Question

Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.﻿{x+y−z=−22x−y+z=5−x+2y+2z=1\begin{cases}
x + y - z = -2 \
2x - y + z = 5 \
-x + 2y + 2z = 1
\end{cases}⎩⎨⎧​x+y−z=−22x−y+z=5−x+2y+2z=1​

Accepted Answer

Write the system of equations as an augmented matrix. The system is:

$$x + y - z = -2$$

$$2x - y + z = 5$$

$$-x + 2y + 2z = 1$$

The augmented matrix is:

$$\begin{bmatrix} 1 & 1 & -1 & | & -2 \ 2 & -1 & 1 & | & 5 \ -1 & 2 & 2 & | & 1 \end{bmatrix}$$ Use Gaussian elimination to transform the matrix into an upper triangular form. Start by using the first row to eliminate the $$x-$$terms in the second and third rows:

- For row 2: Replace row 2 with (row 2) - 2*(row 1).

- For row 3: Replace row 3 with (row 3) + (row 1). After these operations, the matrix will have zeros below the leading 1 in the first column. Next, use the second row to eliminate the $$y-$$term in the third row by replacing row 3 with (row 3) - (appropriate multiple of row 2) to get a zero in the second column of row 3. Once the matrix is in upper triangular form, use back-substitution to solve for the variables starting from the last row. Solve for $$z$$ from the third row, then substitute back into the second row to find $$y$$, and finally substitute $$y$$ and $$z$$ into the first row to find $$x. $$Alternatively, you can continue with Gauss-Jordan elimination by making the matrix into reduced row echelon form (RREF) by creating leading 1s and zeros above and below each pivot, which directly gives the solution for $$x$$, $$y$$, and $$z.$$

Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
$\begin{cases}\)x + y - z = -2 \\2x - y + z = 5 \\-x + 2y + 2z = 1\(\end{cases}$

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주요 개념

Systems of Linear Equations

Matrix Representation of Systems

Gaussian Elimination and Gauss-Jordan Elimination

Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.{x+y−z=−22x−y+z=5−x+2y+2z=1\(\begin{cases}\)x + y - z = -2 \\2x - y + z = 5 \\-x + 2y + 2z = 1\(\end{cases}\)⎩⎨⎧x+y−z=−22x−y+z=5−x+2y+2z=1