If the system of equations 3x + y + 4z = 3, 2 | vedclass

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(A) For a system of linear equations to have no solution, the determinant of the coefficient matrix $\Delta$ must be $0$ and at least one of the Crame

Vedclass · Accepted Answer

$19$

Vedclass · Answer

(A) For a system of linear equations to have no solution, the determinant of the coefficient matrix $\Delta$ must be $0$ and at least one of the Cramer's determinants $\Delta_x, \Delta_y, \Delta_z$ must be non-zero.First, we calculate $\Delta$:$\Delta = \begin{vmatrix} 3 & 1 & 4 \ 2 & a & -1 \ 1 & 2 & 1 \end{vmatrix} = 0$$3(a + 2) - 1(2 + 1) + 4(4 - a) = 0$$3a + 6 - 3 + 16 - 4a = 0$$19 - a = 0 \Rightarrow a = 19$Now, we verify for $a = 19$ by checking $\Delta_x$:$\Delta_x = \begin{vmatrix} 3 & 1 & 4 \ -3 & 19 & -1 \ 4 & 2 & 1 \end{vmatrix} = 3(19 + 2) - 1(-3 + 4) + 4(-6 - 76)$$= 3(21) - 1(1) + 4(-82) = 63 - 1 - 328 = -266 
eq 0$Since $\Delta = 0$ and $\Delta_x 
eq 0$, the system has no solution for $a = 19$.

If the system of equations $3x + y + 4z = 3$, $2x + ay - z = -3$, $x + 2y + z = 4$ has no solution, then the value of $a$ is equal to:

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