- Home
- Standard 12
- Mathematics
સમીકરણો : $x + ay = 0$, $y + az = 0$ and $z + ax = 0$ આપેલ છે તો $'a'$ ની વાસ્તવિક કિમંતો નો ગણ મેળવો કે જેથી સમીકરણો ને અનન્ય ઉકેલ હોય.
$R - \left\{ 1 \right\}$
$R - \left\{ -1 \right\}$
$\left\{ {1, - 1} \right\}$
$\left\{ {1,0, - 1} \right\}$
Solution
Given system of equations is homogeneous which is
$x + ay = 0$
$y + az = 0$
$z + ax = 0$
It can be written inmatrix from as
$A = \left[ {\begin{array}{*{20}{c}}
1&a&0\\
0&1&a\\
a&0&1
\end{array}} \right]$
Now, $\left| A \right| = \left[ {1 – a\left( { – {a^2}} \right)} \right] = 1 + {a^3} \ne 0$
So,system has only trivial solution.
Now, $\left| A \right| = 0$ only when $a=-1$
So, system of equations has infinitely many solutions which is not possible because it is given that system has a unique solutioin.
Hence set of all real value of $'a'$ is
$R-{-1}$.
