If the system of linear equations x + ay + z | vedclass

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(D) For a system of linear equations to have no solution,the determinant of the coefficient matrix $\Delta$ must be zero,and at least one of the Crame

Vedclass · Accepted Answer

$a = -1, b 
eq 9$

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(D) For a system of linear equations to have no solution,the determinant of the coefficient matrix $\Delta$ must be zero,and at least one of the Cramer's rule determinants $(\Delta_1, \Delta_2, \Delta_3)$ must be non-zero.First,calculate $\Delta$:$\Delta = \begin{vmatrix} 1 & a & 1 \ 1 & 2 & 2 \ 1 & 5 & 3 \end{vmatrix} = 1(6 - 10) - a(3 - 2) + 1(5 - 2) = -4 - a + 3 = -a - 1$.Setting $\Delta = 0$,we get $-a - 1 = 0$,which implies $a = -1$.Next,calculate $\Delta_2$ (replacing the second column with the constants $3, 6, b$):$\Delta_2 = \begin{vmatrix} 1 & 3 & 1 \ 1 & 6 & 2 \ 1 & b & 3 \end{vmatrix} = 1(18 - 2b) - 3(3 - 2) + 1(b - 6) = 18 - 2b - 3 + b - 6 = 9 - b$.For the system to have no solution,we require $\Delta_2 
eq 0$,so $9 - b 
eq 0$,which implies $b 
eq 9$.Thus,the condition is $a = -1$ and $b 
eq 9$.

If the system of linear equations $x + ay + z = 3$,$x + 2y + 2z = 6$,and $x + 5y + 3z = b$ has no solution,then:

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