Question

For a non-trial solution $\left| A \right|$ is
a) $\left| A \right| > 0$
b) $\left| A \right| < 0$
c) $\left| A \right| = 0$
d) $\left| A \right| \ne 0$

Answer 1

Hint: Solutions of three variable equations i.e. $\begin{align}

  & {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z={{d}_{1}}, \\

 & {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z={{d}_{2}}, \\

 & {{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}z={{d}_{3}} \\

\end{align}$, will be given by relation

$\left[ \begin{matrix}

   x \\

   y \\

   z \\

\end{matrix} \right]={{A}^{-1}}B$

Where, $A=\left| \begin{matrix}

   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\

   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\

   {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\

\end{matrix} \right|,B=\left| \begin{matrix}

   {{d}_{1}} \\

   {{d}_{2}} \\

   {{d}_{3}} \\

\end{matrix} \right|$

And ${{A}^{-1}}$ will be determined by relation ${{A}^{-1}}=\dfrac{1}{\left| A \right|}adjA$.

Complete step-by-step answer:
As we know the value of (x, y, z) from the following equations can be calculated with the help of the matrix.

Equations are

$\begin{align}

  & {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z={{d}_{1}}, \\

 & {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z={{d}_{2}}, \\

 & {{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}z={{d}_{3}} \\

\end{align}$

Values of x, y, z can be given by the following relations as

$\left[ \begin{matrix}

   x \\

   y \\

   z \\

\end{matrix} \right]={{A}^{-1}}B$

Where matrix A and B are given as

$A=\left| \begin{matrix}

   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\

   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\

   {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\

\end{matrix} \right|,B=\left| \begin{matrix}

   {{d}_{1}} \\

   {{d}_{2}} \\

   {{d}_{3}} \\

\end{matrix} \right|$

And we know ${{A}^{-1}}$can be calculated with the help of relation

${{A}^{-1}}=\dfrac{1}{\left| A \right|}adjA$

Now, we can observe that ${{A}^{-1}}$ can only be calculated for a definite value if value of $\left| A \right|$ is not 0. Hence, ${{A}^{-1}}$ cannot be calculated if $\left| A \right|=0$ and we can observe that values of x, y, z can be calculated by calculating the value of ${{A}^{-1}}$. So, if $\left| A \right|=0$, then we will not be able to get the solutions, which is the case for getting non trivial solutions. As trivial means solution is existing and non-trivial means solutions of the given equations are not existing.
It means for a non-trivial solution $\left| A \right|=0$. So, option (c) is correct.

Note: One may think that, what happens if ${{d}_{1}},{{d}_{2}},{{d}_{3}}=0$. Then solution of three equations of this type will (0, 0, 0), and will not depend on the value of $\left| A \right|$. So, it is the exceptional case, where we do not need to think about $\left| A \right|$.

Don’t confuse with the identity of

$X={{A}^{-1}}B$

Where $x=\left[ \begin{matrix}

   x \\

   y \\

   z \\

\end{matrix} \right]$

One may use identity as $x=B{{A}^{-1}}$, which is wrong. So, be clear with the formulae for future reference.