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Matrices and Determinants question

2023 · 25 Jan · Shift 1 · Q36
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Matrices and Determinants question

2023 · 25 Jan · Shift 1 · Q36

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let S 1_11​ and S 2_22​ be respectively the sets of all a∈R−{0}a \in \mathbb{R} - \{ 0\}a∈R−{0} for which the system of linear equations ax+2ay−3az=1(2a+1)x+(2a+3)y+(a+1)z=2(3a+5)x+(a+5)y+(a+2)z=3ax + 2ay - 3az = 1(2a + 1)x + (2a + 3)y + (a + 1)z = 2(3a + 5)x + (a + 5)y + (a + 2)z = 3ax+2ay−3az=1(2a+1)x+(2a+3)y+(a+1)z=2(3a+5)x+(a+5)y+(a+2)z=3 has unique solution and infinitely many solutions. Then
  1. A
    n(S1)=2\mathrm{n({S_1}) = 2}n(S1​)=2 and S 2_22​ is an infinite set
  2. B
    S1=Φ\mathrm{{S_1} = \Phi}S1​=Φ and S2=R−{0}\mathrm{{S_2} = \mathbb{R} - \{ 0\}}S2​=R−{0}
  3. C
    S1=R−{0}\mathrm{{S_1} = \mathbb{R} - \{ 0\}}S1​=R−{0} and S2=Φ\mathrm{{S_2} = \Phi}S2​=Φ
  4. D
    S 1_11​ is an infinite set and n(S 2_22​) = 2
View written solutionFree

Correct answer: C

We are given the system

{ax+2ay−3az=1(2a+1)x+(2a+3)y+(a+1)z=2(3a+5)x+(a+5)y+(a+2)z=3\begin{cases} ax+2ay-3az=1 \\ (2a+1)x+(2a+3)y+(a+1)z=2 \\ (3a+5)x+(a+5)y+(a+2)z=3 \end{cases}⎩⎨⎧​ax+2ay−3az=1(2a+1)x+(2a+3)y+(a+1)z=2(3a+5)x+(a+5)y+(a+2)z=3​

with a∈R∖{0}a\in \mathbb R\setminus\{0\}a∈R∖{0}.

Let

  • S1S_1S1​ be the set of all aaa for which the system has a unique solution,
  • S2S_2S2​ be the set of all aaa for which the system has infinitely many solutions.

We must determine the correct option.


1. Coefficient matrix

The coefficient matrix is

A=(a2a−3a2a+12a+3a+13a+5a+5a+2).A= \begin{pmatrix} a & 2a & -3a\\ 2a+1 & 2a+3 & a+1\\ 3a+5 & a+5 & a+2 \end{pmatrix}.A=​a2a+13a+5​2a2a+3a+5​−3aa+1a+2​​.

A system of 3 linear equations in 3 unknowns has:

  1. a unique solution if det⁡(A)≠0\det(A)\neq 0det(A)=0,
  2. infinitely many solutions if det⁡(A)=0\det(A)=0det(A)=0 and the system is consistent with rank <3<3<3.

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


2. Compute the determinant

det⁡(A)=∣a2a−3a2a+12a+3a+13a+5a+5a+2∣\det(A)= \begin{vmatrix} a & 2a & -3a\\ 2a+1 & 2a+3 & a+1\\ 3a+5 & a+5 & a+2 \end{vmatrix}det(A)=​a2a+13a+5​2a2a+3a+5​−3aa+1a+2​​

Expand along the first row:

det⁡(A)=a∣2a+3a+1a+5a+2∣−2a∣2a+1a+13a+5a+2∣+(−3a)∣2a+12a+33a+5a+5∣\det(A)=a \begin{vmatrix} 2a+3 & a+1\\ a+5 & a+2 \end{vmatrix} -2a \begin{vmatrix} 2a+1 & a+1\\ 3a+5 & a+2 \end{vmatrix} +(-3a) \begin{vmatrix} 2a+1 & 2a+3\\ 3a+5 & a+5 \end{vmatrix}det(A)=a​2a+3a+5​a+1a+2​​−2a​2a+13a+5​a+1a+2​​+(−3a)​2a+13a+5​2a+3a+5​​

Now compute each minor.

