Question

19．設(shè)二階矩陣A，B滿足A^-1=$[\begin{array}{l}{1}&{2}\\{3}&{4}\end{array}]$，BA=$[\begin{array}{l}{1}&{0}\\{0}&{1}\end{array}]$，求B^-1．

Answer 1

分析 由A^-1=$[\begin{array}{l}{1}&{2}\\{3}&{4}\end{array}]$，求得A^-1的行列式丨A^-1丨及隨矩陣（A^-1）*，即可求得矩陣A，BA=$[\begin{array}{l}{1}&{0}\\{0}&{1}\end{array}]$=E，矩陣A和B互為逆矩陣，B^-1=A，即可求得矩陣B^-1．

解答 解：A^-1=$[\begin{array}{l}{1}&{2}\\{3}&{4}\end{array}]$，
丨A^-1丨=1×4-2×3=-2，
A^-1的伴隨矩陣（A^-1）*=$[\begin{array}{l}{4}&{-2}\\{-3}&{1}\end{array}]$，
∴A=$\frac{1}{丨{A}^{-1}丨}$•（A^-1）*=$[\begin{array}{l}{-2}&{1}\\{\frac{3}{2}}&{-\frac{1}{2}}\end{array}]$，
∵BA=$[\begin{array}{l}{1}&{0}\\{0}&{1}\end{array}]$=E，
∴B與A互為逆矩陣，
∴B^-1=A，
B^-1=$[\begin{array}{l}{-2}&{1}\\{\frac{3}{2}}&{-\frac{1}{2}}\end{array}]$．

點評 本題考查逆變換與逆矩陣，考查矩陣乘法的運算，屬于基礎(chǔ)題．