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Matrices and Determinants question

2021 · 26 Feb · Shift 2 · Q28
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  5. /2021 · 26 Feb · Shift 2 · Q28

Matrices and Determinants question

2021 · 26 Feb · Shift 2 · Q28

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Consider the following system of equations : x + 2y −-− 3z = a 2x + 6y −-− 11z = b x −-− 2y + 7z = c, where a, b and c are real constants. Then the system of equations :
  1. A
    has no solution for all a, b and c
  2. B
    has a unique solution when 5a = 2b + c
  3. C
    has infinite number of solutions when 5a = 2b + c
  4. D
    has a unique solution for all a, b and c
View written solutionFree

Correct answer: C

  1. Write the coefficient matrix

The system is

{x+2y−3z=a2x+6y−11z=bx−2y+7z=c\begin{cases} x+2y-3z=a\\ 2x+6y-11z=b\\ x-2y+7z=c \end{cases}⎩⎨⎧​x+2y−3z=a2x+6y−11z=bx−2y+7z=c​

Its coefficient matrix is

A=(12−326−111−27).A=\begin{pmatrix} 1&2&-3\\ 2&6&-11\\ 1&-2&7 \end{pmatrix}.A=​121​26−2​−3−117​​.
  1. Check whether the system has a unique solution

A linear system has a unique solution iff det⁡(A)≠0\det(A)\neq 0det(A)=0.

Compute:

det⁡(A)=∣12−326−111−27∣\det(A)= \begin{vmatrix} 1&2&-3\\ 2&6&-11\\ 1&-2&7 \end{vmatrix}det(A)=​121​26−2​−3−117​​

Expanding along the first row,

det⁡(A)=1∣6−11−27∣−2∣2−1117∣+(−3)∣261−2∣\det(A)=1\begin{vmatrix}6&-11\\-2&7\end{vmatrix} -2\begin{vmatrix}2&-11\\1&7\end{vmatrix} +(-3)\begin{vmatrix}2&6\\1&-2\end{vmatrix}det(A)=1​6−2​−117​​−2​21​−117​​+(−3)​21​6−2​​

Now,

∣6−11−27∣=42−22=20,\begin{vmatrix}6&-11\\-2&7\end{vmatrix}=42-22=20,​6−2​−117​​=42−22=20, ∣2−1117∣=14+11=25,\begin{vmatrix}2&-11\\1&7\end{vmatrix}=14+11=25,​21​−117​​=14+11=25, ∣261−2∣=−4−6=−10.\begin{vmatrix}2&6\\1&-2\end{vmatrix}=-4-6=-10.​21​6−2​​=−4−6=−10.

So,

det⁡(A)=1(20)−2(25)+(−3)(−10)=20−50+30=0.\det(A)=1(20)-2(25)+(-3)(-10)=20-50+30=0.det(A)=1(20)−2(25)+(−3)(−10)=20−50+30=0.

Hence, the system never has a unique solution.

So options B and D are false.

  1. Check consistency condition

Since det⁡(A)=0\det(A)=0det(A)=0, the equations are linearly dependent. Let us find that dependence.

Let

α(1,2,−3)+β(2,6,−11)+γ(1,−2,7)=(0,0,0).\alpha(1,2,-3)+\beta(2,6,-11)+\gamma(1,-2,7)=(0,0,0).α(1,2,−3)+β(2,6,−11)+γ(1,−2,7)=(0,0,0).

This gives

α+2β+γ=0,\alpha+2\beta+\gamma=0,α+2β+γ=0, 2α+6β−2γ=0,2\alpha+6\beta-2\gamma=0,2α+6β−2γ=0, −3α−11β+7γ=0.-3\alpha-11\beta+7\gamma=0.−3α−11β+7γ=0.

A simple relation is obtained by noticing:

−5(x+2y−3z)+2(2x+6y−11z)+(x−2y+7z)=0.-5(x+2y-3z)+2(2x+6y-11z)+(x-2y+7z)=0.−5(x+2y−3z)+2(2x+6y−11z)+(x−2y+7z)=0.

Check:

−5x−10y+15z+4x+12y−22z+x−2y+7z=0.-5x-10y+15z+4x+12y-22z+x-2y+7z=0.−5x−10y+15z+4x+12y−22z+x−2y+7z=0.

So indeed,

−5(Eq.1)+2(Eq.2)+(Eq.3)=0.-5\text{(Eq.1)}+2\text{(Eq.2)}+\text{(Eq.3)}=0.−5(Eq.1)+2(Eq.2)+(Eq.3)=0.

Therefore, for consistency, the constants on the right must satisfy the same relation:

−5a+2b+c=0-5a+2b+c=0−5a+2b+c=0

or

5a=2b+c.5a=2b+c.5a=2b+c.
  1. Interpret the result
  • If 5a≠2b+c5a\ne 2b+c5a=2b+c, the system is inconsistent, so it has no solution.
  • If 5a=2b+c5a=2b+c5a=2b+c, then one equation is dependent on the other two, so rank <3<3<3 and the system is consistent. Hence there are infinitely many solutions.

Thus the correct option is:

C: has infinite number of solutions when 5a=2b+c\boxed{\text{C: has infinite number of solutions when } 5a=2b+c}C: has infinite number of solutions when 5a=2b+c​
  1. Compare with stored correct answer

Stored correct answer: C

This matches the derived answer.

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