First minor

∣2a+3a+1a+5a+2∣=(2a+3)(a+2)−(a+1)(a+5)\begin{vmatrix} 2a+3 & a+1\\ a+5 & a+2 \end{vmatrix} =(2a+3)(a+2)-(a+1)(a+5)​2a+3a+5​a+1a+2​​=(2a+3)(a+2)−(a+1)(a+5) =(2a2+7a+6)−(a2+6a+5)=a2+a+1=(2a^2+7a+6)-(a^2+6a+5)=a^2+a+1=(2a2+7a+6)−(a2+6a+5)=a2+a+1

Second minor

∣2a+1a+13a+5a+2∣=(2a+1)(a+2)−(a+1)(3a+5)\begin{vmatrix} 2a+1 & a+1\\ 3a+5 & a+2 \end{vmatrix} =(2a+1)(a+2)-(a+1)(3a+5)​2a+13a+5​a+1a+2​​=(2a+1)(a+2)−(a+1)(3a+5) =(2a2+5a+2)−(3a2+8a+5)=−(a2+3a+3)=(2a^2+5a+2)-(3a^2+8a+5)=-(a^2+3a+3)=(2a2+5a+2)−(3a2+8a+5)=−(a2+3a+3)

Third minor

∣2a+12a+33a+5a+5∣=(2a+1)(a+5)−(2a+3)(3a+5)\begin{vmatrix} 2a+1 & 2a+3\\ 3a+5 & a+5 \end{vmatrix} =(2a+1)(a+5)-(2a+3)(3a+5)​2a+13a+5​2a+3a+5​​=(2a+1)(a+5)−(2a+3)(3a+5) =(2a2+11a+5)−(6a2+19a+15)=−(4a2+8a+10)=(2a^2+11a+5)-(6a^2+19a+15)=-(4a^2+8a+10)=(2a2+11a+5)−(6a2+19a+15)=−(4a2+8a+10)

Substitute back:

det⁡(A)=a(a2+a+1)−2a(−(a2+3a+3))+(−3a)(−(4a2+8a+10))\det(A)=a(a^2+a+1)-2a\bigl(-(a^2+3a+3)\bigr)+(-3a)\bigl(-(4a^2+8a+10)\bigr)det(A)=a(a2+a+1)−2a(−(a2+3a+3))+(−3a)(−(4a2+8a+10)) =a(a2+a+1)+2a(a2+3a+3)+3a(4a2+8a+10)= a(a^2+a+1)+2a(a^2+3a+3)+3a(4a^2+8a+10)=a(a2+a+1)+2a(a2+3a+3)+3a(4a2+8a+10) =(a3+a2+a)+(2a3+6a2+6a)+(12a3+24a2+30a)= (a^3+a^2+a)+(2a^3+6a^2+6a)+(12a^3+24a^2+30a)=(a3+a2+a)+(2a3+6a2+6a)+(12a3+24a2+30a) =15a3+31a2+37a=15a^3+31a^2+37a=15a3+31a2+37a det⁡(A)=a(15a2+31a+37)\det(A)=a(15a^2+31a+37)det(A)=a(15a2+31a+37)

3. When is the determinant zero?

Since a∈R∖{0}a\in \mathbb R\setminus\{0\}a∈R∖{0}, we only need to check

15a2+31a+37=0.15a^2+31a+37=0.15a2+31a+37=0.

Its discriminant is

Δ=312−4⋅15⋅37=961−2220=−1259<0.\Delta=31^2-4\cdot 15\cdot 37=961-2220=-1259<0.Δ=312−4⋅15⋅37=961−2220=−1259<0.

So the quadratic has no real root.

Therefore, for every real a≠0a\neq 0a=0,

det⁡(A)≠0.\det(A)\neq 0.det(A)=0.

Hence the system has a unique solution for every a∈R∖{0}a\in \mathbb R\setminus\{0\}a∈R∖{0}.

So,

S1=R∖{0}.S_1=\mathbb R\setminus\{0\}.S1​=R∖{0}.

4. Determine S2S_2S2​

For infinitely many solutions, we need det⁡(A)=0\det(A)=0det(A)=0 first. But we found that for all real a≠0a\neq 0a=0,

det⁡(A)≠0.\det(A)\neq 0.det(A)=0.

Thus there is no value of a∈R∖{0}a\in \mathbb R\setminus\{0\}a∈R∖{0} for which the system has infinitely many solutions.

So,

S2=∅.S_2=\varnothing.S2​=∅.

5. Match with the options

We obtained:

S1=R∖{0},S2=∅.S_1=\mathbb R\setminus\{0\}, \qquad S_2=\varnothing.S1​=R∖{0},S2​=∅.

This matches Option C.


Final Answer

C\boxed{\text{C}}C​
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