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Matrices and Determinants question

2021 · 31 Aug · Shift 1 · Q31
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Matrices and Determinants question

2021 · 31 Aug · Shift 1 · Q31

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
If the following system of linear equations 2x + y + z = 5 x −-− y + z = 3 x + y + az = b has no solution, then :
  1. A
    a=−13,be73a = - {1 \over 3},b e {7 \over 3}a=−31​,be37​
  2. B
    ae13,b=73a e {1 \over 3},b = {7 \over 3}ae31​,b=37​
  3. C
    ae−13,b=73a e - {1 \over 3},b = {7 \over 3}ae−31​,b=37​
  4. D
    a=13,be73a = {1 \over 3},b e {7 \over 3}a=31​,be37​
View written solutionFree

Correct answer: D

  1. Write the system in matrix form

The system is:

{2x+y+z=5x−y+z=3x+y+az=b\begin{cases} 2x+y+z=5 \\ x-y+z=3 \\ x+y+az=b \end{cases}⎩⎨⎧​2x+y+z=5x−y+z=3x+y+az=b​

Its coefficient matrix is

A=(2111−1111a)A=\begin{pmatrix} 2&1&1\\ 1&-1&1\\ 1&1&a \end{pmatrix}A=​211​1−11​11a​​

For a system to have no solution, we need:

  • det⁡(A)=0\det(A)=0det(A)=0 so that the coefficient matrix is singular, and
  • the augmented system must be inconsistent.
  1. Find when det⁡(A)=0\det(A)=0det(A)=0

Compute:

det⁡(A)=∣2111−1111a∣\det(A)= \begin{vmatrix} 2&1&1\\ 1&-1&1\\ 1&1&a \end{vmatrix}det(A)=​211​1−11​11a​​

Expanding along the first row:

det⁡(A)=2∣−111a∣−1∣111a∣+1∣1−111∣\det(A)=2\begin{vmatrix}-1&1\\1&a\end{vmatrix}-1\begin{vmatrix}1&1\\1&a\end{vmatrix}+1\begin{vmatrix}1&-1\\1&1\end{vmatrix}det(A)=2​−11​1a​​−1​11​1a​​+1​11​−11​​

Now,

∣−111a∣=−a−1\begin{vmatrix}-1&1\\1&a\end{vmatrix}=-a-1​−11​1a​​=−a−1 ∣111a∣=a−1\begin{vmatrix}1&1\\1&a\end{vmatrix}=a-1​11​1a​​=a−1 ∣1−111∣=2\begin{vmatrix}1&-1\\1&1\end{vmatrix}=2​11​−11​​=2

So,

det⁡(A)=2(−a−1)−(a−1)+2\det(A)=2(-a-1)-(a-1)+2det(A)=2(−a−1)−(a−1)+2 =−2a−2−a+1+2= -2a-2-a+1+2=−2a−2−a+1+2 =1−3a=1-3a=1−3a

For singularity,

1−3a=0  ⟹  a=131-3a=0 \implies a=\frac131−3a=0⟹a=31​
  1. Check consistency when a=13a=\frac13a=31​

Substitute a=13a=\frac13a=31​:

{2x+y+z=5x−y+z=3x+y+13z=b\begin{cases} 2x+y+z=5 \\ x-y+z=3 \\ x+y+\frac13 z=b \end{cases}⎩⎨⎧​2x+y+z=5x−y+z=3x+y+31​z=b​

Now observe the relation among the left-hand sides.

Add the first two equations:

(2x+y+z)+(x−y+z)=5+3(2x+y+z)+(x-y+z)=5+3(2x+y+z)+(x−y+z)=5+3 3x+2z=83x+2z=83x+2z=8

But instead, let us check whether the third row is a linear combination of the first two rows.

Suppose

λ(2,1,1)+μ(1,−1,1)=(1,1,13)\lambda(2,1,1)+\mu(1,-1,1)=(1,1,\tfrac13)λ(2,1,1)+μ(1,−1,1)=(1,1,31​)

Then:

2λ+μ=12\lambda+\mu=12λ+μ=1 λ−μ=1\lambda-\mu=1λ−μ=1 λ+μ=13\lambda+\mu=\frac13λ+μ=31​

From λ−μ=1\lambda-\mu=1λ−μ=1, we get μ=λ−1\mu=\lambda-1μ=λ−1. Substitute into 2λ+μ=12\lambda+\mu=12λ+μ=1:

2λ+(λ−1)=12\lambda+(\lambda-1)=12λ+(λ−1)=1 3λ=2  ⟹  λ=233\lambda=2 \implies \lambda=\frac233λ=2⟹λ=32​

Then

μ=−13\mu=-\frac13μ=−31​

And indeed,

λ+μ=23−13=13\lambda+\mu=\frac23-\frac13=\frac13λ+μ=32​−31​=31​

So,

(1,1,13)=23(2,1,1)−13(1,−1,1)(1,1,\tfrac13)=\frac23(2,1,1)-\frac13(1,-1,1)(1,1,31​)=32​(2,1,1)−31​(1,−1,1)

Hence for consistency, the constants must satisfy the same relation:

b=23(5)−13(3)=103−33=73b=\frac23(5)-\frac13(3)=\frac{10}{3}-\frac{3}{3}=\frac73b=32​(5)−31​(3)=310​−33​=37​

Therefore:

  • if a=13a=\frac13a=31​ and b=73b=\frac73b=37​, the system has infinitely many solutions,
  • if a=13a=\frac13a=31​ and b≠73b\ne\frac73b=37​, the system is inconsistent and has no solution.
  1. Match with options

The required condition is:

a=13,b≠73a=\frac13,\quad b\ne \frac73a=31​,b=37​

So the correct option is D.

  1. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They agree.

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