For the system of equations x+y+z=6,x+2y+alph | vedclass

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(A) The system of equations is given by:$x+y+z=6$$x+2y+\alpha z=10$$x+3y+5z=\beta$The determinant of the coefficient matrix $D$ is:$D = \begin{vmatrix

Vedclass · Accepted Answer

System has a unique solution for $\alpha=3, \beta 
eq 14$.

Vedclass · Answer

(A) The system of equations is given by:$x+y+z=6$$x+2y+\alpha z=10$$x+3y+5z=\beta$The determinant of the coefficient matrix $D$ is:$D = \begin{vmatrix} 1 & 1 & 1 \ 1 & 2 & \alpha \ 1 & 3 & 5 \end{vmatrix} = 1(10-3\alpha) - 1(5-\alpha) + 1(3-2) = 10-3\alpha-5+\alpha+1 = 6-2\alpha$.For a unique solution,$D 
eq 0$,which implies $6-2\alpha 
eq 0$,so $\alpha 
eq 3$.If $\alpha=3$,then $D=0$. The system becomes:$x+y+z=6$$x+2y+3z=10$$x+3y+5z=\beta$Subtracting the first equation from the second: $y+2z=4$.Subtracting the second equation from the third: $y+2z=\beta-10$.For the system to have solutions,we must have $4 = \beta-10$,which means $\beta=14$. If $\beta=14$,there are infinitely many solutions. If $\beta 
eq 14$,there is no solution.Option $A$ states the system has a unique solution for $\alpha=3$,which is false because $D=0$ when $\alpha=3$.

For the system of equations $x+y+z=6$,$x+2y+\alpha z=10$,and $x+3y+5z=\beta$,which one of the following is $NOT$ true?

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