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

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(D) The given system of equations is:$3x + 2y + z = 6$$3x + 4y + 3z = 14$$6x + 10y + 8z = a$Let $A = \begin{bmatrix} 3 & 2 & 1 \ 3 & 4 & 3 \ 6 & 10

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$36$

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(D) The given system of equations is:$3x + 2y + z = 6$$3x + 4y + 3z = 14$$6x + 10y + 8z = a$Let $A = \begin{bmatrix} 3 & 2 & 1 \ 3 & 4 & 3 \ 6 & 10 & 8 \end{bmatrix}$ and $B = \begin{bmatrix} 6 \ 14 \ a \end{bmatrix}$.First,we calculate the determinant of $A$:$|A| = 3(32 - 30) - 2(24 - 18) + 1(30 - 24) = 3(2) - 2(6) + 6 = 6 - 12 + 6 = 0$.Since $|A| = 0$,the system has either no solution or infinite solutions.For infinite solutions,we must have $(	ext{adj } A) \cdot B = 0$.The cofactor matrix of $A$ is:$C_{11} = 2, C_{12} = -6, C_{13} = 6$$C_{21} = -6, C_{22} = 18, C_{23} = -18$$C_{31} = 2, C_{32} = -6, C_{33} = 6$Thus,$	ext{adj } A = \begin{bmatrix} 2 & -6 & 2 \ -6 & 18 & -6 \ 6 & -18 & 6 \end{bmatrix}$.Now,$(	ext{adj } A) \cdot B = \begin{bmatrix} 2 & -6 & 2 \ -6 & 18 & -6 \ 6 & -18 & 6 \end{bmatrix} \begin{bmatrix} 6 \ 14 \ a \end{bmatrix} = \begin{bmatrix} 12 - 84 + 2a \ -36 + 252 - 6a \ 36 - 252 + 6a \end{bmatrix} = \begin{bmatrix} 2a - 72 \ 216 - 6a \ 6a - 216 \end{bmatrix}$.Setting this to zero vector,we get $2a - 72 = 0$,which implies $a = 36$.

The system of equations $3x + 2y + z = 6$,$3x + 4y + 3z = 14$ and $6x + 10y + 8z = a$ has an infinite number of solutions if $a$ is equal to

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