Question

4．設(shè)矩陣A=$[\begin{array}{l}{1}&{-2}\\{3}&{-7}\end{array}]$的逆矩陣為A^-1，矩陣B滿足AB=$[\begin{array}{l}{3}\\{1}\end{array}]$，求 A^-1，B．

Answer 1

分析 由逆矩陣的公式A^-1=$\frac{1}{丨A丨}$×A*，求得其伴隨矩陣和|A|，即可求得 A^-1，由AB=$[\begin{array}{l}{1}&{-2}\\{3}&{-7}\end{array}]$×$[\begin{array}{l}{a}\\\end{array}]$=$[\begin{array}{l}{3}\\{1}\end{array}]$，列二元一次方程組，求得a和b的值，即可求得矩陣B．

解答 解：|A|=ad-bc=-7+6=-1，
A^-1=$\frac{1}{丨A丨}$×A*=$[\begin{array}{l}{7}&{-2}\\{3}&{-1}\end{array}]$，
∴A^-1=$[\begin{array}{l}{7}&{-2}\\{3}&{-1}\end{array}]$，
設(shè)B=$[\begin{array}{l}{a}\\\end{array}]$
AB=$[\begin{array}{l}{1}&{-2}\\{3}&{-7}\end{array}]$×$[\begin{array}{l}{a}\\\end{array}]$=$[\begin{array}{l}{3}\\{1}\end{array}]$，
$\left\{\begin{array}{l}{a-2b=3}\\{3a-7b=1}\end{array}\right.$，解得：$\left\{\begin{array}{l}{a=19}\\{b=8}\end{array}\right.$，
∴B=$[\begin{array}{l}{19}\\{8}\end{array}]$．

點(diǎn)評(píng) 本題考查逆變換與逆矩陣，矩陣與矩陣的乘法的意義，考查學(xué)生的計(jì)算能力，屬于基礎(chǔ)題．