Exercise · ZW-GAUSSIAN-ELIMINATION-005 · ★★☆ Intermediate
Solve \(\begin{cases}x+y+z=6\\2y+z=7\\x-z=-2\end{cases}\).
Choose one option.
Show answer
- A. \((1,2,3)\)
Explanation
From the third, \(x=z-2\). In the first: \(y+2z=8\); with \(2y+z=7\) we get \(y=2\), \(z=3\), \(x=1\).
Review the concept: Gaussian elimination
Gaussian elimination solves a linear system by reducing its augmented matrix, with coefficients \(A\) and vector \(\mathbf b\), to an echelon form \(U\) using elementary row operations.
\[[A\mid \mathbf b]\ \longrightarrow\ [U\mid \mathbf c]\]
More practice problems: gaussian elimination
- ZW-GAUSSIAN-ELIMINATION-001 Which of these is NOT an elementary row operation?
- ZW-GAUSSIAN-ELIMINATION-002 Solve begincasesx+y=3\\2x-y=0endcases.
- ZW-GAUSSIAN-ELIMINATION-003 In the augmented matrix [1 2 5; 2 3 8], apply R_2arrow R_2-2R_1. What is the new row 2?
- ZW-GAUSSIAN-ELIMINATION-004 In [0 2 1; 3 1 0], the (1,1) entry is zero. What should you do to get a pivot in the first column?
- ZW-GAUSSIAN-ELIMINATION-006 After elimination, a system has the row (0\ \ 0\ \ |\ \ 4). What can you conclude?
- ZW-GAUSSIAN-ELIMINATION-007 A consistent system with 3 equations and 4 unknowns has 2 pivots in echelon form. How many free var…
- ZW-GAUSSIAN-ELIMINATION-008 Which of these matrices is in row echelon form?
- ZW-GAUSSIAN-ELIMINATION-009 Use elimination to find the rank of [1 2 3; 2 4 6; 1 0 1].