Question

The following system of linear equations has:

:$$\begin{gathered} 7\mathrm{x}+\; 6\mathrm{y}-\; 2\mathrm{z}=\; 0 \\ 3\mathrm{x}+\; 4\mathrm{y}+2\mathrm{z}=\; 0 \\ \mathrm{x}-\; 2\mathrm{y}-\; 6\mathrm{z}=\; 0 \end{gathered}$$

• A. infinitely many solutions, $(x,y,z)$ satisfying $y=2z$
• B. infinitely many solutions $(x,y,z)$ satisfying $x=2z$
• C. no solution
• D. only the trivial solution

Answer 1

The correct option is B

infinitely many solutions $(x,y,z)$ satisfying $x=2z$

Explanation for the correct option:

Solve the given set of equations

Given equations:

$$\begin{gathered} 7\mathrm{x}+\; 6\mathrm{y}-\; 2\mathrm{z}=\; 0 \\ 3\mathrm{x}+\; 4\mathrm{y}+2\mathrm{z}=\; 0 \\ \mathrm{x}-\; 2\mathrm{y}-\; 6\mathrm{z}=\; 0 \end{gathered}$$

As the system of equations are homogeneous the system is consistent.

$\left|\begin{array}{ccc}7& 6& -2\\ 3& 4& 2\\ 1& -2& -6\end{array}\right|$

$$\begin{gathered} =7\left|\begin{array}{cc}4& 2\\ -2& -6\end{array}\right|-6\left|\begin{array}{cc}3& 2\\ 1& -6\end{array}\right|+(-2)\left|\begin{array}{cc}3& 4\\ 1& -2\end{array}\right| \\ =7(-20)-6(-20)-2\left(70\right) \\ =-140+120+20 \\ =0 \end{gathered}$$

$\Rightarrow$ Infinite solutions exist (both trivial and non-trivial solutions)

For option$\left(A\right):$-

When $y=2z$,

Let’s take $y=2,z=1$

When $(x,2,1)$ is substituted in the system of equations

$\Rightarrow$$7\mathrm{x}+10=0,3\mathrm{x}+10=0,\mathrm{x}-10=0$(which is not possible)

So, for$y=2z\Rightarrow$ Infinitely many solutions does not exist.

For option $\left(B\right)$:-

For $x=2z$, lets take $x=2,z=1,y=y$

Substitute $(2,y,1)$ in system of equations $\Rightarrow y=-2$

So, for each pair of $(x,z),$ we get a value of $y$.

Thus, for $x=2z$ infinitely many solutions exists.

Therefore, option (B) is the correct answer.