c) $\begin{array}{r}4 x_{2}+x_{3}=0 \\ x_{1}+x_{2}+x_{3}=0 \\ 2 x_{1}+4 x_{2}+3 x_{3}=0\end{array}$

Full solution

Q. c)

\begin{array}{r}4 x_{2}+x_{3}=0 \\ x_{1}+x_{2}+x_{3}=0 \\ 2 x_{1}+4 x_{2}+3 x_{3}=0\end{array}

Given System of Equations: We are given a system of linear equations: $\newline$ $\begin{align*} 4x_2 + x_3 &= 0 \quad \text{(Equation 1)} \\ x_1 + x_2 + x_3 &= 0 \quad \text{(Equation 2)} \\ 2x_1 + 4x_2 + 3x_3 &= 0 \quad \text{(Equation 3)} \end{align*}$ $\newline$ We will use the method of substitution or elimination to solve for $x_1, x_2,$ and $x_3$ .
Solving Equation $1$ : First, we can solve Equation $1$ for $x_3$ in terms of $x_2$ : $\newline$ $x_3 = -4x_2$ $\newline$ We will use this expression for $x_3$ to substitute into the other equations.
Substitute into Equation $2$ : Substitute $x_3 = -4x_2$ into Equation $2$ : $\newline$ $x_1 + x_2 - 4x_2 = 0$ $\newline$ Simplify the equation: $\newline$ $x_1 - 3x_2 = 0$ $\newline$ Now, we can solve for $x_1$ in terms of $x_2$ : $\newline$ $x_1 = 3x_2$
Solving for x $1$ : Substitute $x_3 = -4x_2$ and $x_1 = 3x_2$ into Equation $3$ : $\newline$ $2(3x_2) + 4x_2 + 3(-4x_2) = 0$ $\newline$ Simplify the equation: $\newline$ $6x_2 + 4x_2 - 12x_2 = 0$ $\newline$ $-2x_2 = 0$ $\newline$ Now, solve for $x_2$ : $\newline$ $x_2 = 0$
Substitute into Equation $3$ : Since $x_2 = 0$ , we can substitute back into the expressions for $x_1$ and $x_3$ : $\newline$ $x_1 = 3(0) = 0$ $\newline$ $x_3 = -4(0) = 0$
Solving for x $2$ : We have found the values for $x_1, x_2,$ and $x_3$ : $\newline$ $x_1 = 0$ $\newline$ $x_2 = 0$ $\newline$ $x_3 = 0$ $\newline$ These values satisfy all three equations in the system.