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Matrices and Determinants question

2020 · 9 Jan · Shift 2 · Q36
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Matrices and Determinants question

2020 · 9 Jan · Shift 2 · Q36

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
The following system of linear equations 7x + 6y – 2z = 0 3x + 4y + 2z = 0 x – 2y – 6z = 0, has
  1. A
    no solution
  2. B
    infinitely many solutions, (x, y, z) satisfying y = 2z
  3. C
    infinitely many solutions, (x, y, z) satisfying x = 2z
  4. D
    only the trivial solution
View written solutionFree

Correct answer: C

  1. Write the homogeneous system in matrix form

The system is

{7x+6y−2z=03x+4y+2z=0x−2y−6z=0\begin{cases} 7x+6y-2z=0\\ 3x+4y+2z=0\\ x-2y-6z=0 \end{cases}⎩⎨⎧​7x+6y−2z=03x+4y+2z=0x−2y−6z=0​

Since it is a homogeneous system, it always has at least the trivial solution (x,y,z)=(0,0,0)(x,y,z)=(0,0,0)(x,y,z)=(0,0,0).

To determine whether there are non-trivial solutions, check the determinant of the coefficient matrix:

A=(76−23421−2−6)A=\begin{pmatrix} 7 & 6 & -2\\ 3 & 4 & 2\\ 1 & -2 & -6 \end{pmatrix}A=​731​64−2​−22−6​​
  1. Compute the determinant
det⁡(A)=7∣42−2−6∣−6∣321−6∣+(−2)∣341−2∣\det(A)=7\begin{vmatrix}4&2\\-2&-6\end{vmatrix}-6\begin{vmatrix}3&2\\1&-6\end{vmatrix}+(-2)\begin{vmatrix}3&4\\1&-2\end{vmatrix}det(A)=7​4−2​2−6​​−6​31​2−6​​+(−2)​31​4−2​​

Now evaluate each minor:

∣42−2−6∣=4(−6)−2(−2)=−24+4=−20\begin{vmatrix}4&2\\-2&-6\end{vmatrix}=4(-6)-2(-2)=-24+4=-20​4−2​2−6​​=4(−6)−2(−2)=−24+4=−20 ∣321−6∣=3(−6)−2(1)=−18−2=−20\begin{vmatrix}3&2\\1&-6\end{vmatrix}=3(-6)-2(1)=-18-2=-20​31​2−6​​=3(−6)−2(1)=−18−2=−20 ∣341−2∣=3(−2)−4(1)=−6−4=−10\begin{vmatrix}3&4\\1&-2\end{vmatrix}=3(-2)-4(1)=-6-4=-10​31​4−2​​=3(−2)−4(1)=−6−4=−10

So,

det⁡(A)=7(−20)−6(−20)+(−2)(−10)\det(A)=7(-20)-6(-20)+(-2)(-10)det(A)=7(−20)−6(−20)+(−2)(−10) =−140+120+20=0=-140+120+20=0=−140+120+20=0

Since det⁡(A)=0\det(A)=0det(A)=0, the system has infinitely many solutions or dependent equations.

  1. Solve the system to find the relation among variables

From the third equation:

x−2y−6z=0  ⟹  x=2y+6zx-2y-6z=0 \implies x=2y+6zx−2y−6z=0⟹x=2y+6z

Substitute into the second equation:

3x+4y+2z=03x+4y+2z=03x+4y+2z=0 3(2y+6z)+4y+2z=03(2y+6z)+4y+2z=03(2y+6z)+4y+2z=0 6y+18z+4y+2z=06y+18z+4y+2z=06y+18z+4y+2z=0 10y+20z=010y+20z=010y+20z=0 y+2z=0  ⟹  y=−2zy+2z=0 \implies y=-2zy+2z=0⟹y=−2z

Now substitute into x=2y+6zx=2y+6zx=2y+6z:

x=2(−2z)+6z=−4z+6z=2zx=2(-2z)+6z=-4z+6z=2zx=2(−2z)+6z=−4z+6z=2z

Thus the solutions are of the form

(x,y,z)=(2z,−2z,z)=z(2,−2,1)(x,y,z)=(2z,-2z,z)=z(2,-2,1)(x,y,z)=(2z,−2z,z)=z(2,−2,1)
  1. Match with the options
  • A: no solution — false, since homogeneous systems always have at least the trivial solution.
  • B: infinitely many solutions, satisfying y=2zy=2zy=2z — false, because actually y=−2zy=-2zy=−2z.
  • C: infinitely many solutions, satisfying x=2zx=2zx=2z — true.
  • D: only the trivial solution — false, because there are infinitely many non-trivial solutions.

Hence, the correct option is C.

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