants on the right side:

3k+αℓ=33k+\alpha \ell=33k+αℓ=3
  1. Solve for k,ℓk,\ellk,ℓ from the first two equations:

From

k−ℓ=3⇒k=3+ℓk-\ell=3 \Rightarrow k=3+\ellk−ℓ=3⇒k=3+ℓ

Substitute into 2k+ℓ=32k+\ell=32k+ℓ=3:

2(3+ℓ)+ℓ=32(3+\ell)+\ell=32(3+ℓ)+ℓ=3 6+3ℓ=36+3\ell=36+3ℓ=3 3ℓ=−33\ell=-33ℓ=−3 ℓ=−1\ell=-1ℓ=−1

Hence,

k=3+(−1)=2k=3+(-1)=2k=3+(−1)=2
  1. Now find β\betaβ:
β=−k−ℓ=−(2)−(−1)=−1\beta=-k-\ell=-(2)-(-1)=-1β=−k−ℓ=−(2)−(−1)=−1
  1. Use the constant term condition:
3k+αℓ=33k+\alpha \ell=33k+αℓ=3 3(2)+α(−1)=33(2)+\alpha(-1)=33(2)+α(−1)=3 6−α=36-\alpha=36−α=3 α=3\alpha=3α=3
  1. Compute the required value:
α+β−αβ=3+(−1)−(3)(−1)\alpha+\beta-\alpha\beta=3+(-1)-(3)(-1)α+β−αβ=3+(−1)−(3)(−1) =2+3=5=2+3=5=2+3=5

So the required integer is

5\boxed{5}5​
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