QuestionJune 18, 2025

A) } x+2 y-3 z & =-2 x-y+z & =-1 3 x+4 y-4 z & =4

A) } x+2 y-3 z & =-2 x-y+z & =-1 3 x+4 y-4 z & =4
A) } x+2 y-3 z & =-2 x-y+z & =-1 3 x+4 y-4 z & =4

Solution
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Answer

x = \frac{-9}{11}, y = \frac{20}{11}, z = \frac{28}{11} Explanation 1. Write the system in matrix form The system can be written as AX = B, where A = \begin{bmatrix} 1 & 2 & -3 \\ 1 & -1 & 1 \\ 3 & 4 & -4 \end{bmatrix}, X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, and B = \begin{bmatrix} -2 \\ -1 \\ 4 \end{bmatrix}. 2. Use Gaussian elimination Perform row operations to transform A into an upper triangular matrix. Start with R_2 = R_2 - R_1: \[ \begin{bmatrix} 1 & 2 & -3 & | & -2 \\ 0 & -3 & 4 & | & 1 \\ 3 & 4 & -4 & | & 4 \end{bmatrix} \] 3. Eliminate first column in third row Use R_3 = R_3 - 3R_1: \[ \begin{bmatrix} 1 & 2 & -3 & | & -2 \\ 0 & -3 & 4 & | & 1 \\ 0 & -2 & 5 & | & 10 \end{bmatrix} \] 4. Eliminate second column in third row Use R_3 = R_3 - \frac{2}{3}R_2: \[ \begin{bmatrix} 1 & 2 & -3 & | & -2 \\ 0 & -3 & 4 & | & 1 \\ 0 & 0 & \frac{11}{3} & | & \frac{28}{3} \end{bmatrix} \] 5. Back substitution Solve for z from the last equation: z = \frac{28}{11}. Substitute z into the second equation to find y: -3y + 4(\frac{28}{11}) = 1 \Rightarrow y = \frac{20}{11}. Substitute y and z into the first equation to find x: x + 2(\frac{20}{11}) - 3(\frac{28}{11}) = -2 \Rightarrow x = \frac{-9}{11}.

Explanation

1. Write the system in matrix form<br /> The system can be written as $AX = B$, where $A = \begin{bmatrix} 1 & 2 & -3 \\ 1 & -1 & 1 \\ 3 & 4 & -4 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, and $B = \begin{bmatrix} -2 \\ -1 \\ 4 \end{bmatrix}$.<br /><br />2. Use Gaussian elimination<br /> Perform row operations to transform $A$ into an upper triangular matrix. Start with $R_2 = R_2 - R_1$: <br />\[ \begin{bmatrix} 1 & 2 & -3 & | & -2 \\ 0 & -3 & 4 & | & 1 \\ 3 & 4 & -4 & | & 4 \end{bmatrix} \]<br /><br />3. Eliminate first column in third row<br /> Use $R_3 = R_3 - 3R_1$: <br />\[ \begin{bmatrix} 1 & 2 & -3 & | & -2 \\ 0 & -3 & 4 & | & 1 \\ 0 & -2 & 5 & | & 10 \end{bmatrix} \]<br /><br />4. Eliminate second column in third row<br /> Use $R_3 = R_3 - \frac{2}{3}R_2$: <br />\[ \begin{bmatrix} 1 & 2 & -3 & | & -2 \\ 0 & -3 & 4 & | & 1 \\ 0 & 0 & \frac{11}{3} & | & \frac{28}{3} \end{bmatrix} \]<br /><br />5. Back substitution<br /> Solve for $z$ from the last equation: $z = \frac{28}{11}$. Substitute $z$ into the second equation to find $y$: $-3y + 4(\frac{28}{11}) = 1 \Rightarrow y = \frac{20}{11}$. Substitute $y$ and $z$ into the first equation to find $x$: $x + 2(\frac{20}{11}) - 3(\frac{28}{11}) = -2 \Rightarrow x = \frac{-9}{11}$.
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