Question

10．二元一次方程組的增廣矩陣為$（{\begin{array}{l}1&{-2}&5\\ 3&1&8\end{array}}）$，通過(guò)矩陣的變換，得方程組解的增廣矩陣為$[\begin{array}{l}{1}&{0}&{3}\\{0}&{1}&{-1}\end{array}]$．

Answer 1

分析 根據(jù)方程程增廣矩陣的變換法則變換即可得到方程組解的增廣矩陣．

解答 解：二元一次方程組的增廣矩陣為$（{\begin{array}{l}1&{-2}&5\\ 3&1&8\end{array}}）$，
設(shè)①、②分別表示矩陣的第1、2行，
對(duì)矩陣進(jìn)行下列變換$[\begin{array}{l}{1}&{-2}&{5}\\{3}&{1}&{8}\end{array}]$$\stackrel{①×（-3）+②得，①不變}{→}$$[\begin{array}{l}{1}&{-2}&{5}\\{0，}&{7}&{-7}\end{array}]$$\stackrel{②×\frac{1}{7}，①不變}{→}$$[\begin{array}{l}{1}&{-2}&{5}\\{0}&{1}&{-1}\end{array}]$$\stackrel{②×2+①，②不變}{→}$$[\begin{array}{l}{1}&{0}&{3}\\{0}&{1}&{-1}\end{array}]$，
∴方程組解的增廣矩陣為$[\begin{array}{l}{1}&{0}&{3}\\{0}&{1}&{-1}\end{array}]$，
故答案為：$[\begin{array}{l}{1}&{0}&{3}\\{0}&{1}&{-1}\end{array}]$．

點(diǎn)評(píng) 本題考查了二元線性方程組的增廣矩陣，屬于基礎(chǔ)題．