The system of linear equations x + y + z = 6, | vedclass

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(B) The system of equations is given by:$x + y + z = 6$$2x + 5y + az = 36$$x + 2y + 3z = b$First, calculate the determinant $D$ of the coefficient mat

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infinitely many solutions for $a = 8$ and $b = 14$

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(B) The system of equations is given by:$x + y + z = 6$$2x + 5y + az = 36$$x + 2y + 3z = b$First, calculate the determinant $D$ of the coefficient matrix:$D = \begin{vmatrix} 1 & 1 & 1 \ 2 & 5 & a \ 1 & 2 & 3 \end{vmatrix} = 1(15 - 2a) - 1(6 - a) + 1(4 - 5) = 15 - 2a - 6 + a - 1 = 8 - a$.For the system to have infinitely many solutions or no solution, we must have $D = 0$, which implies $a = 8$.Now, calculate $D_3$ with $a = 8$:$D_3 = \begin{vmatrix} 1 & 1 & 6 \ 2 & 5 & 36 \ 1 & 2 & b \end{vmatrix} = 1(5b - 72) - 1(2b - 36) + 6(4 - 5) = 5b - 72 - 2b + 36 - 6 = 3b - 42$.For $D_3 = 0$, we get $3b = 42$, so $b = 14$.When $a = 8$ and $b = 14$, we check $D_1$ and $D_2$:$D_1 = \begin{vmatrix} 6 & 1 & 1 \ 36 & 5 & 8 \ 14 & 2 & 3 \end{vmatrix} = 6(15 - 16) - 1(108 - 112) + 1(72 - 70) = -6 + 4 + 2 = 0$.$D_2 = \begin{vmatrix} 1 & 6 & 1 \ 2 & 36 & 8 \ 1 & 14 & 3 \end{vmatrix} = 1(108 - 112) - 6(6 - 8) + 1(28 - 36) = -4 + 12 - 8 = 0$.Since $D = D_1 = D_2 = D_3 = 0$ for $a = 8$ and $b = 14$, the system has infinitely many solutions.

The system of linear equations $x + y + z = 6$, $2x + 5y + az = 36$, and $x + 2y + 3z = b$ has:

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