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Matrices and Determinants question

2021 · 27 Jul · Shift 1 · Q39
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Matrices and Determinants question

2021 · 27 Jul · Shift 1 · Q39

JEE MainMathematicsMatrices and DeterminantsNumerical+4 / −1
For real numbers α\alphaα and β\betaβ, consider the following system of linear equations : x + y −-− z = 2, x + 2y +α\alphaα z = 1, 2x −-− y + z =β\betaβ. If the system has infinite solutions, then α\alphaα+β\betaβ is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 5

  1. Write the system in matrix form:
{x+y−z=2x+2y+αz=12x−y+z=β\begin{cases} x+y-z=2 \\ x+2y+\alpha z=1 \\ 2x-y+z=\beta \end{cases}⎩⎨⎧​x+y−z=2x+2y+αz=12x−y+z=β​

The coefficient matrix is

A=(11−112α2−11)A=\begin{pmatrix} 1&1&-1\\ 1&2&\alpha\\ 2&-1&1 \end{pmatrix}A=​112​12−1​−1α1​​

and the augmented matrix is

(11−1212α12−11β).\left(\begin{array}{ccc|c} 1&1&-1&2\\ 1&2&\alpha&1\\ 2&-1&1&\beta \end{array}\right).​112​12−1​−1α1​21β​​.
  1. For the system to have infinite solutions, we need:
  • det⁡(A)=0\det(A)=0det(A)=0 so that the system is not uniquely solvable.
  • The system must be consistent, i.e. the rank of the coefficient matrix must equal the rank of the augmented matrix, and this common rank must be less than 3.

So first compute det⁡(A)\det(A)det(A).

det⁡(A)=∣11−112α2−11∣\det(A)=\begin{vmatrix} 1&1&-1\\ 1&2&\alpha\\ 2&-1&1 \end{vmatrix}det(A)=​112​12−1​−1α1​​

Expanding along the first row:

det⁡(A)=1∣2α−11∣−1∣1α21∣+(−1)∣122−1∣\det(A)=1\begin{vmatrix}2&\alpha\\-1&1\end{vmatrix}-1\begin{vmatrix}1&\alpha\\2&1\end{vmatrix}+(-1)\begin{vmatrix}1&2\\2&-1\end{vmatrix}det(A)=1​2−1​α1​​−1​12​α1​​+(−1)​12​2−1​​ =1(2⋅1−α(−1))−(1⋅1−2α)−(1⋅(−1)−2⋅2)=1(2\cdot1-\alpha(-1))-(1\cdot1-2\alpha)-\big(1\cdot(-1)-2\cdot2\big)=1(2⋅1−α(−1))−(1⋅1−2α)−(1⋅(−1)−2⋅2) =(2+α)−(1−2α)−(−5)=(2+\alpha)-(1-2\alpha)-(-5)=(2+α)−(1−2α)−(−5) =2+α−1+2α+5=6+3α.=2+\alpha-1+2\alpha+5=6+3\alpha.=2+α−1+2α+5=6+3α.

For infinite solutions,

6+3α=0  ⟹  α=−2.6+3\alpha=0 \implies \alpha=-2.6+3α=0⟹α=−2.
  1. Substitute α=−2\alpha=-2α=−2 into the system:
{x+y−z=2x+2y−2z=12x−y+z=β\begin{cases} x+y-z=2 \\ x+2y-2z=1 \\ 2x-y+z=\beta \end{cases}⎩⎨⎧​x+y−z=2x+2y−2z=12x−y+z=β​

Now check whether the third equation is dependent on the first two, since for infinite solutions one equation must be a linear combination of the others, and the constants must match accordingly.

Let

λ(x+y−z)+μ(x+2y−2z)=2x−y+z.\lambda(x+y-z)+\mu(x+2y-2z)=2x-y+z.λ(x+y−z)+μ(x+2y−2z)=2x−y+z.

Comparing coefficients:

λ+μ=2\lambda+\mu=2λ+μ=2 λ+2μ=−1\lambda+2\mu=-1λ+2μ=−1 −λ−2μ=1-\lambda-2\mu=1−λ−2μ=1

From the first two equations:

Subtracting,

(λ+2μ)−(λ+μ)=−1−2  ⟹  μ=−3.(\lambda+2\mu)-(\lambda+\mu)=-1-2 \implies \mu=-3.(λ+2μ)−(λ+μ)=−1−2⟹μ=−3.

Then

λ+(−3)=2  ⟹  λ=5.\lambda+(-3)=2 \implies \lambda=5.λ+(−3)=2⟹λ=5.

Check the zzz-coefficient:

−λ−2μ=−5−2(−3)=−5+6=1,-\lambda-2\mu=-5-2(-3)=-5+6=1,−λ−2μ=−5−2(−3)=−5+6=1,

which is correct.

So the third left-hand side is

5(Eq.1)−3(Eq.2).5(\text{Eq.1})-3(\text{Eq.2}).5(Eq.1)−3(Eq.2).

Therefore, for consistency, the right-hand side must satisfy

β=5(2)−3(1)=10−3=7.\beta=5(2)-3(1)=10-3=7.β=5(2)−3(1)=10−3=7.
  1. Hence,
α+β=−2+7=5.\alpha+\beta=-2+7=5.α+β=−2+7=5.

So the required integer is

5.\boxed{5}.5​.
  1. Comparison with stored answer:

Stored correct answer = 555, which matches our result.

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