SATHEE: Matrices and Determinants 4 Question 4

Matrices and Determinants 4 Question 4

4. If the system of equations $2 x + 3 y - z = 0, x + k y - 2 z = 0$ and $2 x - y + z = 0$ has a non-trivial solution $(x, y, z)$ , then $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} + k$ is equal to

(2019 Main, 9 April II)

(a) -4

(b) $\frac{1}{2}$

(d) $\frac{3}{4}$

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Answer:

Correct Answer: 4. (b)

Solution:

Given system of linear equations

$\begin{matrix} 2 x + 3 y - z = 0, \\ x + k y - 2 z = 0 \end{matrix}$

and $2 x - y + z = 0$ has a non-trivial solution $(x, y, z)$ .

$∴ Δ = 0 \Rightarrow | \begin{matrix} 2 & 3 & - 1 \\ 1 & k & - 2 \\ 2 & - 1 & 1 \end{matrix} | = 0$

$\begin{aligned} 2 (k - 2) - 3 (1 + 4) - 1 (- 1 - 2 k) = 0 \\ \Rightarrow & 2 k - 4 - 15 + 1 + 2 k = 0 \\ \Rightarrow & 4 k = 18 \Rightarrow k = \frac{9}{2} \end{aligned}$

So, system of linear equations is

and

$\begin{aligned} 2 x + 3 y - z & = 0 \\ 2 x + 9 y - 4 z & = 0 \end{aligned}$

From Eqs. (i) and (ii), we get

$6 y - 3 z = 0, \frac{y}{z} = \frac{1}{2}$

From Eqs. (i) and (iii), we get

$\begin{aligned} 4 x + 2 y = 0 \Rightarrow \frac{x}{y} = - \frac{1}{2} \\ So, \frac{x}{z} = \frac{x}{y} \times \frac{y}{z} = - \frac{1}{4} \Rightarrow \frac{z}{x} = - 4 [∵ \frac{y}{z} = \frac{1}{2} and \frac{x}{y} = - \frac{1}{2}] \\ ∴ \frac{x}{y} + \frac{y}{z} + \frac{z}{x} + k = - \frac{1}{2} + \frac{1}{2} - 4 + \frac{9}{2} = \frac{1}{2} . \end{aligned}$

Matrices and Determinants 4 Question 4

4. If the system of equations $2 x + 3 y - z = 0, x + k y - 2 z = 0$ and $2 x - y + z = 0$ has a non-trivial solution $(x, y, z)$ , then $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} + k$ is equal to

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Matrices and Determinants 4 Question 4

4. If the system of equations 2x+3y−z=0,x+ky−2z=0 and 2x−y+z=0 has a non-trivial solution (x,y,z), then xy+yz+zx+k is equal to

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4. If the system of equations $2 x + 3 y - z = 0, x + k y - 2 z = 0$ and $2 x - y + z = 0$ has a non-trivial solution $(x, y, z)$ , then $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} + k$ is equal to