Question

Find all solutions of this system.

4x + 2y + 3z = 0

3x − y + 2z = 0

x + 2y − z = 0







Compare the value of the determinant of the coefficient matrix to zero using the nature

of the solution.

Answer 1

$\mathbf{To\;find: Solutions\;of\;the\;system:-}$

4x + 2y + 3z = 0

3x − y + 2z = 0

x + 2y − z = 0

$\mathbf{The \;augmented\;matrix\;associated\;with\;above\;system\;is-}$

$\left[ \begin{array}{ccc|c} 4 & 2 & 3 & 0 \\ 3 & -1 & 2 & 0 \\ 1 & 2 & -1 & 0 \\ \end{array} \right]$

$\mathbf{Row\;reducing\;above\;matrix\;R_1 \to R_1-4R_3\;and\;R_2\to R_2-3R_3}$

$\left[ \begin{array}{ccc|c} 0 & -6 & 7 & 0 \\ 0 & -7 & 5 & 0 \\ 1 & 2 & -1 & 0 \\ \end{array} \right]$

$\mathbf{Again\;R_1\to R_1-R_2\;we\;get-}$

$\left[ \begin{array}{ccc|c} 0 & 1 & 2 & 0 \\ 0 & -7 & 5 & 0 \\ 1 & 2 & -1 & 0 \\ \end{array} \right]$

$\mathbf{R_3\to R_3-2R_1\;gives-}$

$\left[ \begin{array}{ccc|c} 0 & 1 & 2 & 0 \\ 0 & -7 & 5 & 0 \\ 1 & 0 & -5 & 0 \\ \end{array} \right]$

$\mathbf{R_2\to R_2+7R_1\;gives-}$

$\left[ \begin{array}{ccc|c} 0 & 1 & 2 & 0 \\ 0 & 0 & 19 & 0 \\ 1 & 0 & -5 & 0 \\ \end{array} \right]$

$\mathbf{R_2\to \dfrac{1}{19} R_2\;gives-}$

$\left[ \begin{array}{ccc|c} 0 & 1 & 2 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & -5 & 0 \\ \end{array} \right]$

$\mathbf{So,\;the\;system\;of\;equations\;gives-}$

$\mathbf{y+2z=0 \implies y=-2z}$

$\mathbf{z=0}$

$\mathbf{x-5z=0 \implies x=5z \implies x=0}$

$\mathbf{Similarly, y=-2z => y=0}$

$\mathbf{Hence,\;(0,0,0)\;is\;the\;only\;solution\;to\;this\;system.}$

$\mathbf{The\;coefficient\;matrix\;is\;A-}$

$\left[ \begin{array}{ccc} 4 & 2 & 3\\ 3 & -1 & 2\\ 1 & 2 & -1\\ \end{array} \right]$

$\mathbf{and\;the\;determinant\;is\;given\;by-}$

$\mathbf{det(A)=4}\begin{vmatrix} -1 & 2 \\ 2 & -1 \end{vmatrix}-2\begin{vmatrix} 3 & 2 \\ 1 & -1 \end{vmatrix}+3\begin{vmatrix} 3 & -1 \\ 1 & 2 \end{vmatrix}$

$\mathbf{=4(1-4)-2(-3-2)+3(6+1)=4(-4)-2(-5)+3(7)}$

$\mathbf{=-16+10+21=15}$

$\mathbf{Observe\;that\;det(A) \neq 0 \;which \;implies\;that\;the\;solution }$

$\mathbf{ \;of\;this\;system\;is\;unique.}$

Result :- A n x n non-homogeneous system of linear equations has a unique non-trivial solution if and only if its determinant is non-zero.