JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
If the system of linear equations \(x + ky + 3z = 0;3x + ky - 2z = 0\) ; \(2x + 4y - 3z = 0\) has a non-zero solution \(\left( {x,y,z} \right)\) then \(\frac{{xz}}{{{y^2}}} = \). . . . .
- A \(10\)
- B \(-30\)
- C \(30\)
- D \(-10\)
Answer & Solution
Correct Answer
(A) \(10\)
Step-by-step Solution
Detailed explanation
For non zero solution of the system of linear equation; \(\left| {\begin{array}{*{20}{c}} 1&k&3\\ 3&k&{ - 2}\\ 2&4&{ - 3} \end{array}} \right| = 0\) \( \Rightarrow k = 11\) Now equations become \(x+11y+3z=0\) ......\((1)\) \(3x+11y-2z=0\) ......\((2)\) \(2x+4y-3z=0\)…
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