QJEEM20Mab1be: JEE Main 2020 Mathematics - Class 12 JEEM20Mab1be Solution: If the system of linear equations x + y + 3z = 0 x + 3y + k 2 z …

Question

Verified Solution • 2025–26 Session moderate

If the system of linear equations
x + y + 3z = 0
x + 3y + k²z = 0
3x + y + 3z = 0
has a non-zero solution (x, y, z) for some k

\in

R, then x +

(\frac{y}{z})

is equal to :

Accepted Answer

x + y + 3z = 0 .....(i)
x + 3y + k²z = 0 .........(ii)
3x + y + 3z = 0 ......(iii)

| \begin{matrix} 1 & 1 & 3 \\ 1 & 3 & k^{2} \\ 3 & 1 & 3 \end{matrix} |

= 0

\Rightarrow

9 + 3 + 3k² – 27 – k² – 3 = 0

\Rightarrow

k² = 9

Perform (i) – (iii),

–2x = 0

\Rightarrow

x = 0

Now from (i), y + 3z = 0

\Rightarrow

\frac{y}{z} = - 3

∴

x +

(\frac{y}{z})

= -3

Step-by-Step Verified Solution