The system 2x + 3y + z = 5,3x + y + 5z = 7 an | vedclass

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(D) Given the system of equations: $2x + 3y + z = 5$ $3x + y + 5z = 7$ $x + 4y - 2z = 3$ The system can be written as $AX = B$,where $A = \begin{bmatr

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(D) Given the system of equations: $2x + 3y + z = 5$ $3x + y + 5z = 7$ $x + 4y - 2z = 3$ The system can be written as $AX = B$,where $A = \begin{bmatrix} 2 & 3 & 1 \ 3 & 1 & 5 \ 1 & 4 & -2 \end{bmatrix}$,$X = \begin{bmatrix} x \ y \ z \end{bmatrix}$,$B = \begin{bmatrix} 5 \ 7 \ 3 \end{bmatrix}$ First,calculate the determinant $|A|$: $|A| = 2(-2 - 20) - 3(-6 - 5) + 1(12 - 1)$ $|A| = 2(-22) - 3(-11) + 1(11) = -44 + 33 + 11 = 0$ Since $|A| = 0$,the system is either inconsistent (no solution) or has infinitely many solutions. We check $(	ext{adj } A)B$: $	ext{adj } A = \begin{bmatrix} -22 & 10 & 14 \ 11 & -5 & -7 \ 11 & -5 & -7 \end{bmatrix}$ $(	ext{adj } A)B = \begin{bmatrix} -22 & 10 & 14 \ 11 & -5 & -7 \ 11 & -5 & -7 \end{bmatrix} \begin{bmatrix} 5 \ 7 \ 3 \end{bmatrix} = \begin{bmatrix} -110 + 70 + 42 \ 55 - 35 - 21 \ 55 - 35 - 21 \end{bmatrix} = \begin{bmatrix} 2 \ -1 \ -1 \end{bmatrix} 
eq 0$ Since $(	ext{adj } A)B 
eq 0$,the system has no solution.

The system $2x + 3y + z = 5$,$3x + y + 5z = 7$ and $x + 4y - 2z = 3$ has

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