Question

The solution set the given set of equations will be

$x+y+z=6$

$x+2y+3z=10$

$x+2y+z=1$

• A. Unique trivial solutions
• B. Unique non trivial solution
• C. No Solutions
• D. Infinite solutions

Answer 1

The correct option is B

Unique non trivial solution

Here, A = $\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 1& 2& 1\end{array}\right]$ B = $\left[\begin{array}{c}6\\ 10\\ 1\end{array}\right]$ C = $\left[\begin{array}{cccc}1& 1& 1& 6\\ 1& 2& 3& 10\\ 1& 2& 1& 1\end{array}\right]$ Rank of A will be: On A , $R_2\to R_2-R_3$$\left[\begin{array}{ccc}1& 1& 1\\ 0& 0& 1\\ 1& 2& 1\end{array}\right]$ $R_3\to R_3–R_1$$\left[\begin{array}{ccc}1& 1& 1\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]$ $R_1\to R_1–R_2$; $R_1\to R_1–R_3$ $\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]$ $R_2\leftrightarrow R_3$ $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$ $\to$ Rank = 3 Rank of C will be $\left[\begin{array}{cccc}1& 1& 1& 6\\ 1& 2& 3& 10\\ 1& 2& 1& 1\end{array}\right]$ $R_2\to R_2–R_1,R_3\to R_3–R_1$$\left[\begin{array}{cccc}1& 1& 1& 6\\ 0& 1& 2& 4\\ 0& 1& 0& -5\end{array}\right]$ $R_3\to R_3–R_2$ $\left[\begin{array}{cccc}1& 1& 1& 6\\ 0& 1& 2& 4\\ 0& 0& -2& -9\end{array}\right]$ This is now in row Echelon form and Rank of C is 3. $\therefore$ Rank of A = Rank of C = number of unknowns $\therefore$ unique nontrivial solution.