Question

The number of non- trivial solution of the system $x - y + z = 0,\;x + 2y - z = 0$ and $2x + y + 3z = 0$ is:
A) 0. B) 1. C) 2. D) 3

Answer 1

Hint: For the non- trivial solution for the system of equations. The determinant of the coefficient of the matrix is zero.
$\boxed{\Delta = 0}$

Complete step-by-step answer:
In the question given above, we are asked for the number of non- trivial solutions for the given equation.
The equation given in the question are as follows:
$x - y + z = 0$ ①
$x + 2y - z = 0$ ②
And $2x + y + 3z = 0$ ③
Now, if we are asked for the non-trivial solution for the given system of linear equations, then we find the determinant of the coefficient of the matrix which must be equal to zero (0).
So, coefficient of equation ①; of ${\text{x,y,}}\;{\text{and}}\;{\text{'z'}}$ respectively are 1,-1, and 1 for $x - y + z = 0$ similarly, the coefficient of equation ②; for $x,y,\;{\text{and}}\;{\text{z}}$ respectively are $1,2, - 1$ for equation $x + 2y - z = 0$
In the similar manner,
the coefficient of equation ③; for $x,y\;{\text{and}}\;{\text{z}}$ respectively are $2,1,3$ for equation $2x + y + 3z = 0$
finding determinant:

1&{ - 1}&1 \\\ 1&2&{ - 1} \\\ 2&1&3 \end{array}} \right|$$ For determinant of ; $\begin{gathered} \Delta = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}} \\\ {{a_2}}&{{b_2}}&{{c_2}} \\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right| = \\\ \\\ \end{gathered} $ $\begin{gathered} {a_1}\left( {{b_2}{C_3} - {b_3}{C_2}} \right) - \\\ {b_1}\left( {{a_2}{c_3} - {a_3}{c_2}} \right) + {c_1} \\\ \;\;\;\;\left( {{a_2}{b_3} - {a_3}{b_2}} \right) \\\ \end{gathered} $ So, $\Delta = \left| {\begin{array}{*{20}{c}} 1&{ - 1}&1 \\\ 1&2&{ - 1} \\\ 2&1&3 \end{array}} \right| = $ $\begin{gathered} 1\left( {3 \times 2 - \left( { - 1 \times 1} \right)} \right) - \\\ \left( { - 1} \right)\left( {1 \times 3 - \left( { - 1 \times 2} \right)} \right) + \\\ 1\left( {1 \times 1 - 2 \times 2} \right) \\\ \end{gathered} $ $ \Rightarrow 1\left( {6 - \left( { - 1} \right)} \right) + 1\left( {3 - \left( { - 2} \right)} \right) + 1\left( {1 - 4} \right)$ $ \Rightarrow 1\left( {6 + 1} \right) + 1\left( {3 + 2} \right) + 1\left( { - 3} \right)$ $ \Rightarrow 1\left( 7 \right) + 1\left( 5 \right) - 3$ $ \Rightarrow 7 + 5 - 3$ $ \Rightarrow 12 - 3 = 9$ So, here determinant value in 9 not 0 Hence, the given system of linear equations has no trivial solution. Therefore the number of trivial solutions is 0. Note : The given system of linear equations will have a non-trivial solution only if the determinant of the coefficient of the matrix is zero.