If the system of linear equations
x + ky + 3z = 0
3x + ky - 2z = 0
2x + 4y - 3z = 0
has a non-zero solution (x, y, z), then \[{{xz} \over {{y^2}}}\] is equal to
Solutions · 1
Official solution
System of equations has non-zero solution when determinant of coefficient = 0.
So, in this questions,
\[\left| {132KK43−2−3 } \right| = 0\]
\[ \Rightarrow \,\,\,\,\] (- 3K + 8) - K (-9 + 4) + 3(12 - 2K) = 0
\[ \Rightarrow \,\,\,\,\] - 3K + 8 + 9K - 4K + 36 - 6K = 0
\[ \Rightarrow \,\,\,\,\] - 4K + 44 = 0
\[ \Rightarrow \,\,\,\,\] K = 11
Now the equations become
x + 11y + 3z = 0 . . . (1)
3x + 11y - 2z = 0 . . . (2)
2x + 4y - 3z = 0 . . . (3)
By adding equation (1) and (3) we get,
3x + 15y = 0
\[ \Rightarrow \,\,\,\,\] x = - 5y
Putting x = - 5y in equation (1) we get
- 5y + 11y + 3z = 0
\[ \Rightarrow \,\,\,\,\] 6y + 3z = 0
\[ \Rightarrow \,\,\,\,\] z = - 2y
\[\therefore\,\,\,\,\] \[{{xz} \over {{y^2}}}\]
\[ = {{\left( { - 5y} \right)\left( { - 2y} \right)} \over {{y^2}}}\]
\[ = {{10{y^2}} \over {{y^2}}}\]
=10