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Matrices and Determinants question

2020 · 8 Jan · Shift 1 · Q30
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  5. /2020 · 8 Jan · Shift 1 · Q30

Matrices and Determinants question

2020 · 8 Jan · Shift 1 · Q30

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
For which of the following ordered pairs (μ\muμ, δ\deltaδ), the system of linear equations x + 2y + 3z = 1 3x + 4y + 5z = μ\muμ 4x + 4y + 4z =δ\deltaδ is inconsistent ?
  1. A
    (1, 0)
  2. B
    (4, 3)
  3. C
    (4, 6)
  4. D
    (3, 4)
View written solutionFree

Correct answer: B

  1. Consider the system
{x+2y+3z=13x+4y+5z=μ4x+4y+4z=δ\begin{cases} x+2y+3z=1 \\ 3x+4y+5z=\mu \\ 4x+4y+4z=\delta \end{cases}⎩⎨⎧​x+2y+3z=13x+4y+5z=μ4x+4y+4z=δ​

We need the ordered pair (μ,δ)(\mu,\delta)(μ,δ) for which the system is inconsistent.

  1. Write the coefficient matrix:
A=(123345444)A=\begin{pmatrix} 1&2&3\\ 3&4&5\\ 4&4&4 \end{pmatrix}A=​134​244​354​​

Check whether the rows are linearly dependent:

Observe that

R3=2R1+R2?R_3 = 2R_1 + R_2?R3​=2R1​+R2​?

Compute:

2R1+R2=2(1,2,3)+(3,4,5)=(2,4,6)+(3,4,5)=(5,8,11)2R_1 + R_2 = 2(1,2,3) + (3,4,5) = (2,4,6)+(3,4,5)=(5,8,11)2R1​+R2​=2(1,2,3)+(3,4,5)=(2,4,6)+(3,4,5)=(5,8,11)

Not equal to R3R_3R3​.

Let us compute the determinant:

det⁡(A)=∣123345444∣\det(A)= \begin{vmatrix} 1&2&3\\ 3&4&5\\ 4&4&4 \end{vmatrix}det(A)=​134​244​354​​

Expanding along the first row,

det⁡(A)=1∣4544∣−2∣3544∣+3∣3444∣\det(A)=1\begin{vmatrix}4&5\\4&4\end{vmatrix} -2\begin{vmatrix}3&5\\4&4\end{vmatrix} +3\begin{vmatrix}3&4\\4&4\end{vmatrix}det(A)=1​44​54​​−2​34​54​​+3​34​44​​ =1(16−20)−2(12−20)+3(12−16)=1(16-20)-2(12-20)+3(12-16)=1(16−20)−2(12−20)+3(12−16) =−4−2(−8)+3(−4)=-4-2(-8)+3(-4)=−4−2(−8)+3(−4) =−4+16−12=0=-4+16-12=0=−4+16−12=0

So the coefficient matrix is singular. Therefore, the system may be either consistent or inconsistent depending on the constants.

  1. Find the linear relation among the equations.

Apply row operations to the augmented system:

[1231345μ444δ]\left[ \begin{array}{ccc|c} 1&2&3&1\\ 3&4&5&\mu\\ 4&4&4&\delta \end{array} \right]​134​244​354​1μδ​​

Do

R2→R2−3R1,R3→R3−4R1R_2 \to R_2-3R_1, \qquad R_3 \to R_3-4R_1R2​→R2​−3R1​,R3​→R3​−4R1​

Then

[12310−2−4μ−30−4−8δ−4]\left[ \begin{array}{ccc|c} 1&2&3&1\\ 0&-2&-4&\mu-3\\ 0&-4&-8&\delta-4 \end{array} \right]​100​2−2−4​3−4−8​1μ−3δ−4​​

Now do

R3→R3−2R2R_3 \to R_3-2R_2R3​→R3​−2R2​

Then

[12310−2−4μ−3000δ−4−2(μ−3)]\left[ \begin{array}{ccc|c} 1&2&3&1\\ 0&-2&-4&\mu-3\\ 0&0&0&\delta-4-2(\mu-3) \end{array} \right]​100​2−20​3−40​1μ−3δ−4−2(μ−3)​​

For consistency, the last row must satisfy

δ−4−2(μ−3)=0\delta-4-2(\mu-3)=0δ−4−2(μ−3)=0 δ−4−2μ+6=0\delta-4-2\mu+6=0δ−4−2μ+6=0 δ−2μ+2=0\delta-2\mu+2=0δ−2μ+2=0 δ=2μ−2\delta=2\mu-2δ=2μ−2

Thus, the system is inconsistent when

δ≠2μ−2\delta \ne 2\mu-2δ=2μ−2
  1. Check each option.

  • A: (μ,δ)=(1,0)(\mu,\delta)=(1,0)(μ,δ)=(1,0) 2μ−2=2(1)−2=02\mu-2=2(1)-2=02μ−2=2(1)−2=0 Here δ=0\delta=0δ=0, so consistent.

  • B: (μ,δ)=(4,3)(\mu,\delta)=(4,3)(μ,δ)=(4,3) 2μ−2=2(4)−2=62\mu-2=2(4)-2=62μ−2=2(4)−2=6 Here δ=3≠6\delta=3\ne 6δ=3=6, so inconsistent.

  • C: (μ,δ)=(4,6)(\mu,\delta)=(4,6)(μ,δ)=(4,6) 2μ−2=62\mu-2=62μ−2=6 Here δ=6\delta=6δ=6, so consistent.

  • D: (μ,δ)=(3,4)(\mu,\delta)=(3,4)(μ,δ)=(3,4) 2μ−2=2(3)−2=42\mu-2=2(3)-2=42μ−2=2(3)−2=4 Here δ=4\delta=4δ=4, so consistent.

  1. Therefore, the system is inconsistent only for:
(μ,δ)=(4,3)(\mu,\delta)=(4,3)(μ,δ)=(4,3)

So the correct option is B.

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