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13. Solve the system using the elimination method. x + 3y - z = 2x + y - z = 0 3x + 2y - 3z =-1

User Alvins
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1 Answer

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We have the following:


\begin{gathered} x+3y-z=2 \\ x+y-z=0 \\ 3x+2y-3z=-1 \end{gathered}

solving for elimination:


\begin{bmatrix}{1} & {3} & -1 \\ {1} & {1} & {1} \\ {3} & {2} & {-3}\end{bmatrix}=\begin{aligned}2 \\ 0 \\ -1\end{aligned}

R1 <-> R3


\begin{bmatrix}{3} &amp; {2} &amp; -3 \\ {1} &amp; {1} &amp; {1} \\ {1} &amp; {3} &amp; {-1}\end{bmatrix}=\begin{aligned}-1 \\ 0 \\ 2\end{aligned}

R2 - 1/3*R1


\begin{bmatrix}{3} &amp; {2} &amp; -3 \\ {0} &amp; {(1)/(3)} &amp; {0} \\ {1} &amp; {3} &amp; {-1}\end{bmatrix}=\begin{aligned}-1 \\ (1)/(3) \\ 2\end{aligned}

R3 - 1/3 * R1


\begin{bmatrix}{3} &amp; {2} &amp; -3 \\ {0} &amp; {(1)/(3)} &amp; {0} \\ {0} &amp; {(7)/(3)} &amp; {0}\end{bmatrix}=\begin{aligned}-1 \\ (1)/(3) \\ (7)/(3)\end{aligned}

R2 <-> R3


\begin{bmatrix}{3} &amp; {2} &amp; -3 \\ {0} &amp; {(7)/(3)} &amp; {0} \\ {0} &amp; {(1)/(3)} &amp; {0}\end{bmatrix}=\begin{aligned}-1 \\ (7)/(3) \\ (1)/(3)\end{aligned}

R3 - 1/7*R2


\begin{bmatrix}{3} &amp; {2} &amp; -3 \\ {0} &amp; {(7)/(3)} &amp; {0} \\ {0} &amp; {0} &amp; {0}\end{bmatrix}=\begin{aligned}-1 \\ (7)/(3) \\ 0\end{aligned}

Zero row in reduced matrix indicates infinite solutions

User Last Warrior
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