For the system of linear equations a x+y+z=1, | vedclass

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(A) The determinant of the coefficient matrix is $D = \begin{vmatrix} a & 1 & 1 \ 1 & a & 1 \ 1 & 1 & a \end{vmatrix} = a(a^2-1) - 1(a-1) + 1(1-a) =

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It has infinitely many solutions if $a=2$ and $\beta=-1$

Vedclass · Answer

(A) The determinant of the coefficient matrix is $D = \begin{vmatrix} a & 1 & 1 \ 1 & a & 1 \ 1 & 1 & a \end{vmatrix} = a(a^2-1) - 1(a-1) + 1(1-a) = a^3 - 3a + 2 = (a-1)^2(a+2)$.For $a=1$,the equations become $x+y+z=1$,$x+y+z=1$,$x+y+z=\beta$. If $\beta=1$,there are infinitely many solutions. Thus,option $D$ is correct.For $a=-2$,$D=0$. The augmented matrix for $\beta=1$ is $\begin{bmatrix} -2 & 1 & 1 & 1 \ 1 & -2 & 1 & 1 \ 1 & 1 & -2 & 1 \end{bmatrix}$. Adding the rows gives $0=3$,which is impossible. Thus,there is no solution. Option $B$ is correct.For $a=2$ and $\beta=1$,$D = (2-1)^2(2+2) = 4 
eq 0$. The system has a unique solution. Using Cramer's rule,$x=y=z = \frac{1}{4}$. Thus $x+y+z = \frac{3}{4}$. Option $C$ is correct.For $a=2$ and $\beta=-1$,$D=4 
eq 0$,so the system has a unique solution,not infinitely many. Thus,option $A$ is incorrect.

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?

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