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Matrices and Determinants question

2019 · 9 Apr · Shift 2 · Q42
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Matrices and Determinants question

2019 · 9 Apr · Shift 2 · Q42

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
If the system of equations 2x + 3y – z = 0, x + ky – 2z = 0 and 2x – y + z = 0 has a non-trival solution (x, y, z), then xy+yz+zx+k{x \over y} + {y \over z} + {z \over x} + kyx​+zy​+xz​+k is equal to :-
  1. A
    -4
  2. B
    34{3 \over 4}43​
  3. C
    12{1 \over 2}21​
  4. D
    −14-{1 \over 4}−41​
View written solutionFree

Correct answer: C

  1. Write the homogeneous system in matrix form

The system is

{2x+3y−z=0x+ky−2z=02x−y+z=0\begin{cases} 2x+3y-z=0\\ x+ky-2z=0\\ 2x-y+z=0 \end{cases}⎩⎨⎧​2x+3y−z=0x+ky−2z=02x−y+z=0​

For a homogeneous system to have a non-trivial solution, the determinant of the coefficient matrix must be zero.

So consider

A=(23−11k−22−11)A=\begin{pmatrix} 2 & 3 & -1\\ 1 & k & -2\\ 2 & -1 & 1 \end{pmatrix}A=​212​3k−1​−1−21​​

and require

det⁡(A)=0.\det(A)=0.det(A)=0.
  1. Compute the determinant

Expanding along the first row:

det⁡(A)=2∣k−2−11∣−3∣1−221∣+(−1)∣1k2−1∣\det(A)=2\begin{vmatrix}k & -2\\ -1 & 1\end{vmatrix} -3\begin{vmatrix}1 & -2\\ 2 & 1\end{vmatrix} +(-1)\begin{vmatrix}1 & k\\ 2 & -1\end{vmatrix}det(A)=2​k−1​−21​​−3​12​−21​​+(−1)​12​k−1​​

Now,

∣k−2−11∣=k−2\begin{vmatrix}k & -2\\ -1 & 1\end{vmatrix}=k-2​k−1​−21​​=k−2 ∣1−221∣=1+4=5\begin{vmatrix}1 & -2\\ 2 & 1\end{vmatrix}=1+4=5​12​−21​​=1+4=5 ∣1k2−1∣=−1−2k\begin{vmatrix}1 & k\\ 2 & -1\end{vmatrix}=-1-2k​12​k−1​​=−1−2k

Therefore,

det⁡(A)=2(k−2)−3(5)+(−1)(−1−2k)\det(A)=2(k-2)-3(5)+(-1)(-1-2k)det(A)=2(k−2)−3(5)+(−1)(−1−2k) =2k−4−15+1+2k=4k−18=2k-4-15+1+2k=4k-18=2k−4−15+1+2k=4k−18

For non-trivial solution,

4k−18=0  ⟹  k=92.4k-18=0 \implies k=\frac{9}{2}.4k−18=0⟹k=29​.
  1. Find the ratios x:y:zx:y:zx:y:z

Substitute k=92k=\frac{9}{2}k=29​ into the equations:

x+92y−2z=0x+\frac{9}{2}y-2z=0x+29​y−2z=0

Use the first and third equations:

2x+3y−z=0...(1)2x+3y-z=0 \quad ...(1)2x+3y−z=0...(1) 2x−y+z=0...(2)2x-y+z=0 \quad ...(2)2x−y+z=0...(2)

Add (1) and (2):

4x+2y=0  ⟹  2x+y=0  ⟹  y=−2x4x+2y=0 \implies 2x+y=0 \implies y=-2x4x+2y=0⟹2x+y=0⟹y=−2x

Put in (2):

2x−(−2x)+z=02x-(-2x)+z=02x−(−2x)+z=0 4x+z=0  ⟹  z=−4x4x+z=0 \implies z=-4x4x+z=0⟹z=−4x

Hence,

x:y:z=x:−2x:−4x=1:−2:−4.x:y:z = x:-2x:-4x = 1:-2:-4.x:y:z=x:−2x:−4x=1:−2:−4.
  1. Evaluate the required expression

We need

xy+yz+zx+k.\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+k.yx​+zy​+xz​+k.

Using x:y:z=1:−2:−4x:y:z=1:-2:-4x:y:z=1:−2:−4,

xy=1−2=−12,\frac{x}{y}=\frac{1}{-2}=-\frac12,yx​=−21​=−21​, yz=−2−4=12,\frac{y}{z}=\frac{-2}{-4}=\frac12,zy​=−4−2​=21​, zx=−41=−4.\frac{z}{x}=\frac{-4}{1}=-4.xz​=1−4​=−4.

So,

xy+yz+zx=−12+12−4=−4.\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=-\frac12+\frac12-4=-4.yx​+zy​+xz​=−21​+21​−4=−4.

Now add k=92k=\frac{9}{2}k=29​:

−4+92=−8+92=12.-4+\frac{9}{2}=\frac{-8+9}{2}=\frac12.−4+29​=2−8+9​=21​.
  1. Final answer

Thus,

xy+yz+zx+k=12.\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+k=\frac12.yx​+zy​+xz​+k=21​.

So the correct option is C.

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