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 + $\left( {{y \over z}} \right)$ is equal to :
Solution
x + y + 3z = 0 .....(i)
<br>x + 3y + k<sup>2</sup>z = 0 .........(ii)
<br>3x + y + 3z = 0 ......(iii)
<br><br>$$\left| {\matrix{
1 & 1 & 3 \cr
1 & 3 & {{k^2}} \cr
3 & 1 & 3 \cr
} } \right|$$ = 0
<br><br>$\Rightarrow$ 9 + 3 + 3k<sup>2</sup>
– 27 – k<sup>2</sup>
– 3 = 0
<br><br>$\Rightarrow$ k<sup>2</sup> = 9
<br><br>Perform (i) – (iii),
<br><br>–2x = 0 $\Rightarrow$ x = 0
<br><br>Now from (i), y + 3z = 0
<br><br>$\Rightarrow$ ${y \over z} = - 3$
<br><br>$\therefore$ x + $\left( {{y \over z}} \right)$ = -3
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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