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Matrices and Determinants question

2021 · 25 Jul · Shift 1 · Q34
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Matrices and Determinants question

2021 · 25 Jul · Shift 1 · Q34

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
The values of a and b, for which the system of equations 2x + 3y + 6z = 8 x + 2y + az = 5 3x + 5y + 9z = b has no solution, are :
  1. A
    a = 3, b eee 13
  2. B
    a eee 3, b eee 13
  3. C
    a eee 3, b = 3
  4. D
    a = 3, b = 13
View written solutionFree

Correct answer: A

  1. Write the system in matrix form

The system is

{2x+3y+6z=8x+2y+az=53x+5y+9z=b\begin{cases} 2x+3y+6z=8 \\ x+2y+az=5 \\ 3x+5y+9z=b \end{cases}⎩⎨⎧​2x+3y+6z=8x+2y+az=53x+5y+9z=b​

Coefficient matrix:

A=(23612a359)A=\begin{pmatrix} 2&3&6\\ 1&2&a\\ 3&5&9 \end{pmatrix}A=​213​325​6a9​​

For no solution, the system must be inconsistent. This happens when:

  • det⁡(A)=0\det(A)=0det(A)=0, and
  • rank of augmented matrix >>> rank of coefficient matrix.
  1. Compute the determinant
det⁡(A)=∣23612a359∣\det(A)= \begin{vmatrix} 2&3&6\\ 1&2&a\\ 3&5&9 \end{vmatrix}det(A)=​213​325​6a9​​

Expand along the first row:

det⁡(A)=2∣2a59∣−3∣1a39∣+6∣1235∣\det(A)=2\begin{vmatrix}2&a\\5&9\end{vmatrix}-3\begin{vmatrix}1&a\\3&9\end{vmatrix}+6\begin{vmatrix}1&2\\3&5\end{vmatrix}det(A)=2​25​a9​​−3​13​a9​​+6​13​25​​

Now,

∣2a59∣=18−5a,∣1a39∣=9−3a,∣1235∣=5−6=−1\begin{vmatrix}2&a\\5&9\end{vmatrix}=18-5a, \qquad \begin{vmatrix}1&a\\3&9\end{vmatrix}=9-3a, \qquad \begin{vmatrix}1&2\\3&5\end{vmatrix}=5-6=-1​25​a9​​=18−5a,​13​a9​​=9−3a,​13​25​​=5−6=−1

So,

det⁡(A)=2(18−5a)−3(9−3a)+6(−1)\det(A)=2(18-5a)-3(9-3a)+6(-1)det(A)=2(18−5a)−3(9−3a)+6(−1) =36−10a−27+9a−6=36-10a-27+9a-6=36−10a−27+9a−6 =3−a=3-a=3−a

Thus,

det⁡(A)=0  ⟺  a=3\det(A)=0 \iff a=3det(A)=0⟺a=3

So only when a=3a=3a=3 can the system fail to have a unique solution.

  1. Substitute a=3a=3a=3 and check consistency

Then the system becomes

{2x+3y+6z=8x+2y+3z=53x+5y+9z=b\begin{cases} 2x+3y+6z=8 \\ x+2y+3z=5 \\ 3x+5y+9z=b \end{cases}⎩⎨⎧​2x+3y+6z=8x+2y+3z=53x+5y+9z=b​

Now observe the left-hand sides:

(2x+3y+6z)+(x+2y+3z)=3x+5y+9z(2x+3y+6z)+(x+2y+3z)=3x+5y+9z(2x+3y+6z)+(x+2y+3z)=3x+5y+9z

So equation (3) has LHS equal to the sum of the LHS of equations (1) and (2). Hence for consistency, its RHS must also be the sum:

b=8+5=13b=8+5=13b=8+5=13
  • If b=13b=13b=13, the third equation is dependent on the first two, so the system is consistent (infinitely many solutions).
  • If b≠13b\ne 13b=13, then
3x+5y+9z=b3x+5y+9z=b3x+5y+9z=b

contradicts the sum of the first two equations, so the system is inconsistent, i.e. no solution.

  1. Conclusion

For no solution,

a=3,b≠13a=3,\quad b\ne 13a=3,b=13

So the correct option is A.

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