The system of equations x+2y+3z=6,x+3y+5z=9,a | vedclass

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(D) For a system of linear equations to have no solution,the determinant of the coefficient matrix $D$ must be $0$,and the system must be inconsistent

Vedclass · Accepted Answer

$8$

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(D) For a system of linear equations to have no solution,the determinant of the coefficient matrix $D$ must be $0$,and the system must be inconsistent.First,we calculate the determinant $D$ of the coefficient matrix:$D = \left|\begin{array}{lll}1 & 2 & 3 \ 1 & 3 & 5 \ 2 & 5 & a\end{array}ight|$$D = 1(3a - 25) - 2(a - 10) + 3(5 - 6)$$D = 3a - 25 - 2a + 20 - 3$$D = a - 8$For the system to have no solution,we set $D = 0$,which gives $a = 8$.Now,check for consistency at $a = 8$ using the augmented matrix $[A|B]$:$\left[\begin{array}{ccc|c}1 & 2 & 3 & 6 \ 1 & 3 & 5 & 9 \ 2 & 5 & 8 & 12\end{array}ight]$Applying $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - 2R_1$:$\left[\begin{array}{ccc|c}1 & 2 & 3 & 6 \ 0 & 1 & 2 & 3 \ 0 & 1 & 2 & 0\end{array}ight]$Applying $R_3 	o R_3 - R_2$:$\left[\begin{array}{ccc|c}1 & 2 & 3 & 6 \ 0 & 1 & 2 & 3 \ 0 & 0 & 0 & -3\end{array}ight]$Since the last row represents $0 = -3$,which is a contradiction,the system has no solution when $a = 8$.

The system of equations $x+2y+3z=6$,$x+3y+5z=9$,and $2x+5y+az=12$ has no solution when $a=$

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