Determinants Question 8
Question 8 - 01 February - Shift 2
For the system of linear equations $a x+y+z=1$, $x+a y+z=1, x+y+a z=\beta$, which one of the following statements is NOT correct ?
(1) It has infinitely many solutions if $\alpha=2$ and $\beta=-1$
(2) It has no solution if $\alpha=-2$ and $\beta=1$
(3) $x+y+z=\frac{3}{4}$ if $\alpha=2$ and $\beta=1$
(4) It has infinitely many solutions if $\alpha=1$ and $\beta=1$
Show Answer
Answer: (1)
Solution:
Formula: System of equations with 3 variables, consistency of solutions, unique solution
$ \begin{vmatrix} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha\end{vmatrix} =0$
$\alpha(\alpha^{2}-1)-1(\alpha-1)+1(1-\alpha)=0$
$\alpha^{3}-3 \alpha+2=0$
$\alpha^{2}(\alpha-1)+\alpha(\alpha-1)-2(\alpha-1)=0$
$(\alpha-1)(\alpha^{2}+\alpha-2)=0$
$\alpha=1, \alpha=-2,1$
For $\alpha=1, \beta=1$
$.\begin{matrix} x+y+z=1 \\ x+y+z=b\end{matrix} $ infinite solution
For $\alpha=2, \beta=1$
$\Delta=4$
$$ \begin{array}{ll} \Delta_1=\left|\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{array}\right| \begin{array}{c} =3-1-1 \\ =1 \end{array} & \Rightarrow x=\frac{1}{4} \\ \Delta_2=\left|\begin{array}{lll} 2 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 2 \end{array}\right|=2-1=1 & \Rightarrow y=\frac{1}{4} \\ \Delta_3=\left|\begin{array}{lll} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 1 \end{array}\right|=2-1=1 & \Rightarrow z=\frac{1}{4} \end{array} $$
For $\alpha=2 \Rightarrow$ unique solution
